A vector is a quantity with both magnitude and direction, and this unit builds its algebra first in the plane and then in three-dimensional space with the unit vectors i^, j^, k^. You learn to add, subtract and scale vectors, to divide a segment in a given ratio, and to prove geometry results by vector methods. The dot product gives angles, projections and work done, the cross product gives areas and moments, and the scalar triple product gives volumes and the test for coplanar vectors. These tools are used in physics and engineering for forces, torque, motion and navigation.
Learning outcomes (22)
Recognise the rectangular coordinate system in space.
Recognise the unit vectors and the x, y and z components of a vector.
Find the magnitude of a vector.
Repeat all fundamental mathematical operations for vectors in space which, in the plane, have already been discussed.
Demonstrate and prove the properties of vector addition: the commutative law, the associative law, 0 as the identity for vector addition, −r as the inverse for r, and the dot or scalar product.
Explain the dot or scalar product of two vectors and give the mathematical interpretation.
Express the dot or scalar product of two vectors and give its geometrical interpretation.
Express the dot product in terms of components.
Find the condition for the orthogonality of two vectors.
Find the projection of a vector along the other vector.
Find the work done by a constant force in moving an object along a given vector.
Solve daily life problems based on vectors, cross or vector product.
Explain the cross or vector product of two vectors and give its geometrical interpretation.
Apply the cross product to find the angle between two vectors.
Solve situations in daily life based on the cross or dot vector product.
Describe the scalar triple product of vectors.
Express the scalar triple product of vectors in terms of components (determinant form).
Prove that i^⋅j^×k^=j^⋅k^×i^=k^⋅i^×j^=1 and i^⋅k^×j^=j^⋅i^×k^=k^⋅j^×i^=−1.
Prove that dot and cross are interchangeable in the scalar triple product.
Find the volume of a parallelepiped and a tetrahedron determined by three given vectors.
Define coplanar vectors and find the condition for the coplanarity of three vectors.
Apply the concept of vectors in space to real-world problems such as designing and executing optimal navigation paths in transportation and logistics, graphing complex 3D motion, vector operations in engineering and computer graphics, and practical proficiency for work, flux and circulation.
3.1Vectors
Vectors describe quantities that have a size and also a direction. Newton's three laws are relations between vectors, which is why vectors are used to describe anything from a ball in free fall to a rocket on its way to the moon. This unit builds the algebra of vectors in the plane and in space.
Scalar and vector
Scalar
A physical quantity which can be completely specified by its magnitude only. Examples: mass, time, distance and volume.
Vector
A physical quantity which is completely specified by its magnitude as well as its direction. Examples: weight, displacement, velocity and acceleration.
Geometrical representation of a vector (3.1.1)
Geometrically a vector is represented by a line segment with an arrow head at one end. The length of the segment describes the magnitude and the arrow head shows the direction. The end A is called the tail or initial point and the end B is called the terminal point (the head). The vector in the figure is written AB.
Figure 3.1: the vector AB is a line segment from the tail A to the head B; the arrow shows the direction.
Vectors are usually written in bold face letters a,b,c, or as a,b,c. Other notations such as a,b,c are also used. This book mostly writes a.
Some fundamental definitions and terms related to vectors (3.1.2)
Magnitude of a vector
The magnitude (length or norm) of the vector OA=a is written ∣OA∣ or ∣a∣. It is a non-negative number: the length of the arrow.
Equal vectors
Two vectors a and b are equal, a=b, if they have the same magnitude and the same direction. Equal vectors need not have the same position: geometrically, two vectors are equal if one is a translation of the other.
Figure 3.2: a=b. The two arrows are parallel and equally long, although they sit in different places.
Negative of a vector
A vector with the same magnitude as a but the opposite direction is called the negative of a and is written −a.
Figure 3.3: a and its negative −a have the same length and opposite directions.
Zero or null vector
If the initial and terminal points of a vector coincide, the vector has zero length. It is called the zero vector and is written 0. The zero vector has no direction; a convenient direction may be assigned to it as the situation needs.
Unit vector
A vector which has the direction of a non-zero vector a and magnitude 1 is called the unit vector of a and is written a^. For a non-zero vector a of arbitrary length ∣a∣,
a^=∣a∣a⟹a=∣a∣a^a^=∣a∣a⟹a=∣a∣a^
So any vector is obtained by multiplying its magnitude by its unit vector. Finding the unit vector of a is called normalising the vector a.
Parallel vectors
Two non-zero vectors a and b are parallel if a=λb, where λ is a scalar. If λ>0 both vectors have the same direction; if λ<0 they have opposite directions. Vectors in opposite directions are called antiparallel vectors.
Figure 3.4: a and c point the same way (λ>0); b points the opposite way (λ<0), so b is antiparallel to a.
Position vector
The vector used to specify the position of a point P with respect to the origin O is the position vector of P. Its tail is at the origin and its tip is at P, so OP is the position vector of P with respect to O.
Figure 3.5: OP is the position vector of the point P; its tail is at the origin O.
3.1.3Addition of vectorsOften examined
Head to tail rule or triangle law of addition
To add two non-zero vectors a and b, join the tail of the second vector to the head of the first. The vector from the tail of the first to the head of the second is the resultanta+b. This is called the head to tail rule. Since a, b and a+b are the sides of a triangle ABC, it is also called the triangle law of addition.
Figure 3.6: triangle law, AB=a, BC=b, so AB+BC=AC=a+b.
Parallelogram law of addition
Consider a parallelogram ABCD with AB=a and AD=b. The side BC has the same magnitude and direction as AD, so BC=AD=b. Likewise DC=AB=a. By the triangle law,
AB+BC=ACi.e.a+b=ACAB+BC=ACi.e.a+b=AC
So the diagonal AC of the parallelogram is the sum of the vectors a and b drawn from the same vertex. This is the parallelogram law of addition.
Figure 3.7: parallelogram law, the diagonal AC is a+b.
Adding two vectors in space (Example 1)
Let a=2i^+3j^−5k^ and b=i^−5j^+2k^. Add them by the head to tail rule, in both orders.
Add a+b by adding the components of i^, j^ and k^ separately.
So IE, the vector from the tail of the first to the head of the last, is the sum of all five vectors. The same method adds any number of vectors and is called the polygon law of addition.
Figure 3.8: polygon law, the closing vector IE is a+b+c+d+e.
Subtraction of two vectors
For non-zero vectors a and b, a−b=a+(−b). To find a−b, draw AB=a and BC=−b. Then
AB+BC=AC⟹a+(−b)=ACAB+BC=AC⟹a+(−b)=AC
Thus AC represents a−b.
Figure 3.9: subtraction, AB=a, BC=−b and AC=a−b.
Subtracting two vectors in space (Example 2)
Let a=2i^+7j^+4k^ and b=6i^−4j^−3k^. Find a−b.
Write the difference with brackets so the signs of b are all reversed.
If λ is a non-zero scalar and a a non-zero vector, the scalar multiple λa is a vector whose magnitude is ∣λ∣ times the magnitude of a. It has the same direction as a if λ>0 and the opposite direction if λ<0. If λa=0 then either λ=0 or a=0.
Figure 3.10: 21a is half as long as a, 2a twice as long, and −2a is twice as long and reversed.
3.1.4Position vector of a point dividing the line segment in a given ratioOften examined
Let \overrightarrow{AB} be a line segment and P a point on it dividing it in the ratio m:n, that is |\overrightarrow{AP}|:|\overrightarrow{PB}|=m:n. Let the position vectors of A, B and P be \vec a, \vec b and \vec r. We want \vec r in terms of \vec a, \vec b, m and n.
Figure 3.11: P divides AB internally in the ratio m:n; \overrightarrow{OP}=\vec r is the position vector of P.
Derivation Position vector of a point dividing a line segment internally
Aim. Show that if P divides AB internally in the ratio m:n, then \vec r=\dfrac{m\vec b+n\vec a}{m+n}.
Assumptions.
P lies on the segment AB between A and B (internal division)
When P divides AB externally in the ratio m:n, the point P lies on the line AB outside the segment (beyond B in the figure), and |\overrightarrow{AP}|:|\overrightarrow{BP}|=m:n.
Figure 3.12: P divides AB externally in the ratio m:n, so P lies on the line AB beyond B.
Find the vector which divides \overrightarrow{OA}=\vec a=3\hat i+\hat j+10\hat k and \overrightarrow{OB}=\vec b=3\hat i+7\hat j+\hat k in the ratio 3:2 (internally).
3\vec b=9\hat i+21\hat j+3\hat k,\qquad 2\vec a=6\hat i+2\hat j+20\hat k
Add them and divide by 5.
\vec r=\frac{15\hat i+23\hat j+23\hat k}{5}=3\hat i+\frac{23}{5}\hat j+\frac{23}{5}\hat k
\vec r=3\hat i+\dfrac{23}{5}\hat j+\dfrac{23}{5}\hat k
External division (Example 5)
Find the vector which divides \overrightarrow{OA}=\vec a=3\hat i+\hat j+10\hat k and \overrightarrow{OB}=\vec b=3\hat i+7\hat j+\hat k in the ratio 3:2 (externally).
Take m=3, n=2 in the external division formula.
\vec r=\frac{n\vec a-m\vec b}{n-m}=\frac{2\vec a-3\vec b}{2-3}=3\vec b-2\vec a
Vector methods give short proofs of simple geometrical results. The book proves three theorems.
Theorem 1: the diagonals of a parallelogram bisect each other
Figure 3.13: parallelogram ABCD with the position vectors of its vertices; the diagonals AC and BD meet at one point M.
Proof. Let \vec a,\vec b,\vec c,\vec d be the position vectors of the vertices A,B,C,D of the parallelogram ABCD. The midpoint of a segment has the average of the end position vectors (Case I above).
The midpoint M_1 of the diagonal AC.
\text{p.v. of }M_1=\frac{\vec a+\vec c}{2}\qquad(1)
The midpoint M_2 of the diagonal BD.
\text{p.v. of }M_2=\frac{\vec b+\vec d}{2}\qquad(2)
In a parallelogram the opposite sides are equal vectors, \overrightarrow{AB}=\overrightarrow{DC}.
\vec b-\vec a=\vec c-\vec d
Rearrange so that the vertices A,C and B,D are paired.
\vec b+\vec d=\vec a+\vec c
Divide by 2, and compare with (1) and (2).
\text{p.v. of }M_2=\frac{\vec b+\vec d}{2}=\frac{\vec a+\vec c}{2}=\text{p.v. of }M_1
The two midpoints have the same position vector, so they are the same point. Hence the diagonals bisect each other.
Theorem 2: the line joining the midpoints of two sides of a triangle
Statement. The line joining the midpoints of any two sides of a triangle is parallel to the third side and half its length.
Figure 3.14: M_1 and M_2 are the midpoints of CA and BC; M_1M_2 is parallel to AB.
Proof. Let \vec a,\vec b,\vec c be the position vectors of A,B,C and let M_1,M_2 be the midpoints of CA and BC.
Write the position vectors of the midpoints.
\text{p.v. of }M_1=\frac{\vec a+\vec c}{2},\qquad \text{p.v. of }M_2=\frac{\vec b+\vec c}{2}
Subtract, tail position vector from head position vector, to get \overrightarrow{M_1M_2}.
So M_1M_2 is parallel to AB and its length is half the length of AB.
Theorem 3: the line joining the midpoints of the non-parallel sides of a trapezium
Statement. The line joining the midpoints of the two non-parallel sides of a trapezium is parallel to its parallel sides, and its length is half the sum of the lengths of the parallel sides.
Figure 3.15: trapezium ABCD with AB\parallel DC; M_1 and M_2 are the midpoints of the non-parallel sides DA and BC.
Proof. Let \vec a,\vec b,\vec c,\vec d be the position vectors of A,B,C,D, with AB\parallel DC. Let M_1 and M_2 be the midpoints of the non-parallel sides DA and BC (the labels follow the figure).
Write the position vectors of the midpoints.
\text{p.v. of }M_1=\frac{\vec d+\vec a}{2},\qquad \text{p.v. of }M_2=\frac{\vec b+\vec c}{2}
Form \overrightarrow{M_1M_2} and group the terms into \vec b-\vec a and \vec c-\vec d.
Hence \overrightarrow{M_1M_2} is parallel to \overrightarrow{DC} and therefore also to \overrightarrow{AB}. For the length, note that \overrightarrow{AB} and \overrightarrow{DC} point the same way in the trapezium, so in (1) the lengths simply add:
Common mistake: using the midpoint of AB or BC instead of the diagonals. The vertex D is opposite B, so the diagonals are AC and BD, not the sides.
D(4,\,1,\,6)
3.2Vectors in space (three-dimensional space)Often examined
Rectangular coordinate system (3.2.1)
To represent a vector in space we need a three-dimensional coordinate system. Take three mutually perpendicular lines meeting at a common point O, the origin. They are the coordinate axes, called the x-axis, y-axis and z-axis. The distance of a point along the x-axis is its x-coordinate, along the y-axis its y-coordinate, and along the z-axis its z-coordinate. A general point of space has coordinates (x,y,z).
This system of locating points is called the rectangular or Cartesian coordinate system and is denoted \mathbb R\times\mathbb R\times\mathbb R or \mathbb R^3. The set of all points of space is
\mathbb R^3=\{(x,y,z):x,y,z\in\mathbb R\}
Figure 3.16: the three mutually perpendicular coordinate axes meet at the origin O (dotted: the negative halves).
Unit vectors \hat i, \hat j and \hat k (3.2.2)
To represent a vector in space we need unit vectors along the coordinate axes. There are three fundamental unit vectors: \hat i along the x-axis, \hat j along the y-axis and \hat k along the z-axis.
Components of a vector (3.2.3)
Let \overrightarrow{OP} be the position vector of the point P(x,y,z). Drop the perpendicular PQ to the xy-plane, so that Q=(x,y,0), and let A, B, C be the feet of the coordinates on the three axes. Then \overrightarrow{OA}=x\hat i, \overrightarrow{OB}=y\hat j and \overrightarrow{OC}=z\hat k.
Derivation Components of a position vector
Aim. Express \overrightarrow{OP} in terms of \hat i, \hat j, \hat k.
Figure 3.17: \overrightarrow{OA}=x\hat i, \overrightarrow{OB}=y\hat j, \overrightarrow{OC}=z\hat k; \overrightarrow{OQ}=x\hat i+y\hat j and \overrightarrow{OP} is the position vector of P(x,y,z).
Go from O to P by first going to Q in the xy-plane and then straight up to P.
In the rectangle OAQB the opposite sides are equal, \overrightarrow{AQ}=\overrightarrow{OB}, so by the triangle law \overrightarrow{OA}+\overrightarrow{AQ}=\overrightarrow{OQ} gives
\overrightarrow{QP} is equal to \overrightarrow{OC}, which runs along the z-axis.
\overrightarrow{QP}=z\hat k
Put both into equation (1).
\overrightarrow{OP}=x\hat i+y\hat j+z\hat k
Result\overrightarrow{OP}=x\hat i+y\hat j+z\hat k
In \vec r=x\hat i+y\hat j+z\hat k the numbers x, y, z are called the components of the vector along the x-axis, y-axis and z-axis respectively.
Analytical representation of a vector (3.2.4)
Writing a vector in space in component form \vec r=x\hat i+y\hat j+z\hat k is called the analytic representation of the vector \vec r.
Magnitude of a vector (3.2.5)
Derivation Magnitude of a vector in space
Aim. Find |\vec r| for \vec r=\overrightarrow{OP}=x\hat i+y\hat j+z\hat k.
Figure 3.18: |\overrightarrow{OP}|^2=|\overrightarrow{OQ}|^2+|\overrightarrow{QP}|^2, with |\overrightarrow{OQ}|^2=x^2+y^2 and |\overrightarrow{QP}|=z.
OQP is a right-angled triangle (the right angle is at Q, since QP is parallel to the z-axis and so perpendicular to the xy-plane). Use Pythagoras.
Take the positive square root, as a length is not negative.
|\vec r|=\sqrt{x^2+y^2+z^2}
Result|\vec r|=\sqrt{x^2+y^2+z^2}
This is the Pythagoras theorem applied twice, first in the xy-plane and then upwards.
Magnitude of vectors (Example 6)
Find the magnitude of the following vectors: (i) \vec a=2\hat i+3\hat j-5\hat k, (ii) \vec b=2\hat i-3\hat j+7\hat k, (iii) \vec c=7\hat i+4\hat j-\hat k.
(i) Use |\vec r|=\sqrt{x^2+y^2+z^2} with x=2, y=3, z=-5.
This is the condition for two vectors to be parallel: the corresponding components are in the same ratio. For positive \lambda the vectors have the same direction, for negative \lambda opposite directions.
Addition of vectors
The sum of \vec r_1 and \vec r_2 is found by adding the corresponding components:
\vec r_1+\vec r_2=(x_1+x_2)\hat i+(y_1+y_2)\hat j+(z_1+z_2)\hat k
Subtraction of two vectors
\vec r_1-\vec r_2=(x_1-x_2)\hat i+(y_1-y_2)\hat j+(z_1-z_2)\hat k
3.3Properties of vector addition
In each proof the vectors are written in components. The vector law then follows from the same law for real numbers, which is applied to each component.
Commutative law for vector addition (3.3.1)
Statement. For any two vectors \vec r_1 and \vec r_2 in space, \vec r_1+\vec r_2=\vec r_2+\vec r_1.
Proof. Let \vec r_1=x_1\hat i+y_1\hat j+z_1\hat k and \vec r_2=x_2\hat i+y_2\hat j+z_2\hat k.
Add component by component.
\vec r_1+\vec r_2=(x_1+x_2)\hat i+(y_1+y_2)\hat j+(z_1+z_2)\hat k
The components are real numbers, and addition of real numbers is commutative, so each sum can be turned round.
=(x_2+x_1)\hat i+(y_2+y_1)\hat j+(z_2+z_1)\hat k
Split into the two vectors again, in the new order.
Multiplication of real numbers is commutative, so \lambda x=x\lambda and so on.
=(x\lambda)\hat i+(y\lambda)\hat j+(z\lambda)\hat k
Factor \lambda out on the right.
=(x\hat i+y\hat j+z\hat k)\lambda=\vec r\lambda
3.4Distance between two points in R³ (distance formula)Often examined
Take two points P(x_1,y_1,z_1) and Q(x_2,y_2,z_2) in \mathbb R^3. The distance between P and Q is the magnitude of the vector \overrightarrow{PQ}. Let \overrightarrow{OP}=x_1\hat i+y_1\hat j+z_1\hat k and \overrightarrow{OQ}=x_2\hat i+y_2\hat j+z_2\hat k be the position vectors.
Figure 3.19: the distance PQ is the length of \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}.
Derivation Distance formula in space
Aim. Find the distance between P(x_1,y_1,z_1) and Q(x_2,y_2,z_2).
Here \theta is measured from \vec a to \vec b. The value of the dot product is a scalar (a number, not a vector), which is why it is also called the scalar product.
Figure 3.20: \theta is the angle between \vec a and \vec b, measured where the tails (or heads) of the two vectors meet.
The dot product is commutative. For \vec b\cdot\vec a the angle is measured from \vec b to \vec a, that is clockwise, so it is -\theta. Since \cos(-\theta)=\cos\theta,
\vec b\cdot\vec a=|\vec b||\vec a|\cos(-\theta)=|\vec a||\vec b|\cos\theta=\vec a\cdot\vec b
Case III gives |\vec a|^2=\vec a\cdot\vec a, so |\vec a|=\sqrt{\vec a\cdot\vec a}.
Dot product of the fundamental unit vectors \hat i,\hat j,\hat k (3.5.2)
In \mathbb R^3, \hat i lies along the x-axis, \hat j along the y-axis and \hat k along the z-axis, and |\hat i|=|\hat j|=|\hat k|=1. A unit vector with itself makes the angle 0^\circ; two different ones are at 90^\circ. So
This is known as the analytical expression for the dot product: multiply matching components and add.
Condition for orthogonality of two vectors (3.5.4)
Two non-zero vectors \vec a=a_1\hat i+a_2\hat j+a_3\hat k and \vec b=b_1\hat i+b_2\hat j+b_3\hat k are orthogonal (perpendicular) if and only if \vec a\cdot\vec b=0 (because \cos\theta=0 only when \theta=90^\circ). In components:
Statement. For any two vectors, \vec a\cdot\vec b=\vec b\cdot\vec a.
Write both vectors in components, \vec a=a_1\hat i+a_2\hat j+a_3\hat k and \vec b=b_1\hat i+b_2\hat j+b_3\hat k, and use the component form of the dot product.
\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3
Multiplication of real numbers is commutative, so each product can be turned round.
=b_1a_1+b_2a_2+b_3a_3
This is the component form of \vec b\cdot\vec a.
=(b_1\hat i+b_2\hat j+b_3\hat k)\cdot(a_1\hat i+a_2\hat j+a_3\hat k)=\vec b\cdot\vec a
Distributive property of the dot product (3.5.6)
Statement. For any three vectors, \vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c.
Proof. Let \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \vec b=b_1\hat i+b_2\hat j+b_3\hat k and \vec c=c_1\hat i+c_2\hat j+c_3\hat k.
Add \vec b and \vec c component by component.
\vec b+\vec c=(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k
Regroup into the terms with b and the terms with c.
=(a_1b_1+a_2b_2+a_3b_3)+(a_1c_1+a_2c_2+a_3c_3)=\vec a\cdot\vec b+\vec a\cdot\vec c
Direction angles (3.5.7)
The angles which a non-zero vector \vec r makes with the positive directions of the coordinate axes are the direction angles of \vec r. Call them \alpha, \beta and \gamma (with the x-, y- and z-axis). Then
The numbers \cos\alpha, \cos\beta and \cos\gamma are the direction cosines of \vec r. Let \vec r=\overrightarrow{OP}=x\hat i+y\hat j+z\hat k, so |\vec r|=\sqrt{x^2+y^2+z^2}. Let A, B, C be the feet of the perpendiculars from P on the axes, so |\overrightarrow{OA}|=x, |\overrightarrow{OB}|=y, |\overrightarrow{OC}|=z.
Figure 3.21: the direction angles \alpha, \beta, \gamma of \overrightarrow{OP} are the angles it makes with the x-, y- and z-axes.
The three direction cosines
In the right-angled triangle AOP (right angle at A), cosine is adjacent over hypotenuse.
The numbers which are proportional to the direction cosines of a non-zero vector \vec r are called direction ratios. Let a,b,c be proportional to \cos\alpha,\cos\beta,\cos\gamma. Then, with a constant of proportionality k\ne0,
These relations are used to find the direction cosines when the direction ratios are given (all three signs together, either all + or all -, because k is one number).
Coordinates of a point (Example 9)
Find the coordinates of the point P if \overrightarrow{OP} is a vector of magnitude 2 and is parallel to the vector 2\hat i-3\hat j+4\hat k.
Let \vec a=2\hat i-3\hat j+4\hat k. Find its magnitude.
For -\vec r the magnitude is the same but every component changes sign, so every direction cosine changes sign and every angle becomes 180^\circ minus the old one: 107.9^\circ, 76.7^\circ, 157.4^\circ.
Projection of a vector along another vector (3.5.12)
Take non-zero vectors \vec a=\overrightarrow{OA} and \vec b=\overrightarrow{OB} with angle \theta between them, and drop the perpendicular BL from B on the line OA. Then |\overrightarrow{OL}| is the projection of \vec b upon \vec a.
Figure 3.22: |\overrightarrow{OL}| is the projection of \vec b along \vec a.
In the right-angled triangle BOL, \cos\theta=\dfrac{|\overrightarrow{OL}|}{|\overrightarrow{OB}|}.
\text{Projection of }\vec b\text{ along }\vec a=\frac{\vec a\cdot\vec b}{|\vec a|}=\frac{\vec a}{|\vec a|}\cdot\vec b=\hat a\cdot\vec b=\vec b\cdot\hat a
In the same way,
\text{Projection of }\vec a\text{ along }\vec b=\frac{\vec a\cdot\vec b}{|\vec b|}=\vec a\cdot\hat b
Projections (Example 12)
If \vec a=\hat i-\hat j+\hat k and \vec b=-\hat i+\hat j+3\hat k, find the projection of \vec a along \vec b and the projection of \vec b along \vec a.
Projection of \vec b along \vec a is \dfrac{\vec a\cdot\vec b}{|\vec a|}.
\frac{\vec a\cdot\vec b}{|\vec a|}=\frac1{\sqrt3}
\text{proj. of }\vec a\text{ along }\vec b=\dfrac1{\sqrt{11}},\qquad\text{proj. of }\vec b\text{ along }\vec a=\dfrac1{\sqrt3}
The cosine of a difference (Example 13)
Prove that \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.Figure 3.23: the unit vectors \hat u=\overrightarrow{OP} and \hat v=\overrightarrow{OQ}; the angle between them is \alpha-\beta.
Take two unit vectors \hat u=\overrightarrow{OP} and \hat v=\overrightarrow{OQ} making the angles \alpha and \beta with the x-axis. Then \alpha-\beta is the angle between \hat u and \hat v.
Since \hat u is the position vector of P(\cos\alpha,\sin\alpha) and \hat v that of Q(\cos\beta,\sin\beta),
For any triangle ABC, with the usual notations, prove that |\vec a|=|\vec b|\cos\gamma+|\vec c|\cos\beta.Figure 3.24: triangle ABC with a=BC, b=CA, c=AB and the angles \alpha,\beta,\gamma at A,B,C.
Take \vec a=\overrightarrow{BC}, \vec b=\overrightarrow{CA}, \vec c=\overrightarrow{AB}, going round the triangle. The three vectors form a closed path, so their sum is zero.
\vec a+\vec b+\vec c=\vec 0\;\Longrightarrow\;\vec a=-\vec b-\vec c
Take the dot product of both sides with \vec a.
\vec a\cdot\vec a=\vec a\cdot(-\vec b-\vec c)=-\vec a\cdot\vec b-\vec a\cdot\vec c
The angle between \vec a and \vec b (head of \vec a meets tail of \vec b at C) is 180^\circ-\gamma, and between \vec a and \vec c it is 180^\circ-\beta.
A constant force \vec F acts on an object and displaces it from A to B. The force makes the angle \theta with the displacement vector \vec S=\overrightarrow{AB}. The component of \vec F along \vec S is |\vec F|\cos\theta, so the work done by the force is
\text{work done}=|\vec F|\cos\theta\,(|\vec S|)=|\vec F||\vec S|\cos\theta=\vec F\cdot\vec S
Figure 3.25: the force \vec F at angle \theta to the displacement \vec S=\overrightarrow{AB}; only the component |\vec F|\cos\theta along \vec S does work.
Work done by a force (Example 15)
Find the work done by the constant force \vec F=2\hat i+\hat j-\hat k in moving an object from A(0,1,3) to B(-1,2,4).
The displacement is \vec S=\overrightarrow{AB}, the position of B minus the position of A.
\vec S=(-1-0)\hat i+(2-1)\hat j+(4-3)\hat k=-\hat i+\hat j+\hat k
Worked example: the sign of work and the angle between force and displacement
A constant force of magnitude 10\,\mathrm{N} acts on a body while the body is displaced 5\,\mathrm{m}. Find the work done when the angle between \vec F and the displacement \vec d is 0^\circ, 60^\circ, 90^\circ, 120^\circ and 180^\circ, and explain the sign each time.
Work against the angle
W=50\cos\theta for F=10\,\mathrm{N} and d=5\,\mathrm{m}: positive for \theta<90^\circ, zero at 90^\circ, negative beyond.
Work done is the dot product of force and displacement.
Read the signs. For \theta<90^\circ, \cos\theta>0: the force helps the motion. At 90^\circ the force is perpendicular to the displacement (\vec F\cdot\vec d=0), so it does no work. For \theta>90^\circ, \cos\theta<0: the force opposes the motion, as friction does.
Common mistake: thinking negative work is an error. It only says that the force acts against the displacement.
W=50,\ 25,\ 0,\ -25,\ -50\ \mathrm{J}
3.6Cross or vector product of two vectorsOften examined
Cross or vector product of two vectors and its geometrical interpretation (3.6.1)
If \vec a and \vec b are two non-zero vectors and \theta is the angle between them, their cross product is also a vector. It is written \vec a\times\vec b and defined by
where \hat n is the unit vector normal (perpendicular) to the plane containing both \vec a and \vec b. Its direction is given by the right-hand rule: it is the direction in which a right-handed screw advances when turned from \vec a towards \vec b. The angle \theta is positive when measured anticlockwise and negative when measured clockwise.
Figure 3.26: \vec a\times\vec b is perpendicular to the plane of \vec a and \vec b; \vec b\times\vec a points the opposite way.
When computing \vec b\times\vec a the angle is measured from \vec b to \vec a, which is clockwise, so it is -\theta:
Apply the definition with \vec b first.
\vec b\times\vec a=|\vec b||\vec a|\sin(-\theta)\,\hat n
Use \sin(-\theta)=-\sin\theta.
\vec b\times\vec a=-|\vec b||\vec a|\sin\theta\,\hat n=-\vec a\times\vec b
So \vec a\times\vec b and \vec b\times\vec a are opposite in direction: \vec a\times\vec b=-\vec b\times\vec a. The cross product is anti-commutative.
Cross product of the fundamental unit vectors (3.6.2)
From \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n: a unit vector crossed with itself has \theta=0^\circ, and different unit vectors are at 90^\circ with the third axis as the normal.
By anti-commutativity the reversed products change sign: \hat j\times\hat i=-\hat i\times\hat j=-\hat k, \hat k\times\hat j=-\hat j\times\hat k=-\hat i and \hat i\times\hat k=-\hat k\times\hat i=-\hat j.
Row \times column. Going round \hat i\to\hat j\to\hat k\to\hat i gives a plus sign, against that order a minus sign.
\times
\hat i
\hat j
\hat k
\hat i
\vec0
\hat k
-\hat j
\hat j
-\hat k
\vec0
\hat i
\hat k
\hat j
-\hat i
\vec0
Row \times column. Going round \hat i\to\hat j\to\hat k\to\hat i gives a plus sign, against that order a minus sign.
\times
\hat i
\hat j
\hat k
\hat i
\vec0
-\hat k
\hat j
\hat j
\hat k
\vec0
-\hat i
\hat k
-\hat j
\hat i
\vec0
Figure 3.27: going round \hat i\to\hat j\to\hat k\to\hat i (positive sense), the cross product of any two in order is the third, e.g. \hat j\times\hat k=\hat i.
Cross product in terms of components (3.6.3)
Derivation Cross product in components
Aim. Find \vec a\times\vec b for \vec a=a_1\hat i+a_2\hat j+a_3\hat k and \vec b=b_1\hat i+b_2\hat j+b_3\hat k.
Multiply out term by term (the cross product is distributive), keeping the order of each pair.
Write the components of \vec a in the second row and of \vec b in the third; expand along the first row with the signs +,-,+.
Area of a parallelogram (3.6.4)
Let ABCD be a parallelogram with \overrightarrow{AB}=\vec a and \overrightarrow{AD}=\vec b as two adjacent sides, and let \theta be the angle between them. Drop the perpendiculars DE and CF to the line AB. The area of the parallelogram equals the area of the rectangle EFCD.
Figure 3.28: parallelogram ABCD and the rectangle EFCD of the same base DC and height ED=h.
In the right-angled triangle EAD, \sin\theta=\dfrac{|\overrightarrow{ED}|}{|\overrightarrow{AD}|}=\dfrac{|\overrightarrow{ED}|}{|\vec b|}.
|\overrightarrow{ED}|=|\vec b|\sin\theta
Area of the parallelogram is the area of the rectangle EFCD, base |\overrightarrow{EF}|=|\overrightarrow{DC}|=|\vec a| times height |\overrightarrow{ED}|.
Find the area of the triangle with vertices (0,0), (2,9), (3,5).
Let A=(0,0), B=(2,9), C=(3,5). Take the two sides from A: \vec a=\overrightarrow{AC} and \vec b=\overrightarrow{AB}, written as space vectors with z=0.
\vec a=3\hat i+5\hat j+0\hat k,\qquad\vec b=2\hat i+9\hat j+0\hat k
Form the determinant.
\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\3&5&0\\2&9&0\end{vmatrix}=(0-0)\hat i+(0-0)\hat j+(27-10)\hat k=17\hat k
Worked example: area of a triangle from three vertices
Find the area of the triangle with vertices A(1,1,1), B(3,2,4) and C(2,4,0), and check that \vec n=\overrightarrow{AB}\times\overrightarrow{AC} is perpendicular to the plane.
Write two sides from the common vertex A (head minus tail).
\overrightarrow{AB}=2\hat i+\hat j+3\hat k,\qquad\overrightarrow{AC}=\hat i+3\hat j-\hat k
Expand the determinant for the cross product.
\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&1&3\\1&3&-1\end{vmatrix}=(-1-9)\hat i-(-2-3)\hat j+(6-1)\hat k=-10\hat i+5\hat j+5\hat k
The area of a triangle is half the magnitude of the cross product of two sides.
Condition for two non-zero vectors to be parallel (3.6.5)
If \vec a and \vec b are parallel, the angle between them is \theta=0^\circ, so \vec a\times\vec b=|\vec a||\vec b|\sin0^\circ\hat n=|\vec a||\vec b|(0)\hat n=\vec 0. If they are antiparallel, \theta=180^\circ and again \vec a\times\vec b=|\vec a||\vec b|\sin180^\circ\hat n=\vec 0.
Distributive laws for the cross product (3.6.6)
For any three vectors \vec a,\vec b,\vec c: (i) \vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c and (ii) (\vec a+\vec b)\times\vec c=\vec a\times\vec c+\vec b\times\vec c.
Proof of (i). Let \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \vec b=b_1\hat i+b_2\hat j+b_3\hat k, \vec c=c_1\hat i+c_2\hat j+c_3\hat k.
Add \vec b and \vec c.
\vec b+\vec c=(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k
If \vec a=2\hat i-\hat j+\hat k, \vec b=-\hat i+2\hat j-\hat k and \vec c=\hat i+\hat j-3\hat k, find the angle between the vectors \vec a+\vec b and \vec a+\vec c.
Show that \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.Figure 3.29: the unit vectors \hat u=\overrightarrow{OA} and \hat v=\overrightarrow{OB} at angles \alpha and \beta with the x-axis.
Take unit vectors \hat u=\overrightarrow{OA} and \hat v=\overrightarrow{OB} making the angles \alpha and \beta with the x-axis (\alpha>\beta). Then \alpha-\beta is the angle between them.
Moment or torque of a given force about a given point (3.6.8)
The moment of a force is its turning effect about a point. It is the product of the force and d, where d is the perpendicular distance of the point from the line of action of the force. In the figure, the force \vec F acts at the point P and O is the point about which it turns, with \vec r=\overrightarrow{OP}. The perpendicular from O to the line of action of \vec F meets it at A.
Figure 3.30: moment of the force \vec F acting at P about the point O; d=|\overrightarrow{OA}| is the perpendicular distance from O to the line of action of \vec F.
Moment about O = force times the perpendicular distance.
\text{Moment}=|\overrightarrow{OA}||\vec F|
In the right-angled triangle OAP, \sin\theta=\dfrac{|\overrightarrow{OA}|}{|\overrightarrow{OP}|}, where \theta is the angle between \vec r and \vec F.
=(-5+6)\hat i-(5-9)\hat j+(-2+3)\hat k=\hat i+4\hat j+\hat k
\vec M=\hat i+4\hat j+\hat k
Moment on a door (Example 21)
Find the moment of the force \vec F=7\hat i+4\hat j+2\hat k when it is applied at the handle of a door at the point (2,1,4) about the hinge at the point (0,0,1).
Let O=(0,0,1) be the hinge and H=(2,1,4) the handle. Then \vec r=\overrightarrow{OH}.
\vec r=(2-0)\hat i+(1-0)\hat j+(4-1)\hat k=2\hat i+\hat j+3\hat k
=(2-12)\hat i-(4-21)\hat j+(8-7)\hat k=-10\hat i+17\hat j+\hat k
\vec M=-10\hat i+17\hat j+\hat k
3.7Scalar triple productOften examined
Scalar triple product of vectors (3.7.1)
The scalar (dot) product of two vectors in which one vector is already a cross product of two vectors is called a scalar triple product. If one vector is \vec a and the other is \vec b\times\vec c, the scalar triple product is \vec a\cdot(\vec b\times\vec c). It is also written [\vec a\ \vec b\ \vec c], or simply \vec a\cdot\vec b\times\vec c. Its value is a number (a scalar).
Scalar triple product in the form of components (3.7.2)
Derivation Scalar triple product as a determinant
Aim. Show that \vec a\cdot(\vec b\times\vec c)=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}.
Take \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \vec b=b_1\hat i+b_2\hat j+b_3\hat k, \vec c=c_1\hat i+c_2\hat j+c_3\hat k and find \vec b\times\vec c first.
\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=(b_2c_3-b_3c_2)\hat i-(b_1c_3-b_3c_1)\hat j+(b_1c_2-b_2c_1)\hat k
Dot with \vec a: multiply matching components and add.
If any two of the vectors in the scalar triple product are the same or parallel, two rows of the determinant are the same (or proportional), so the determinant is zero: \vec a\cdot\vec b\times\vec c=0.
If any two vectors are interchanged, the corresponding rows of the determinant are interchanged, which changes the sign of the determinant. So the new scalar triple product is the negative of the original: \vec a\cdot\vec b\times\vec c=-\vec b\cdot\vec a\times\vec c=-\vec a\cdot\vec c\times\vec b and so on.
Scalar triple product of \hat i, \hat j, \hat k (3.7.3)
Using \hat j\times\hat k=\hat i, \hat k\times\hat i=\hat j, \hat i\times\hat j=\hat k and \hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1:
Thus \hat i\cdot\hat k\times\hat j=\hat j\cdot\hat i\times\hat k=\hat k\cdot\hat j\times\hat i=-1. Similarly, when two of the vectors are the same the product is zero, for example \hat i\cdot\hat j\times\hat j=\hat i\cdot\vec 0=0 and \hat j\cdot\hat k\times\hat k=\hat j\cdot\vec 0=0.
Dot and cross are interchangeable in a scalar triple product (3.7.4)
Statement.\vec a\cdot\vec b\times\vec c=\vec a\times\vec b\cdot\vec c. In words: the dot and the cross may be swapped without changing the value.
Proof. Let \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \vec b=b_1\hat i+b_2\hat j+b_3\hat k, \vec c=c_1\hat i+c_2\hat j+c_3\hat k.
From 3.7.2 we already know the left side as a determinant.
Volume of a parallelepiped and a tetrahedron (3.7.5)
Volume of a parallelepiped
A parallelepiped has three adjacent edges \vec a, \vec b, \vec c meeting at the vertex O (a box whose faces are parallelograms).
Figure 3.31: parallelepiped with edges \vec a,\vec b,\vec c from O. \overrightarrow{OD} is the perpendicular height, along \vec a\times\vec b.
Derivation Volume of a parallelepiped
Aim. Show that the volume of the parallelepiped with adjacent edges \vec a,\vec b,\vec c is \vec a\cdot\vec b\times\vec c (in magnitude).
Assumptions.
the edges \vec a,\vec b,\vec c are taken so that \vec a\cdot\vec b\times\vec c is positive; otherwise the volume is its numerical value |\vec a\cdot\vec b\times\vec c| (a volume is never negative)
Volume = area of base times perpendicular height. The base is the parallelogram on \vec a and \vec b, with area |\vec a\times\vec b| (3.6.4).
\vec a\times\vec b is perpendicular to both \vec a and \vec b, that is to the base, so the height \overrightarrow{OD} is in the direction of \vec a\times\vec b. Therefore |\overrightarrow{OD}| is the projection of \vec c along \vec a\times\vec b, and by 3.5.12
Result\text{Volume of parallelepiped}=\vec a\cdot\vec b\times\vec c
The volume of the box is the absolute value of the determinant of the three edge vectors.
Volume of a tetrahedron
A tetrahedron has three coterminal edges \vec a,\vec b,\vec c from the vertex O. Its base is the triangle OAB and its apex is C.
Figure 3.32: tetrahedron with coterminal edges \vec a,\vec b,\vec c; \overrightarrow{OD} is the perpendicular height from C on the base OAB.
Derivation Volume of a tetrahedron
Aim. Show that the volume of the tetrahedron with coterminal edges \vec a,\vec b,\vec c is \dfrac16\,\vec a\cdot\vec b\times\vec c.
Volume of a pyramid is one third of base area times height. The base is the triangle OAB, half of the parallelogram on \vec a,\vec b.
\text{Volume}=\frac13(\text{area of }\triangle OAB)(|\overrightarrow{OD}|)=\frac13\left(\frac12|\vec a\times\vec b|\right)|\overrightarrow{OD}|=\frac16|\vec a\times\vec b||\overrightarrow{OD}|\qquad(1)
As for the parallelepiped, \overrightarrow{OD} lies along \vec a\times\vec b and |\overrightarrow{OD}| is the projection of \vec c on \vec a\times\vec b.
A volume cannot be negative; the sign only shows the order of the edges, so take the numerical value.
\text{Volume}=\frac{11}{2}
\text{Volume}=\dfrac{11}{2}\text{ cubic units}
Coplanar vectors and condition for the coplanarity of three vectors (3.7.6)
Two or more vectors lying in the same plane are called coplanar vectors.
Condition for the coplanarity of three vectors. Let \vec a, \vec b, \vec c be three coplanar vectors. \vec a\times\vec b is perpendicular to both \vec a and \vec b, that is, to their plane. Since \vec c lies in the same plane, \vec a\times\vec b is also perpendicular to \vec c, so
Figure 3.33: three coplanar vectors lie in one plane, so \vec a\times\vec b (perpendicular to that plane) is perpendicular to \vec c as well.
Find the value of \lambda so that the vectors \vec a=\hat i-\hat j+\hat k, \vec b=2\hat i+\hat j+\hat k and \vec c=-\hat i+\lambda\hat j+2\hat k are coplanar.
The vectors are coplanar if \vec a\cdot\vec b\times\vec c=0.
Check: for \lambda=-4, \vec c=2\vec a-\vec b, since 2(2,-1,1)-(1,2,-3)=(3,-4,5). So \vec c lies in the plane of \vec a and \vec b.
\lambda=-4
3.8Application of vectors in real world
Vectors can be used by air-traffic controllers when tracking planes, by meteorologists when describing wind conditions, and by computer programmers when they are designing virtual worlds. This section presents some applications of vectors that are commonly used in the study of physics: work, torque and magnetic force.
Projectile motion (3.8.1)
A projectile (a stone) thrown with an initial speed u at an angle \phi with the horizontal has a vertical component u\sin\phi-gt and a horizontal component u\cos\phi of its velocity. This is the resolution of a vector into components.
Figure 3.34: the launch velocity u resolved into the horizontal component u\cos\phi and the vertical component u\sin\phi (which becomes u\sin\phi-gt after time t).
Sharpening a wooden pencil with a blade (3.8.2)
We cut the pencil at an angle. The component of the force in the direction perpendicular to the pencil cuts the pencil. The component of the force in the direction parallel to the pencil removes the thin wooden part.
Earth's magnetic field (3.8.3)
Earth's magnetic field has two components, B and H, which are perpendicular to Earth's surface and parallel to the surface.
Pendulum
The tension in the string has two components: one balances the weight and the other gives the centripetal force.
Digital graphics, programming and GPS (3.8.4)
Digital graphics
Vector art is digital graphics made with mathematical formulas to construct shapes and lines. Vector images keep their quality at any size, which makes vector file formats flexible, resilient and always sharp. Vector artwork is produced with vector design software such as Linearity Curve (formerly Vectornator), Adobe Illustrator and Sketch. These editors generate simple shapes between points instead of pixels.
Programming
A vector in programming is a type of array that is one-dimensional. Vectors are a logical element in programming languages used for storing a sequence of data elements of the same basic type. The members of a vector are called components.
GPS unit
When you use a GPS unit to get from point A to point B, it gives you a distance (magnitude) and a direction. A vector is therefore a directed quantity: a number with a direction.
Aeroplane and wind (Example 28)
An aeroplane is flying with an airspeed of 475\ \mathrm{km\,h^{-1}} on a heading of 70^\circ. If an 80\ \mathrm{km\,h^{-1}} wind is blowing from a true heading of 120^\circ, determine the velocity and direction of the plane relative to the ground.Figure 3.35: the plane's velocity \vec P relative to the air, the wind \vec A, and the ground velocity \overrightarrow{OB}=\vec P+\vec A (not to scale).
Headings are measured clockwise from north, so a heading of 70^\circ is 90^\circ-70^\circ=20^\circ above the east (x) axis. The plane's velocity relative to the air is \vec P, of magnitude 475.
A wind blowing from the heading 120^\circ blows towards 300^\circ, which is 150^\circ measured anticlockwise from the east axis. Its velocity \vec A has magnitude 80.
\vec r: from the point of turning to the point where the force acts
Scalar triple product
\vec a\cdot\vec b\times\vec c=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=\vec a\times\vec b\cdot\vec c
Also written [\vec a\ \vec b\ \vec c]; swapping two vectors changes the sign
Volume of a parallelepiped
V=|\vec a\cdot\vec b\times\vec c|
\vec a,\vec b,\vec c: adjacent edges
Volume of a tetrahedron
V=\dfrac16|\vec a\cdot\vec b\times\vec c|
\vec a,\vec b,\vec c: coterminal edges
Coplanar vectors
\vec a\cdot\vec b\times\vec c=0
Condition for three vectors to lie in one plane
Key terms
Scalar
A physical quantity completely specified by its magnitude only, such as mass, time, distance or volume.
Vector
A physical quantity completely specified by its magnitude and direction, such as weight, displacement, velocity or acceleration.
Magnitude of a vector
The length (norm) of the vector, written |\vec a|.
Equal vectors
Vectors with the same magnitude and direction (in space: the same corresponding components).
Negative of a vector
The vector with the same magnitude and opposite direction, -\vec a.
Zero (null) vector
The vector of zero length, \vec 0=0\hat i+0\hat j+0\hat k; it has no direction.
Unit vector
A vector of magnitude 1; the unit vector of \vec a is \hat a=\vec a/|\vec a|.
Parallel vectors
Non-zero vectors with \vec a=\lambda\vec b; antiparallel when \lambda<0.
Position vector
The vector \overrightarrow{OP} from the origin to the point P.
Triangle law (head to tail rule)
The sum \vec a+\vec b is the vector from the tail of \vec a to the head of \vec b when \vec b is drawn from the head of \vec a.
Parallelogram law
The diagonal of the parallelogram on \vec a and \vec b is \vec a+\vec b.
Polygon law
The closing side of a polygon of head-to-tail vectors is their sum.
Rectangular (Cartesian) coordinate system
Three mutually perpendicular axes meeting at the origin; space is \mathbb R^3=\{(x,y,z)\}.
Components of a vector
The numbers x,y,z in \vec r=x\hat i+y\hat j+z\hat k.
Dot (scalar) product
\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta, a scalar.
Orthogonal vectors
Perpendicular vectors; their dot product is zero.
Projection of a vector
The length \dfrac{\vec a\cdot\vec b}{|\vec a|} of the component of \vec b along \vec a.
Direction angles
The angles \alpha,\beta,\gamma a vector makes with the positive x-, y-, z-axes.
Direction cosines
\cos\alpha,\cos\beta,\cos\gamma, also written l,m,n; their squares add to 1.
Direction ratios
Numbers proportional to the direction cosines; also called direction numbers or components.
Cross (vector) product
\vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n, a vector perpendicular to both \vec a and \vec b.
Anti-commutative
\vec a\times\vec b=-\vec b\times\vec a: the cross product changes sign when the order is reversed.
Moment (torque) of a force
The turning effect of a force about a point; the vector moment is \vec M=\vec r\times\vec F.
Scalar triple product
\vec a\cdot\vec b\times\vec c=[\vec a\ \vec b\ \vec c], the dot product of \vec a with \vec b\times\vec c.
Parallelepiped
A solid with six parallelogram faces; its volume is |\vec a\cdot\vec b\times\vec c|.
Tetrahedron
A solid with four triangular faces; its volume is \frac16|\vec a\cdot\vec b\times\vec c|.
Coplanar vectors
Vectors lying in the same plane; three are coplanar when \vec a\cdot\vec b\times\vec c=0.
Exercise 3.1p. 86–88
Component form of vectors, parallel and collinear vectors, magnitude and unit vectors, the section formula and vector proofs of geometry theorems. A vector \overrightarrow{AB} is the position vector of B minus that of A; parallel vectors have proportional components.
1
In the following find the required vector in its component form. Given that P=(3,-1); Q=(-4,-6); R=(1,4) and S=(2,5).
Solution
Method. In the plane a vector is x\hat{i}+y\hat{j}. The vector \overrightarrow{AB} is found as (coordinates of B) minus (coordinates of A): head minus tail. Scalar multiples and sums are then done component by component.
The four points. \overrightarrow{PQ} runs from P to Q, so its components are (coordinates of Q) minus (coordinates of P); the same holds for \overrightarrow{RS}.
(i)\overrightarrow{PQ}
For two points, the vector from the first to the second is the position vector of the second minus the position vector of the first (the book proves \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP} in the distance formula). Write the points as position vectors.
Reversing a vector changes every sign, so \overrightarrow{SP}=-\overrightarrow{PS} and \overrightarrow{QP}=-\overrightarrow{PQ}. Use \overrightarrow{PS}=-\hat{i}+6\hat{j} and \overrightarrow{PQ}=-7\hat{i}-5\hat{j}.
Book answer. (i) -7\hat{i}-5\hat{j}; (ii) -22\hat{i}-16\hat{j}; (iii) -7\hat{i}+28\hat{j}; (iv) -\tfrac{35}{2}\hat{i}-\tfrac{13}{2}\hat{j}; (v) -18\hat{i}+155\hat{j}
2
Show that:
Solution
(i)the points A(1,0); B(6,0) and C(0,0) are collinear.
Three points are collinear when two of the vectors joining them are parallel, because they then lie on one line through the shared point. Find \overrightarrow{AB} and \overrightarrow{AC} (head minus tail).
(ii)if \vec a and \vec b are the position vectors of points (2,-7) and \left(\dfrac m2,11\right), find the value of m for which \vec a and \vec b are collinear.
Both vectors start at the origin O, so they are collinear exactly when they are parallel: corresponding components are in the same ratio (book, 3.2.6 vi).
\dfrac{2}{m/2}=\dfrac{-7}{11}
Simplify the left side and cross-multiply.
\dfrac4m=\dfrac{-7}{11}\;\Rightarrow\;-7m=44
Solve for m.
m=-\dfrac{44}{7}
Check: with m=-\tfrac{44}{7} we get \vec b=-\tfrac{22}{7}\hat{i}+11\hat{j}, and -\tfrac{11}{7}\vec a=-\tfrac{22}{7}\hat{i}+11\hat{j}=\vec b, so \vec b is a scalar multiple of \vec a.
m=-\dfrac{44}{7}
Note. The printed x-coordinate of A in part (i) is not clearly printed (it looks like a 1 with a small mark under it). All three points have y-coordinate 0, so they lie on the x-axis and are collinear whatever the x-coordinate of A is; the working above uses 1. The book gives no answer for question 2.
3
If \vec u=\langle-1,1\rangle; \vec v=\langle0,1\rangle and \vec w=\langle3,4\rangle then
Solution
The notation \langle x,y\rangle means x\hat{i}+y\hat{j} and \langle x,y,z\rangle means x\hat{i}+y\hat{j}+z\hat{k}. Vectors are added, subtracted and multiplied by scalars component by component.
(i)Find \vec x that satisfies \vec u-2\vec x=\vec x-\vec w+3\vec v.
Treat \vec x as the unknown in an equation. Collect the terms with \vec x on one side by adding 2\vec x and adding \vec w to both sides.
\vec u+\vec w-3\vec v=3\vec x
Substitute \vec u=\langle-1,1\rangle, \vec v=\langle0,1\rangle, \vec w=\langle3,4\rangle and combine the components.
Check: \vec u-2\vec x=\langle-1-\tfrac43,1-\tfrac43\rangle=\langle-\tfrac73,-\tfrac13\rangle and \vec x-\vec w+3\vec v=\langle\tfrac23-3,\tfrac23-4+3\rangle=\langle-\tfrac73,-\tfrac13\rangle. They agree.
\vec x=\tfrac23\hat{i}+\tfrac23\hat{j}
(ii)Find \vec u and \vec v if \vec u+\vec v=\langle2,-3\rangle; 3\vec u+2\vec v=\langle-1,2\rangle.
The two equations are a pair of simultaneous equations in the unknown vectors. Here \vec u,\vec v are new unknowns (they are not the vectors given at the start of the question). Multiply the first equation by 2 so that the \vec v terms match.
2\vec u+2\vec v=\langle4,-6\rangle
Subtract this from the second equation to eliminate \vec v.
(iii)Find initial point of \vec v=\langle-3,1,2\rangle if its terminal point is (5,0,1).
For a vector, (components) = (terminal point) - (initial point). So the initial point is the terminal point minus the components.
\text{initial}=\text{terminal}-\vec v
Subtract coordinate by coordinate.
(5-(-3),\ 0-1,\ 1-2)=(8,-1,-1)
Check: from (8,-1,-1) to (5,0,1) the change is (5-8,\ 0+1,\ 1+1)=(-3,1,2)=\vec v.
(8,-1,-1)\ \text{i.e. the position vector }8\hat{i}-\hat{j}-\hat{k}
Book answer. (i) \tfrac23\hat{i}+\tfrac23\hat{j}; (ii) \vec u=-5\hat{i}+8\hat{j}, \vec v=7\hat{i}-11\hat{j}; (iii) 8\hat{i}-\hat{j}-\hat{k}
4
Solution
(i)Find the value of m for which the vector \vec a=3\hat{i}+4\hat{j}-9\hat{k} is parallel to \vec b=\hat{i}+m\hat{j}-3\hat{k}.
Two vectors are parallel when \vec a=\lambda\vec b, that is, their components are in the same ratio (book, 3.2.6 vi).
\dfrac{3}{1}=\dfrac{4}{m}=\dfrac{-9}{-3}
Two of the ratios are known: \dfrac31=3 and \dfrac{-9}{-3}=3, so \lambda=3. Both agree, which confirms the vectors can be parallel.
\lambda=3
Equate the middle ratio to 3 and solve.
\dfrac4m=3\;\Rightarrow\;m=\dfrac43
Check: 3\vec b=3\hat{i}+4\hat{j}-9\hat{k}=\vec a.
m=\dfrac43
(ii)Find the value of \lambda for which the points P, Q and R are collinear. Given that \hat{i}+2\hat{j}+3\hat{k},\ -2\hat{i}+2\hat{j}+6\hat{k} and \lambda\hat{i}-\lambda\hat{k} are the position vectors of points P, Q and R respectively.
The points are P(1,2,3), Q(-2,2,6) and R(\lambda,0,-\lambda). They are collinear only if \overrightarrow{PQ} and \overrightarrow{PR} are parallel. Find both (head minus tail).
Parallel means \overrightarrow{PR}=t\,\overrightarrow{PQ} for some scalar t. Compare the \hat{j} components.
-2=t\cdot0=0
This is impossible for every t, whatever \lambda is. The \hat{j} component of \overrightarrow{PQ} is 0, so P and Q both lie in the plane y=2 and so does the whole line PQ, but R has y=0 for every \lambda.
Conclusion: no value of \lambda makes P, Q, R collinear.
Cross-check with the other test: \overrightarrow{PQ} and \overrightarrow{PR} are parallel only if the ratios \dfrac{-3}{\lambda-1}=\dfrac{0}{-2}=\dfrac{3}{-(\lambda+3)} are equal. The middle ratio is 0, so we would need -3=0, which is false.
\text{no real value of }\lambda\text{ exists}
Book answer. (i) m=\tfrac43; (ii) the key only repeats the three position vectors and gives no value of \lambda.
Note. Part (ii) as printed has no solution: P and Q both have y=2, while R=(\lambda,0,-\lambda) has y=0 for every \lambda. The book key gives no value either, so one of the printed position vectors is probably a misprint. The method (set \overrightarrow{PR}=t\,\overrightarrow{PQ} and compare components) is what the exam expects.
5
Solution
(i)If \vec a=\hat{i}-2\hat{j}+\hat{k};\ \vec b=\hat{i}-\hat{j}-\hat{k} and \vec c=2\hat{i}+\hat{k} then find a unit vector in the direction of 2\vec a-3\vec b+\vec c.
First find the vector \vec r=2\vec a-3\vec b+\vec c. Multiply \vec a and \vec b by their scalars.
(ii)Use vectors to find the length of diagonals of a parallelogram having adjacent sides \hat{i}+\hat{j} and \hat{i}-2\hat{j}.
Let the adjacent sides be \vec u=\hat{i}+\hat{j} and \vec v=\hat{i}-2\hat{j}. By the parallelogram law, one diagonal is \vec u+\vec v. The other joins the ends of the two sides, so it is \vec u-\vec v.
The parallelogram on \vec u=\hat{i}+\hat{j} and \vec v=\hat{i}-2\hat{j}. One diagonal is \vec u+\vec v (from the origin to (2,-1)), the other is \vec u-\vec v (from (1,-2) to (1,1)).
\text{diagonals: }\sqrt5\ \text{and}\ 3
Book answer. (i) \dfrac{1}{\sqrt{38}}\hat{i}-\dfrac{1}{\sqrt{38}}\hat{j}+\dfrac{6}{\sqrt{38}}\hat{k}; (ii) \sqrt5 ; 3
6
Solution
(i)Show that the points with position vectors \hat{i}-\hat{j},\ 4\hat{i}-3\hat{j}+\hat{k} and 2\hat{i}-4\hat{j}+5\hat{k} are the vertices of a right-angled triangle.
Name the points A(1,-1,0), B(4,-3,1) and C(2,-4,5). Find the three side vectors (head minus tail).
(ii)Show that the triangle with 2\hat{i}+3\hat{j}+\sqrt3\hat{k},\ \sqrt{10}\hat{i}-\hat{j}+\sqrt5\hat{k} and -3\hat{i}+\sqrt3\hat{j}+2\hat{k} are the vertices of an equilateral triangle.
Name the points P(2,3,\sqrt3), Q(\sqrt{10},-1,\sqrt5) and R(-3,\sqrt3,2). An equilateral triangle needs |PQ|=|QR|=|PR|, so find the three side vectors.
The three squared lengths are different, so as printed the three sides are not equal: |PQ|\approx4.20, |QR|\approx6.74, |PR|\approx5.17.
What is true is that all three position vectors have the same magnitude: |\overrightarrow{OP}|^2=4+9+3=16, |\overrightarrow{OQ}|^2=10+1+5=16, |\overrightarrow{OR}|^2=9+3+4=16. So the three points lie on a sphere of radius 4 about the origin, but that does not make the triangle equilateral.
|PQ|\neq|QR|\neq|PR|\ \text{(the printed statement is not true)}
Note. Part (ii) cannot be proved: with the vertices as printed the sides are \approx4.20, 6.74 and 5.17, so the triangle is scalene (changing signs of the components does not give an equilateral triangle either). Every printed position vector has magnitude 4, which suggests the original statement was about equal magnitudes or that a vertex is misprinted. The book key gives no answer to question 6.
7
Solution
(i)Find the value of \lambda for which |\vec a|=|3\vec b| where \vec a=\hat{i}-3\hat{j}+\lambda\hat{k} and \vec b=\hat{i}+2\hat{j}-\hat{k}.
Find the magnitude of \vec a.
|\vec a|^2=1^2+(-3)^2+\lambda^2=10+\lambda^2
For a scalar multiple, |3\vec b|=3|\vec b|. Find |\vec b| first.
Two vectors are equal only if all corresponding components are equal (book, 3.2.6 ii). Here 2\neq-3 and 3\neq2, so the vectors are different. They have the same length but different directions.
|\vec a|=|\vec b|\ \text{but}\ \vec a\neq\vec b
Book answer. (i) \lambda=\pm2\sqrt{11}; (ii) |\vec a|=|\vec b|; \vec a\neq\vec b
8
If \vec a=2\hat{i}-3\hat{j}+\hat{k} and \vec b=\hat{i}-3\hat{j}+5\hat{k} then find:
Solution
(i)A vector of magnitude 5 in the direction \vec a-2\vec b.
Book answer. (i) \dfrac{5}{\sqrt{10}}\hat{j}-\dfrac{15}{\sqrt{10}}\hat{k}; (ii) -\dfrac{3}{\sqrt{257}}\hat{i}+\dfrac{36}{7\sqrt{257}}\hat{j}-\dfrac{24}{7\sqrt{257}}\hat{k}
9
The position vectors of points A and B are \hat{i}-2\hat{j}+\hat{k} and 2\hat{i}+3\hat{j}-\hat{k} respectively.
Solution
(i)Find the position vector of point P dividing the line segment joining A and B in the ratio 2:3 internally.
Write the position vectors, and the ratio m:n=2:3.
(ii)Find the position vector of point Q dividing the line segment \overline{AB} in the ratio 3:2 externally.
For external division in the ratio m:n (book 3.1.4, Case II) we have |\overrightarrow{AQ}|:|\overrightarrow{BQ}|=m:n and the position vector is \vec r=\dfrac{n\vec a-m\vec b}{n-m}. Here m=3, n=2.
\vec r=\dfrac{2\vec a-3\vec b}{2-3}=3\vec b-2\vec a
Substitute \vec a=\hat{i}-2\hat{j}+\hat{k} and \vec b=2\hat{i}+3\hat{j}-\hat{k}.
Check: \overrightarrow{AQ}=3\hat{i}+15\hat{j}-6\hat{k} and \overrightarrow{BQ}=2\hat{i}+10\hat{j}-4\hat{k}, so \overrightarrow{AQ}=\tfrac32\overrightarrow{BQ}. The two vectors point the same way (so Q is outside the segment, beyond B) and their lengths are in the ratio 3:2.
The book's Example 5 uses the other form \dfrac{n\vec b-m\vec a}{n-m}, which here gives 3\vec a-2\vec b=-\hat{i}-12\hat{j}+5\hat{k}. That is the external point beyond A, where |\overrightarrow{AQ}|:|\overrightarrow{BQ}|=2:3. The key's answer is of this form.
\vec r=4\hat{i}+13\hat{j}-5\hat{k}
Book answer. (i) \dfrac75\hat{i}+\dfrac15\hat{k}; (ii) \hat{i}-12\hat{j}+5\hat{k}
Note. Part (i) agrees with the key. For part (ii) the book's own derivation (Case II, |AP|:|BP|=m:n) gives 4\hat{i}+13\hat{j}-5\hat{k}, which is checked above. The key's \hat{i}-12\hat{j}+5\hat{k} matches the book's Example 5 pattern \dfrac{n\vec b-m\vec a}{n-m}=3\vec a-2\vec b, whose correct value is -\hat{i}-12\hat{j}+5\hat{k}; the sign of \hat{i} in the key looks like a misprint. If your teacher follows Example 5, write -\hat{i}-12\hat{j}+5\hat{k}.
10
Solution
(i)The three vertices of a parallelogram ABCD taken in order are A(3,-4); B(-1,-3) and C(-6,2). Find the fourth vertex D.
Let D=(x,y). In a parallelogram taken in order, opposite sides are equal as vectors: \overrightarrow{AB}=\overrightarrow{DC} (equal length, parallel and the same direction).
Find \overrightarrow{AB} and \overrightarrow{DC} (head minus tail).
Check with the diagonals: they bisect each other, so the midpoint of AC is \left(\tfrac{3-6}{2},\tfrac{-4+2}{2}\right)=\left(-\tfrac32,-1\right) and the midpoint of BD is \left(\tfrac{-1-2}{2},\tfrac{-3+1}{2}\right)=\left(-\tfrac32,-1\right). They agree.
ABCD with D(-2,1): opposite sides are parallel and equal.
D=(-2,1)
(ii)Find the values of x and y if A(1,2); B(4,y); C(x,6) and D(3,5) taken in order are the vertices of a parallelogram.
For a parallelogram ABCD in order, \overrightarrow{AB}=\overrightarrow{DC}.
Check: the midpoint of AC is \left(\tfrac{1+6}{2},\tfrac{2+6}{2}\right)=\left(\tfrac72,4\right) and the midpoint of BD is \left(\tfrac{4+3}{2},\tfrac{3+5}{2}\right)=\left(\tfrac72,4\right).
The parallelogram for x=6, y=3.
x=6,\quad y=3
Book answer. (i) D(-2,1); (ii) x=6 and y=3
11
Show that the line segments joining the mid points of the sides of a quadrilateral consecutively form a parallelogram.
Solution
The midpoints P,Q,R,S of the sides of ABCD are joined in order. PQRS is a parallelogram.
Let ABCD be any quadrilateral with position vectors \vec a,\vec b,\vec c,\vec d for A,B,C,D. Let P,Q,R,S be the midpoints of AB,BC,CD,DA.
The position vector of the midpoint of a segment is half the sum of the position vectors of its ends (book 3.1.4, Case I).
Therefore PQRS is a parallelogram. Its sides are parallel to the diagonals AC and BD of ABCD, and are half as long.
\overrightarrow{PQ}=\overrightarrow{SR}\ \Rightarrow\ PQRS\ \text{is a parallelogram}
12
Show that line segments joining the mid points of the diagonals and the mid points of any two opposite sides of a quadrilateral consecutively form a parallelogram.
Solution
E and F are the midpoints of the diagonals AC and BD; P and R are the midpoints of the opposite sides AB and CD. EPFR is the required quadrilateral.
Let ABCD have position vectors \vec a,\vec b,\vec c,\vec d. Let E,F be the midpoints of the diagonals AC,BD and P,R the midpoints of the opposite sides AB,CD. Taken in order the four points are E,P,F,R.
Both pairs of opposite sides are equal and parallel, so EPFR is a parallelogram.
\overrightarrow{EP}=\overrightarrow{RF}\ \text{and}\ \overrightarrow{PF}=\overrightarrow{ER}\ \Rightarrow\ EPFR\ \text{is a parallelogram}
13
Prove that line segments joining the midpoint of the diagonals of a trapezium is parallel to the parallel sides and its length is half the difference of the lengths of the parallel sides.
Solution
In the trapezium ABCD with AB\parallel DC, the segment EF joins the midpoints of the diagonals. It lies along the same direction as the parallel sides.
Let ABCD be a trapezium with AB\parallel DC and AB the longer side. Let the position vectors of A,B,C,D be \vec a,\vec b,\vec c,\vec d. Let E,F be the midpoints of the diagonals AC and BD.
Since AB\parallel DC and they point the same way in this trapezium, \overrightarrow{AB}=\lambda\,\overrightarrow{DC} for a scalar \lambda>1 (parallel vectors, book 3.1.2 vi).
OPQR is a trapezium made from three equilateral triangles with \overrightarrow{OP}=\vec r, \overrightarrow{OS}=\vec s and M is the midpoint of \overline{QR}.
OPQR is a trapezium made of three equilateral triangles OPS, PQS and SQR.
Solution
(i)Write \overrightarrow{PS} in terms of \vec r and \vec s.
Use the triangle law of addition in triangle OPS: going from O to P and then from P to S is the same as going from O to S.
Substitute \overrightarrow{OP}=\vec r and \overrightarrow{OS}=\vec s and solve for \overrightarrow{PS}.
\vec r+\overrightarrow{PS}=\vec s\;\Rightarrow\;\overrightarrow{PS}=\vec s-\vec r
\overrightarrow{PS}=\vec s-\vec r
(ii)Show that \overrightarrow{OQ} is parallel to \overrightarrow{SM}.
The three triangles are equilateral and congruent, so all their sides have the same length as OP. The sides PQ and SR lie along OR in the same direction as OS, so \overrightarrow{PQ}=\overrightarrow{SR}=\vec s.
Also SQ is parallel to OP (both make 60^\circ with OR) and has the same length, so \overrightarrow{SQ}=\vec r.
Find \overrightarrow{OQ} by going O\to P\to Q.
\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}=\vec r+\vec s
M is the midpoint of QR, so \overrightarrow{SM} is the midpoint vector of \overrightarrow{SQ} and \overrightarrow{SR} (taking S as origin, book 3.1.4 Case I).
\overrightarrow{SM} is a scalar multiple of \overrightarrow{OQ} (positive, so the same direction), hence \overrightarrow{OQ}\parallel\overrightarrow{SM}, and |OQ|=2|SM|.
\overrightarrow{OQ} and \overrightarrow{SM} (heavy lines) point the same way, and OQ is twice SM.
Book answer. (i) \overrightarrow{PS}=\vec s-\vec r
15
ABCD is a trapezium with \overrightarrow{AB} parallel to \overrightarrow{DC}. E is the point on the diagonal \overrightarrow{DB} such that DE=\dfrac13DB. Show that \overrightarrow{BC} is parallel to \overrightarrow{AE}.
ABCD is a trapezium with \overrightarrow{AB}=12\vec a, \overrightarrow{DC}=18\vec a and \overrightarrow{AD}=3\vec b. E lies on DB with DE=\tfrac13DB.
Solution
Read the vectors from the figure: \overrightarrow{AB}=12\vec a, \overrightarrow{DC}=18\vec a and \overrightarrow{AD}=3\vec b.
Find \overrightarrow{DB}: go from D back to A and then on to B.
\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}=-3\vec b+12\vec a
E is on DB with DE=\tfrac13DB (same direction), so \overrightarrow{DE} is one third of \overrightarrow{DB}.
\overrightarrow{DE}=\tfrac13\left(12\vec a-3\vec b\right)=4\vec a-\vec b
Find \overrightarrow{AE}: go from A to D and then from D to E.
\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}=3\vec b+4\vec a-\vec b=4\vec a+2\vec b
Find \overrightarrow{BC}: go from B to A, then A to D, then D to C.
\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DC}=-12\vec a+3\vec b+18\vec a=6\vec a+3\vec b
Compare. \overrightarrow{BC} is a scalar multiple of \overrightarrow{AE}.
ABCDEF is a regular hexagon as shown in the figure. If \overrightarrow{AB}=\vec a and \overrightarrow{BC}=\vec b, then express \overrightarrow{AC}, \overrightarrow{CD}, \overrightarrow{EF}, \overrightarrow{DA}, \overrightarrow{EB}, \overrightarrow{FA} and \overrightarrow{FC} in terms of \vec a and \vec b.
The regular hexagon ABCDEF with \overrightarrow{AB}=\vec a and \overrightarrow{BC}=\vec b.
Solution
Plan. Let O be the centre of the hexagon. In a regular hexagon OA=OB=OC=\dots equals the side, so the six triangles at O are equilateral. We first find the position vectors of all six vertices from O in terms of \vec a and \vec b, and then subtract (head minus tail).
ABCO has all four sides equal (each is the side of the hexagon), so it is a rhombus, hence a parallelogram. Opposite sides of a parallelogram are equal vectors.
\overrightarrow{OC}=\overrightarrow{AB}=\vec a,\qquad \overrightarrow{AO}=\overrightarrow{BC}=\vec b
In the same way BCDO is a rhombus, so \overrightarrow{OD}=\overrightarrow{BC}=\vec b.
\overrightarrow{OD}=\vec b
Now find the remaining position vectors. \overrightarrow{OA} is the opposite of \overrightarrow{AO}; \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}; E and F are opposite to B and C through the centre.
Book answer.\overrightarrow{AC}=\vec a+\vec b, \overrightarrow{CD}=\vec b-\vec a, \overrightarrow{EF}=-\vec b, \overrightarrow{DA}=-2\vec a, \overrightarrow{EB}=2(\vec a-\vec b), \overrightarrow{FA}=\vec a-\vec b, \overrightarrow{FC}=2\vec a
Note. Six of the seven agree with the key. The key prints \overrightarrow{DA}=-2\vec a, but DA is parallel to BC (not AB), so \overrightarrow{DA}=-2\vec b. With coordinates, \vec a=(1,0) and \vec b=\left(\tfrac12,\tfrac{\sqrt3}{2}\right) give \overrightarrow{DA}=(-1,-\sqrt3)=-2\vec b, not -2\vec a=(-2,0).
Exercise 3.2p. 97–98
The dot (scalar) product: angle between vectors, perpendicularity, projections, direction cosines and work done. Multiply matching components and add; \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} and W=\vec F\cdot\vec S.
1
If \vec a=2\hat{i}-3\hat{j}+\hat k; \vec b=\hat{i}-3\hat{j}+4\hat k and \vec c=-\hat{i}-2\hat{j}+5\hat k, then evaluate the followings.
Solution
(i)\vec a\cdot\vec b
Dot product in components: multiply the matching components and add (book 3.5.3), \vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3.
\vec a\cdot\vec b=(2)(1)+(-3)(-3)+(1)(4)
Evaluate.
=2+9+4=15
15
(ii)2\vec a\cdot3\vec b
Scalars come out of a dot product, (m\vec a)\cdot(n\vec b)=mn(\vec a\cdot\vec b). Use \vec a\cdot\vec b=15 from part (i).
(v)\hat\imath\cdot\vec a+\hat\jmath\cdot\vec b+\hat k\cdot\vec c
Dotting with \hat{i}, \hat{j}, \hat{k} picks out the \hat{i}, \hat{j}, \hat{k} component, because \hat{i}\cdot\hat{i}=1 and \hat{i}\cdot\hat{j}=0 (book 3.5.2).
Book answer. (i) 15; (ii) 90; (iii) -16; (iv) 147; (v) 4
2
If \vec a=\hat{j}-\hat{k}; \vec b=3\hat{i}-4\hat{j}+\hat{k} and \vec c=-\hat{i}+2\hat{j}-4\hat{k}, then find the angles between the vectors:
Solution
Method. For two non-zero vectors, \cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|} with 0^\circ\le\theta\le180^\circ (book 3.5.11). The sign of the dot product tells acute (positive) or obtuse (negative).
(i)\vec a and 3\vec b
Write the vectors. A positive multiple of \vec b points the same way, so the angle between \vec a and 3\vec b equals the angle between \vec a and \vec b. Still, we use 3\vec b directly.
Book answer. (i) \cos^{-1}\left(\dfrac{-5}{2\sqrt{13}}\right); (ii) \cos^{-1}\left(\dfrac{57}{\sqrt{6342}}\right); (iii) \cos^{-1}\left(\dfrac{-30}{\sqrt{1870}}\right); (iv) \cos^{-1}\left(\dfrac{-15}{\sqrt{357}}\right); (v) \cos^{-1}\left(\dfrac{18}{\sqrt{438}}\right)
Note. Parts (i), (ii), (iv) and (v) agree with the key. In part (iii) the denominator \sqrt{1870} agrees but the key prints -30 where we get -40: (-1)(5)+(1)(-8)+(-3)(9)=-40. Substituting back confirms -40 (the key value -30 would need a different vector), so the key looks misprinted.
3
Solution
(i)If \vec a,\vec b and \vec c are three vectors such that \vec a+\vec b+\vec c=\vec 0 and |\vec a|=2; |\vec b|=3 and |\vec c|=4 then find angle between \vec a and \vec b.
Isolate \vec c: from \vec a+\vec b+\vec c=\vec0 we get \vec c=-(\vec a+\vec b).
Take the dot product of each side with itself, using \vec x\cdot\vec x=|\vec x|^2 (book 3.5.1, Case III).
For non-zero vectors, |\vec a||\vec b|\cos\theta=0 gives \cos\theta=0.
\theta=90^\circ
\theta=90^\circ\ \text{(the vectors are perpendicular)}
Book answer. (i) \cos^{-1}\left(\dfrac14\right); (ii) 90^\circ
4
Solution
(i)If \vec a=\hat{i}-3\hat{j}+4\hat{k}; \vec b=7\hat{i}-9\hat{j}+\hat{k} and \vec c=3\hat{i}-2\hat{j}+5\hat{k}, find the value of \lambda so that \vec a-\lambda\vec b is perpendicular to \vec c.
Two vectors are perpendicular when their dot product is zero (book 3.5.4).
(\vec a-\lambda\vec b)\cdot\vec c=0
Distribute the dot product (distributive property, book 3.5.6) and take \lambda outside.
(ii)Show that the angle between any two diagonals of a cube is \cos^{-1}\left(\dfrac13\right).
Place the cube of side a with one corner at the origin and edges along the axes. A cube has four diagonals, each joining opposite corners. As direction vectors they are \vec d_1=(a,a,a) from O, and \vec d_2=(-a,a,a), \vec d_3=(a,-a,a), \vec d_4=(a,a,-a) along the other three.
Magnitude of each diagonal.
|\vec d|=\sqrt{a^2+a^2+a^2}=a\sqrt3
Dot product of two different diagonals, for example \vec d_1 and \vec d_2.
Every other pair gives a dot product of a^2 or -a^2 (one or three sign changes), because exactly one product (\pm a)(\pm a) changes sign each time, for example \vec d_2\cdot\vec d_3=-a^2-a^2+a^2=-a^2. So |\cos\theta|=\tfrac13 for every pair. Taking the acute angle between the lines (the usual angle between two diagonals), the angle is the same for all pairs.
A cube of side a with one corner at the origin. Two of its four diagonals (heavy), from O to (a,a,a) and from A to (0,a,a), join opposite corners.
\theta=\cos^{-1}\!\left(\dfrac13\right)
Book answer. (i) \lambda=\dfrac{29}{44}
5
Solution
(i)If \vec a=2\hat{i}-3\hat{j}+4\hat{k}, then find the direction cosine of \vec a.
Magnitude of \vec a.
|\vec a|=\sqrt{4+9+16}=\sqrt{29}
Direction cosines are the components divided by the magnitude: \cos\alpha=\dfrac{x}{|\vec r|}, \cos\beta=\dfrac{y}{|\vec r|}, \cos\gamma=\dfrac{z}{|\vec r|} (book 3.5.8).
(ii)If \vec a=\hat{i}-2\hat{j}+3\hat{k}; \vec b=3\hat{i}-2\hat{j}+\hat{k} and \vec c=7\hat{i}-\hat{j}+8\hat{k}, then find the projection of \vec a-\vec b along \vec c and projection of \vec b along \vec c-\vec a. Also find their vector projections.
Projection of \vec u along \vec v is \dfrac{\vec u\cdot\vec v}{|\vec v|}, and the vector projection is this times the unit vector \dfrac{\vec v}{|\vec v|}, that is \dfrac{\vec u\cdot\vec v}{|\vec v|^2}\vec v (book 3.5.12).
First pair: \vec a-\vec b=-2\hat{i}+2\hat{k} along \vec c.
Book answer. (i) \cos\alpha=\dfrac2{\sqrt{29}}, \cos\beta=\dfrac{-3}{\sqrt{29}}, \cos\gamma=\dfrac4{\sqrt{29}}; (ii) \dfrac{2}{\sqrt{114}}; \dfrac{21}{\sqrt{62}} (vector projections not printed)
6
Solution
(i)Three vertices of triangle are A(0,-1,-2); B(3,1,4) and C(5,7,1). Show that ABC is a right-angled triangle and find the other two angles.
Book answer. (i) 45^\circ, 45^\circ; (ii) \pm\dfrac5{\sqrt3}\hat{i}\pm\dfrac5{\sqrt3}\hat{j}\pm\dfrac5{\sqrt3}\hat{k}
7
Solution
(i)If \vec a,\vec b and \vec c are three vectors such that |\vec a|=2; |\vec b|=5; |\vec c|=4 and \vec a+\vec b+\vec c=\vec0 then find the value of \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a.
Dot the equation \vec a+\vec b+\vec c=\vec0 with itself.
(ii)For any vector \vec r prove that \vec r=(\vec r\cdot\hat{i})\hat{i}+(\vec r\cdot\hat{j})\hat{j}+(\vec r\cdot\hat{k})\hat{k}.
Write \vec r in components.
\vec r=x\hat{i}+y\hat{j}+z\hat{k}
Dot with each unit vector. Because \hat i\cdot\hat i=1 and \hat i\cdot\hat j=\hat i\cdot\hat k=0 (book 3.5.2), each dot product picks out one component.
The dot products of \vec r with the vectors \hat{i}+\hat{j}-3\hat{k}; \hat{i}+3\hat{j}-2\hat{k} and 2\hat{i}+\hat{j}+4\hat{k} are 0, 5 and 8 respectively. Find the vector \vec r.
Solution
Let \vec r=x\hat{i}+y\hat{j}+z\hat{k} and write each dot product as an equation.
(ii)Show that angle in a semi-circle is a right angle.
Let O be the centre, A and B the ends of a diameter, P any point on the circle. Take O as origin and let \overrightarrow{OB}=\vec a. Then \overrightarrow{OA}=-\vec a, and let \overrightarrow{OP}=\vec p with |\vec p|=|\vec a| (both are radii).
Find the vectors \overrightarrow{PA} and \overrightarrow{PB} (head minus tail).
\overrightarrow{PA}=-\vec a-\vec p,\qquad \overrightarrow{PB}=\vec a-\vec p
(i)Prove that altitudes of a triangle are concurrent.
Let two altitudes, from A and from B, meet at H. Take H as the origin and let the position vectors of A,B,C be \vec a,\vec b,\vec c. So \overrightarrow{HA}=\vec a, \overrightarrow{HB}=\vec b, \overrightarrow{HC}=\vec c.
HA is the altitude from A, so it is perpendicular to BC. Use \overrightarrow{BC}=\vec c-\vec b and the zero dot product.
The dot product is commutative, so (1) and (2) give \vec a\cdot\vec c=\vec b\cdot\vec c. Rearrange.
\vec c\cdot(\vec a-\vec b)=0
Since \vec a-\vec b=\overrightarrow{BA}, this says HC\perp AB. So the line HC is the altitude from C, and it passes through H.
All three altitudes pass through H: they are concurrent.
The three altitudes (dashed) of triangle ABC pass through one point H.
\overrightarrow{HC}\cdot\overrightarrow{AB}=0\ \Rightarrow\ \text{the three altitudes meet at }H
(ii)Prove that angle bisectors of triangle are concurrent.
Let a,b,c be the lengths of the sides BC,CA,AB opposite to A,B,C, and \vec A,\vec B,\vec C the position vectors of the vertices from any origin.
The vector \hat u=\dfrac{\overrightarrow{AB}}{c}+\dfrac{\overrightarrow{AC}}{b} is the sum of two unit vectors from A along the two sides. The diagonal of a rhombus bisects its angle, so \hat u lies along the bisector of angle A.
Define the point I by its position vector.
\vec I=\dfrac{a\vec A+b\vec B+c\vec C}{a+b+c}
Find \overrightarrow{AI} (head minus tail). Use a+b+c for the denominator and write a\vec A+b\vec A+c\vec A for (a+b+c)\vec A.
So \overrightarrow{AI} is a scalar multiple of \hat u: I lies on the bisector of angle A.
The expression for \vec I is symmetric in the three vertices, so the same argument at B and at C shows I lies on the other two bisectors. The three bisectors are concurrent at I.
\vec I=\dfrac{a\vec A+b\vec B+c\vec C}{a+b+c}\ \text{lies on all three bisectors}
12
Solution
(i)Prove that \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
Use the unit vectors of book Example 13: \hat u makes angle \alpha with the x-axis and \hat v makes angle -\beta (measured clockwise). The angle between them is \alpha-(-\beta)=\alpha+\beta.
(ii)With usual notations for a triangle ABC; prove that c^2=a^2+b^2-2ab\cos\gamma and b=a\cos\gamma+c\cos\alpha.
Take the sides as vectors head to tail (figure): \vec c=\overrightarrow{AB}, \vec a=\overrightarrow{BC}, \vec b=\overrightarrow{CA}. Going round the triangle returns to the start.
\vec a+\vec b+\vec c=\vec0
First formula. Write \vec c=-(\vec a+\vec b) and square (dot with itself).
c^2=|\vec a+\vec b|^2=a^2+b^2+2\,\vec a\cdot\vec b
The angle between \vec a=\overrightarrow{BC} and \vec b=\overrightarrow{CA}, placed head to tail, is 180^\circ-\gamma, and \cos(180^\circ-\gamma)=-\cos\gamma (as in book Example 14).
Triangle ABC with \vec c=\overrightarrow{AB}, \vec a=\overrightarrow{BC}, \vec b=\overrightarrow{CA} taken head to tail; a,b,c are the side lengths and \alpha,\beta,\gamma the angles at A,B,C.
(i)The resultant of two vectors \vec a and \vec b is perpendicular to \vec a and |\vec a|=\dfrac1{\sqrt7}|\vec b|. Show that resultant of 7\vec a and \vec b is perpendicular to \vec b.
The resultant is \vec a+\vec b. It is perpendicular to \vec a, so the dot product is zero.
The dot product is zero, so 7\vec a+\vec b\perp\vec b.
(7\vec a+\vec b)\cdot\vec b=0
(ii)Prove that \hat a+\hat b is equally inclined with \vec a and \vec b.
Let \theta_1,\theta_2 be the angles of \hat a+\hat b with \vec a and \vec b. The direction of \vec a is \hat a and of \vec b is \hat b, so use the unit vectors. Let \vec s=\hat a+\hat b.
The two cosines are equal and 0^\circ\le\theta\le180^\circ, so \theta_1=\theta_2.
\theta_1=\theta_2
14
A force of \vec F=3\hat{i}-5\hat{j}+7\hat{k} newtons is applied on a body and moves it at a distance of 14 meters in the direction of the vector \hat{i}-3\hat{j}+\hat{k}. How much work is done?
Solution
Work done is W=\vec F\cdot\vec S (book 3.5.13), where \vec S is the displacement vector. Its direction is that of \hat{i}-3\hat{j}+\hat{k} and its length is 14 m.
Find the work done by the forces 2\hat{i}+3\hat{j}-\hat{k} and 3\hat{i}+7\hat{j}+4\hat{k} acting on a particle in moving from point P to Q with position vectors \hat{i}-3\hat{j}+\tfrac32\hat{k} and 2\hat{i}-\hat{j}+\tfrac52\hat{k}.
Solution
Both forces act on the particle, so the total force is their vector sum (the resultant).
Work done W=\vec F\cdot\vec S (the sum of the works of the two forces).
W=(5)(1)+(10)(2)+(3)(1)=5+20+3=28
W=28\ \text{units}
Book answer. 28 units
16
A box is dragged on the surface of a floor by a string that is applying a force of 30\,\mathrm{N} at an angle of 30^\circ with the floor. Find the work done by the force when the box is displaced up-to a distance of 10 meters.
Solution
The force F=30\,\mathrm{N} (heavy) makes 30^\circ with the floor, and the box is displaced s=10\,\mathrm{m} along the floor.
Work done by a constant force is W=\vec F\cdot\vec S=|\vec F||\vec S|\cos\theta (book 3.5.13), where \theta is the angle between force and displacement. The displacement is along the floor, so \theta=30^\circ.
Substitute |\vec F|=30\,\mathrm{N}, |\vec S|=10\,\mathrm{m} and \cos30^\circ=\dfrac{\sqrt3}{2}.
W=(30)(10)\cos30^\circ=300\times\dfrac{\sqrt3}{2}
Simplify.
W=150\sqrt3\ \text{J}\approx259.8\ \text{J}
W=150\sqrt3\ \mathrm{J}
Book answer.150\sqrt3
Exercise 3.3p. 105–107
The cross (vector) product: expand \vec a\times\vec b as a determinant, then use |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta for angles and areas, and \vec M=\vec r\times\vec F for the moment of a force.
1
For the following vectors, find \vec a\times\vec b and \vec b\times\vec a and prove that \vec a\times\vec b=-\vec b\times\vec a.
Solution
Method.\vec a\times\vec b=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}. Swapping two rows of a determinant changes its sign, so \vec b\times\vec a=-\vec a\times\vec b (anti-commutative, book 3.6.1).
(ii)Express the vector 5\hat{i}+2\hat{j}-3\hat{k} as a sum of two vectors one of which is parallel and other is perpendicular to the vector 2\hat{i}-\hat{j}+3\hat{k}.
Let \vec v=5\hat{i}+2\hat{j}-3\hat{k} and \vec d=2\hat{i}-\hat{j}+3\hat{k}. Write \vec v=\vec v_\parallel+\vec v_\perp where \vec v_\parallel is along \vec d.
The parallel part is the vector projection of \vec v on \vec d: \vec v_\parallel=\dfrac{\vec v\cdot\vec d}{|\vec d|^2}\vec d (book 3.5.12).
Book answer. (i) \left(\dfrac{40}{\sqrt{1533}},\dfrac{185}{\sqrt{1533}},\dfrac{50}{\sqrt{1533}}\right); (ii) parallel -\dfrac15\hat{i}+\dfrac1{10}\hat{j}-\dfrac3{10}\hat{k}, perpendicular \dfrac{26}5\hat{i}+\dfrac{19}{10}\hat{j}-\dfrac{27}{10}\hat{k}
Note. Part (ii) disagrees with the key. The parallel part must be \dfrac{\vec v\cdot\vec d}{|\vec d|^2}\vec d and |\vec d|^2=14 (the key effectively divides by 10). The key's two parts do add up to \vec v, but its "perpendicular" part has dot product \tfrac25 with \vec d, not 0, so it is not perpendicular. Our parts add to \vec v and the dot product of the perpendicular part with \vec d is 0.
(ii)For the vectors \vec a=\hat{i}-2\hat{j}, \vec b=2\hat{i}+\hat{k} and \vec c=3\hat{j}+2\hat{k}, find a vector \vec d which is perpendicular to both \vec a and \vec b. It is given that \vec c\cdot\vec d=1.
Write the cross product as a determinant (book 3.6.3) with the components of the first vector in row 2 and of the second in row 3.
Book answer. (ii) \vec d=-\dfrac25\hat{i}-\dfrac15\hat{j}+\dfrac45\hat{k}
6
Solution
(i)Find the vector \vec b such that \vec a\times\vec b=\vec c and \vec a\cdot\vec b=3; where \vec a=\hat{i}-2\hat{j}+3\hat{k} and \vec c=\hat{i}+\hat{j}-\hat{k}.
The vector \vec a\times\vec b is perpendicular to \vec a, so a solution can exist only if \vec c\cdot\vec a=0. Test this.
\vec c\cdot\vec a=(1)(1)+(1)(-2)+(-1)(3)=-4\neq0
So \vec c is not perpendicular to \vec a, and no vector \vec b satisfies \vec a\times\vec b=\vec c.
To see what the usual method gives, take \vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b|\vec a|^2. With \vec a\times\vec c=-\hat{i}+4\hat{j}+3\hat{k} and |\vec a|^2=14 this gives
But then \vec a\times\vec b=\tfrac97\hat{i}+\tfrac37\hat{j}-\tfrac17\hat{k}\neq\vec c, confirming there is no solution.
\text{no such vector }\vec b\text{ exists (as printed)}
(ii)If \vec a\times\vec b=\vec c\times\vec d and \vec a\times\vec c=\vec b\times\vec d, show that \vec a-\vec d is parallel to \vec b-\vec c; where \vec a\neq\vec d and \vec b\neq\vec c.
Two non-zero vectors are parallel when their cross product is zero (book 3.6.5). Find (\vec a-\vec d)\times(\vec b-\vec c) using the distributive law (book 3.6.6).
Book answer. (i) the key prints "\vec a=\hat{i}+2\hat{j}+3\hat{k}"
Note. Part (i) as printed has no solution because \vec a\times\vec b is always perpendicular to \vec a, while \vec c\cdot\vec a=-4. The key's vector \hat{i}+2\hat{j}+3\hat{k} does not work either: it gives \vec a\cdot\vec b=6 and \vec a\times\vec b=-12\hat{i}+4\hat{k}. One of the printed vectors is probably a misprint.
7
Solution
(i)For a non-zero vector \vec a if \vec a\cdot\vec b=\vec a\cdot\vec c and \vec a\times\vec b=\vec a\times\vec c then show that \vec b=\vec c.
Move everything to one side and use the distributive laws for the dot and the cross product.
Suppose \vec b-\vec c\neq\vec0 and let \theta be the angle between \vec a and \vec b-\vec c. The first equation gives |\vec a||\vec b-\vec c|\cos\theta=0 and the second gives |\vec a||\vec b-\vec c|\sin\theta=0. Since |\vec a|\neq0 and |\vec b-\vec c|\neq0, we would need \cos\theta=0 and \sin\theta=0 together.
That is impossible, because \cos^2\theta+\sin^2\theta=1. So the supposition is false: \vec b-\vec c=\vec0.
\vec b=\vec c
\vec b=\vec c
(ii)For three vectors \vec a,\vec b and \vec c if \vec a+\vec b+\vec c=\vec0, then prove that: \vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a.
Take the cross product of \vec a+\vec b+\vec c=\vec0 with \vec a on the left, using \vec a\times\vec a=\vec0.
\vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a
\vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a
8
Solution
(i)If \vec a,\vec b and \vec c are three unit vectors such that \vec a is perpendicular to both \vec b and \vec c and the angle between \vec b and \vec c is \dfrac\pi6, then prove that \vec a=\pm2(\vec b\times\vec c).
\vec b\times\vec c is perpendicular to both \vec b and \vec c, and so is \vec a. Two vectors perpendicular to the same two (non-parallel) vectors are parallel, so \vec a\parallel\vec b\times\vec c.
Find the magnitude of \vec b\times\vec c with |\vec b|=|\vec c|=1 and angle \dfrac\pi6.
Note. In part (ii) the data fix only (\vec a\cdot\vec b)^2=36. The dot product is +6 when the angle between the vectors is acute and -6 when it is obtuse, so both \pm6 are possible; the key lists the positive value.
10
Solution
(i)Find the area of a parallelogram if \vec a=2\hat{i}-3\hat{j}+\hat{k} and \vec b=\hat{i}-2\hat{j}+7\hat{k} are its two adjacent sides.
Write the cross product as a determinant (book 3.6.3) with the components of the first vector in row 2 and of the second in row 3.
(ii)If \vec a,\vec b and \vec c are the position vectors of A,B and C respectively, then show that area of triangle ABC is \dfrac12|\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a|.
The area of the triangle is \tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}| (book 3.6.4).
Write the sides with position vectors (head minus tail).
\overrightarrow{AB}=\vec b-\vec a,\qquad \overrightarrow{AC}=\vec c-\vec a
If both vectors were non-zero, |\vec a||\vec b|\neq0, and we would need \cos\theta=0 and \sin\theta=0 together. That is impossible because \cos^2\theta+\sin^2\theta=1.
So |\vec a||\vec b|=0.
|\vec a|=0\ \text{or}\ |\vec b|=0
Conclusion: at least one of the vectors is the zero vector, that is \vec a=\vec0 or \vec b=\vec0 (or both). (The zero vector is both parallel and perpendicular to every vector, so both conditions hold.)
Book answer. Either \vec a=\vec 0 or \vec b=\vec0 or both are zero
13
Show that the three points with position vectors \vec a-\vec b+3\vec c, 2\vec a+3\vec b-4\vec c and -7\vec b+10\vec c are collinear.
Solution
Name the points P,Q,R with position vectors \vec p=\vec a-\vec b+3\vec c, \vec q=2\vec a+3\vec b-4\vec c and \vec r=-7\vec b+10\vec c. They are collinear if \overrightarrow{PQ} and \overrightarrow{PR} are parallel, that is one is a scalar multiple of the other. Find both (head minus tail).
For \overrightarrow{PR}=t\,\overrightarrow{PQ} the coefficients of \vec a,\vec b,\vec c would have to be in the same ratio (for independent \vec a,\vec b,\vec c).
So with \vec p=\vec a-2\vec b+3\vec c we get \overrightarrow{PR}=-\overrightarrow{PQ}: PQ and PR are parallel and share the point P, so P,Q,R are collinear.
Note. As printed (first point \vec a-\vec b+3\vec c) the points are not collinear. The first point is almost certainly a misprint for \vec a-2\vec b+3\vec c, which makes them collinear, as the question says. The book key gives no answer for question 13.
14
Solution
Method. The moment (torque) of a force \vec F applied at P about a point O is \vec M=\vec r\times\vec F, where \vec r=\overrightarrow{OP} (book 3.6.8). Its magnitude is the moment of the force.
(i)Find the moment of force 2\hat{i}+3\hat{j}+7\hat{k} about the point (1,2,3) when applied at the point (-1,2,0).
Position vector of the point of application relative to the point O(1,2,3) (head minus tail).
(ii)Two forces 2\hat{i}-\hat{j}+3\hat{k} and 3\hat{i}+4\hat{j}-2\hat{k} are applied at the same point (1,-2,4). Find the moment of these concurrent forces about (0,0,0).
Both forces act at the same point, so the moment of the pair is the moment of their resultant (distributive law: \vec r\times\vec F_1+\vec r\times\vec F_2=\vec r\times(\vec F_1+\vec F_2)).
Book answer. (i) \sqrt{181}; (ii) 11\sqrt6 (the key gives the magnitudes)
15
Solution
(i)How much force is required to produce a moment of magnitude \sqrt{57}\ \mathrm{N.m} along the direction 6\hat{i}-21\hat{j}-6\hat{k} when applied at (2,1,-3) about (-1,-1,1).
Position vector of the point of application about O(-1,-1,1).
The moment vector has magnitude \sqrt{57} along 6\hat{i}-21\hat{j}-6\hat{k}=3(2\hat{i}-7\hat{j}-2\hat{k}). Since |2\hat{i}-7\hat{j}-2\hat{k}|=\sqrt{4+49+4}=\sqrt{57}, the moment vector with magnitude \sqrt{57} is exactly this vector.
\vec M=2\hat{i}-7\hat{j}-2\hat{k}
Let the force be \vec F=x\hat{i}+y\hat{j}+z\hat{k} and write \vec r\times\vec F.
Take z=t as a free parameter. From the first, y=\dfrac{1-t}2. From the second, x=\dfrac{7-3t}4. Check the third: 3\cdot\dfrac{1-t}{2}-2\cdot\dfrac{7-3t}{4}=\dfrac{3-3t-7+3t}{2}=-2. It holds for every t.
So the force is not unique: any t works, because adding a multiple of \vec r to \vec F does not change \vec r\times\vec F.
(ii)At what point the force 2\hat{i}+2\hat{j}-3\hat{k} should be applied to produce a vector moment \vec M=3\hat{i}-2\hat{j}+\hat{k} about (-1,2,-3).
Let the point be (x,y,z) and \vec r=\overrightarrow{OP}=(x+1)\hat{i}+(y-2)\hat{j}+(z+3)\hat{k}. Write \vec r=r_1\hat{i}+r_2\hat{j}+r_3\hat{k} with \vec F=2\hat{i}+2\hat{j}-3\hat{k} and expand \vec r\times\vec F.
The second equation fails for every r_3. This is because \vec r\times\vec F must be perpendicular to \vec F, but \vec M\cdot\vec F=6-4-3=-1\neq0. So with the data as printed no point exists.
Using only the first and third equations gives the family in the key, with z=t:
\text{no point satisfies all three equations; the first and third give }\left(-\tfrac{4t+21}{6},-\tfrac{2t+3}{3},t\right)
Book answer. (i) \dfrac14(7-3t)\hat{i}+\dfrac12(1-t)\hat{j}+t\hat{k}; (ii) \left(-\dfrac16(4t+21),-\dfrac13(2t+3),t\right)
Note. Part (i) agrees with the key. In part (ii) the key solves only two of the three component equations. A moment \vec M=\vec r\times\vec F is always perpendicular to \vec F, but here \vec M\cdot\vec F=-1\neq0, so the problem as printed is not consistent. The key's points satisfy the first and third equations only.
16
A toy car is located at a point (2,3,5) relative to origin, such that when a force of 3\hat{i}+2\hat{j}+7\hat{k} is applied on car it starts rotating about the origin. Find the moment produced by the force in the car.
Solution
The moment about the origin is \vec M=\vec r\times\vec F with \vec r=2\hat{i}+3\hat{j}+5\hat{k} and \vec F=3\hat{i}+2\hat{j}+7\hat{k}.
Write the cross product as a determinant (book 3.6.3) with the components of the first vector in row 2 and of the second in row 3.
A seesaw is fixed from its middle point which is at (0,2,3). Two forces F_1[3,4,5] and F_2[9,2,7] are applied at points (4,5,3) and (-4,-1,-3) respectively. Calculate the moment produced by each force about the fixed point in seesaw separately. Also find net moment.
Solution
The fixed point is O(0,2,3). For each force, \vec M=\vec r\times\vec F where \vec r goes from O to the point of application. Brackets [3,4,5] mean 3\hat{i}+4\hat{j}+5\hat{k}.
First force.\vec r_1 from O to (4,5,3) (head minus tail).
\vec r_1=4\hat{i}+3\hat{j}+0\hat{k}
Write the cross product as a determinant (book 3.6.3) with the components of the first vector in row 2 and of the second in row 3.
Book answer.15\hat{i}-20\hat{j}+7\hat{k}; -9\hat{i}-26\hat{j}+19\hat{k}; 6\hat{i}-46\hat{j}+26\hat{k}
Exercise 3.4p. 112
The scalar triple product \vec a\cdot\vec b\times\vec c as a 3\times3 determinant: interchange rules, the coplanarity condition =0, and volumes of a parallelepiped |\vec a\cdot\vec b\times\vec c| and a tetrahedron \tfrac16|\vec a\cdot\vec b\times\vec c|.
12026 Annual I, Set B, Q.6 OR (a)2026 Annual I, Set B, Q.6 OR (b)
For the given vectors \vec a,\vec b and \vec c; prove that \vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a=\vec c\cdot\vec a\times\vec b
Solution
Method.\vec a\cdot\vec b\times\vec c=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}. Moving the first row to the bottom (a cyclic shift) is two interchanges of rows, and each interchange changes the sign, so the value is unchanged.
Now \vec b\cdot\vec c\times\vec a: the rows are \vec b,\vec c,\vec a. This is a cyclic shift of the rows, which is two row interchanges, so the value does not change.
Now \vec b\cdot\vec c\times\vec a: the rows are \vec b,\vec c,\vec a. This is a cyclic shift of the rows, which is two row interchanges, so the value does not change.
-\vec b\cdot\vec a\times\vec c: the rows are \vec b,\vec a,\vec c, which is rows 1 and 2 interchanged, so the determinant changes sign and the minus sign in front restores it.
-\vec b\cdot\vec a\times\vec c: the rows are \vec b,\vec a,\vec c, which is rows 1 and 2 interchanged, so the determinant changes sign and the minus sign in front restores it.
(ii)Find the value of \lambda so that the vectors \vec a=\hat{i}-2\hat{j}+3\hat{k}; \vec b=-2\hat{i}+3\hat{j}-4\hat{k} and \vec c=-\hat{i}+\lambda\hat{j}+2\hat{k} are coplanar.
(ii)If the vectors \vec a=\alpha\hat{i}+\hat{j}+\hat{k}; \vec b=\hat{i}+\beta\hat{j}+\hat{k} and \vec c=\hat{i}+\hat{j}+\gamma\hat{k} are coplanar, then prove that \dfrac{1}{1-\alpha}+\dfrac{1}{1-\beta}+\dfrac{1}{1-\gamma}=1 where \alpha,\beta,\gamma\neq1.
Coplanar means the determinant of the components is zero.
(i)If \vec a,\vec b and \vec c are coplanar then show that \vec a+\vec b, \vec b+\vec c and \vec c+\vec a are also coplanar.
Coplanar means the scalar triple product is zero, so \vec a\cdot\vec b\times\vec c=0. We must show (\vec a+\vec b)\cdot(\vec b+\vec c)\times(\vec c+\vec a)=0.
Expand the cross product with the distributive law and \vec c\times\vec c=\vec0.
Dot with \vec a+\vec b. Any triple product with a repeated vector is zero (book 3.7.2), so \vec a\cdot\vec b\times\vec a, \vec a\cdot\vec c\times\vec a, \vec b\cdot\vec b\times\vec c and \vec b\cdot\vec b\times\vec a vanish.
(ii)If \vec a,\vec b and \vec c are non-zero vectors such that \vec c is a unit vector perpendicular to both \vec a and \vec b and the angle between \vec a and \vec b is \dfrac\pi6; then prove that [\vec a\ \vec b\ \vec c]^2=\dfrac14|\vec a|^2|\vec b|^2.
The notation [\vec a\ \vec b\ \vec c] is the scalar triple product \vec a\cdot\vec b\times\vec c=(\vec a\times\vec b)\cdot\vec c (dot and cross can be interchanged, book 3.7.4).
\vec c is perpendicular to both \vec a and \vec b, so it is parallel to \vec a\times\vec b. Being a unit vector, \vec c=\pm\dfrac{\vec a\times\vec b}{|\vec a\times\vec b|}.
Then the triple product is plus or minus the magnitude of the cross product.
The volume of a parallelepiped with coterminal edges \vec a,\vec b,\vec c is the magnitude of the scalar triple product, V=|\vec a\cdot\vec b\times\vec c| (book 3.7.5).
Write the scalar triple product as a determinant, one vector in each row (book 3.7.2).
The volume of a parallelepiped with coterminal edges \vec a,\vec b,\vec c is the magnitude of the scalar triple product, V=|\vec a\cdot\vec b\times\vec c| (book 3.7.5).
Write the scalar triple product as a determinant, one vector in each row (book 3.7.2).
Volume is non-negative, so take the absolute value and divide by 6.
V=\dfrac16|-18|=3
V=3\ \text{cubic units}
Book answer. (i) \dfrac{27}{6}; (ii) 3
Review Exercise 3p. 114–115
Mixed review of the whole unit: unit vectors, dot and cross products, projections, scalar triple product, the sine rule by vectors and air-velocity problems (\vec G=\vec P+\vec W).
1 (i)
Choose the correct option. The vector in the direction of \hat{i}+2\hat{j}-2\hat{k} and having magnitude 12 is:
A\tfrac1{12}(\hat{i}+2\hat{j}-2\hat{k})
B12(\hat{i}+2\hat{j}-2\hat{k})
C\tfrac14(\hat{i}+2\hat{j}-2\hat{k})
D4(\hat{i}+2\hat{j}-2\hat{k})Answer
Why
Find the magnitude of the given direction vector.
|\hat{i}+2\hat{j}-2\hat{k}|=\sqrt{1+4+4}=3
A vector of magnitude 12 in this direction is 12 times the unit vector, that is \dfrac{12}{3} times the given vector.
The position vectors of three vertices of triangle are 2\hat{i}+\hat{j}-\hat{k}, 3\hat{i}-2\hat{j}+4\hat{k} and \hat{i}+4\hat{j}-3\hat{k}. The triangle is:
Aisosceles
Bright angled
CscaleneAnswer
Dequilateral
Why
Name the points A(2,1,-1), B(3,-2,4), C(1,4,-3). Find the squared side lengths from the side vectors.
All three are different, so the triangle is not isosceles or equilateral. For a right angle the largest square would equal the sum of the other two: 35+14=49\neq89.
So the triangle is scalene.
Book answer. c
1 (iii)
Given two vectors \hat{i}-\hat{j} and \hat{i}+2\hat{j}, then the unit vector coplanar with two vectors and \perp to first is:
A\pm\tfrac1{\sqrt2}(\hat{i}-\hat{j})
B\pm\tfrac1{\sqrt5}(2\hat{i}+\hat{j})
C\pm\tfrac1{\sqrt2}(\hat{i}+\hat{j})Answer
D\pm\tfrac1{\sqrt5}(\hat{i}+2\hat{j})
Why
Both given vectors lie in the xy-plane, so a coplanar vector is x\hat{i}+y\hat{j}. Perpendicular to \hat{i}-\hat{j} means the dot product is zero.
So the vector is along \hat{i}+\hat{j}. Divide by its magnitude \sqrt2 to get a unit vector (two signs).
\pm\dfrac{1}{\sqrt2}(\hat{i}+\hat{j})
Book answer. b
Note. In the book the third option is printed with the labels "b. c.", so its label is a misprint for c. The key says b, which refers to this printed option (the only one that is perpendicular to the first vector).
If \vec a=2\hat{i}+3\hat{j}-\hat{k}; \vec b=-\hat{i}+2\hat{j}-4\hat{k} and \vec c=\hat{i}+\hat{j}+\hat{k} then the value of (\vec a\times\vec b)\cdot(\vec a\times\vec c) is:
A scalar triple product is zero when its three vectors are coplanar, and three vectors with sum zero are coplanar (the third is a combination of the others). So the value is 0 whatever the magnitudes are.
The scalar product of \hat{i}+\hat{j}-\hat{k} with the unit vector along the sum of the vectors 2\hat{i}-3\hat{j}+\hat{k} and \lambda\hat{i}-2\hat{j}+3\hat{k} is -1. Find the value of \lambda.
Check: \lambda-7=-\tfrac{61}{9}<0, as required by \lambda-7=-\sqrt{\cdots}. With \lambda=\tfrac29: \vec s=\tfrac{20}{9}\hat{i}-5\hat{j}+4\hat{k}, |\vec s|=\tfrac{61}{9} and \dfrac{\vec s\cdot(1,1,-1)}{|\vec s|}=\dfrac{-61/9}{61/9}=-1.
\lambda=\dfrac29
Book answer.-\dfrac{11}{2}
Note. The key prints -\tfrac{11}{2}, but substituting \lambda=-\tfrac{11}{2} gives \vec s=-\tfrac72\hat{i}-5\hat{j}+4\hat{k} and a scalar product \dfrac{-12.5}{\sqrt{53.25}}\approx-1.71\neq-1, so it does not satisfy the condition. \lambda=\tfrac29 does, as checked above.
6
For any vector \vec a; prove that |\vec a\times\hat{i}|^2+|\vec a\times\hat{j}|^2+|\vec a\times\hat{k}|^2=2|\vec a|^2.
Solution
Let \vec a=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}. Use the Lagrange identity |\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2 with \vec b equal to each unit vector (so |\vec b|=1).
With usual notations for a triangle ABC; prove that \dfrac{a}{\sin\alpha}=\dfrac{b}{\sin\beta}=\dfrac{c}{\sin\gamma} by vector method.
Solution
Triangle ABC with \vec c=\overrightarrow{AB}, \vec a=\overrightarrow{BC}, \vec b=\overrightarrow{CA} taken head to tail; a,b,c are the side lengths opposite A,B,C and \alpha,\beta,\gamma the angles.
Take the sides as vectors head to tail: \vec c=\overrightarrow{AB}, \vec a=\overrightarrow{BC}, \vec b=\overrightarrow{CA}. Going round the triangle returns to the start.
\vec a+\vec b+\vec c=\vec0
For three vectors with sum zero, the cross products of consecutive pairs are equal (Exercise 3.3, question 7(ii)).
\vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a
Take magnitudes using |\vec x\times\vec y|=|\vec x||\vec y|\sin\theta. The angle between \vec a=\overrightarrow{BC} and \vec b=\overrightarrow{CA} (head to tail) is 180^\circ-\gamma, between \vec b,\vec c it is 180^\circ-\alpha and between \vec c,\vec a it is 180^\circ-\beta. Since \sin(180^\circ-\theta)=\sin\theta:
Suppose an airplane has a velocity relative to the air with a speed of 200 km/h and a direction of 60^\circ. Suppose the wind is blowing from the west at 40 km/h. Calculate the ground speed and the true course for the plane.
Solution
Method (as in the book's Example 28). Directions are headings measured clockwise from north, so heading 60^\circ is the angle 90^\circ-60^\circ=30^\circ from the x-axis (east). The ground velocity is the sum of the plane's velocity relative to the air and the wind velocity.
Book answer. Ground speed \approx235.492 km/h; true course \approx64.872^\circ
9
A pilot wants to have a true course of 100^\circ with a ground speed of 250 km/h. If the wind has a velocity vector (r,\theta)=(20,30^\circ), what should be the speed and direction of airplane with respect to air.
Solution
Method. Ground velocity \vec G=\vec P+\vec W, so the velocity relative to the air is \vec P=\vec G-\vec W. The course 100^\circ is a heading from north, that is the angle 90^\circ-100^\circ=-10^\circ from the x-axis. The wind vector (r,\theta)=(20,30^\circ) is in polar form: magnitude 20 at the angle 30^\circ from the x-axis.
Book answer. Speed \approx237.816 km/h; direction \approx107.980^\circ
Note. We read the wind (r,\theta)=(20,30^\circ) as polar form, as the notation says. The key's values come from treating the wind as the components 20\hat{i}+30\hat{j}: then \vec P=226.20\hat{i}-73.41\hat{j}, giving 237.82 km/h and 107.98^\circ. That reading is not what (r,\theta) means, so our values differ from the key.
FBISE 2025 and 2026 questions
Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.
2025 Annual I, Set B, MCQ 111 mark
If dot product of two vectors is zero then angle between them is:
A180^\circ
B90^\circAnswer
C45^\circ
D0^\circ
Why
\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 gives \cos\theta=0, so \theta=90^\circ.
0^\circ would make the dot product maximum (|\vec a||\vec b|), and 180^\circ makes it -|\vec a||\vec b|; it is the cross product that vanishes at 0^\circ and 180^\circ.
2025 Annual I, Set B, MCQ 121 mark
If \vec a\cdot(\vec b\times\vec c)=12 then volume of parallelepiped with \vec a, \vec b and \vec c as co-terminal edges is:
A6
B12Answer
C24
D36
Why
The volume of the parallelepiped with co-terminal edges \vec a,\vec b,\vec c is |\vec a\cdot\vec b\times\vec c|=|12|=12 (book 3.7.5).
Options A, C and D come from dividing or multiplying 12 by some factor. The \tfrac16 is only for the tetrahedron (which would give 2), so it is not needed here; the parallelepiped volume is the triple product itself.
Note. The printed question reads a.b x c = 12; it is taken as the scalar triple product \vec a\cdot(\vec b\times\vec c).
2025 Annual I, Set B, MCQ 131 mark
A force \vec F=2\hat i+\hat j+3\hat k moves an object along a displacement vector \vec d=3\hat i+4\hat j-2\hat k. What is the work done?
Option B, 10, comes from ignoring the \hat k term (6+4+0), option D, 16, from using +6 instead of -6 for it, and option A, -4, from flipping the overall sign. The \hat k product is (3)(-2)=-6.
2026 Annual I, Set B, MCQ 161 mark
Scalar product of two non-zero vectors is maximum when both vectors are:
AParallelAnswer
BIn opposite direction
CMutually perpendicular
DAt 45^\circ to each other
Why
\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta is largest when \cos\theta=1, that is \theta=0^\circ, so the vectors are parallel (same direction).
Mutually perpendicular vectors give \cos90^\circ=0, the smallest magnitude, not the largest; opposite directions give \cos180^\circ=-1, the minimum value.
2026 Annual I, Set B, MCQ 171 mark
If vector \vec u=2\hat i+3\hat j-4\hat k and \vec v=-2\hat i-\hat k then the dot product of \vec u and \vec v will be:
A6
B5
C0Answer
D-2
Why
\vec u\cdot\vec v=(2)(-2)+(3)(0)+(-4)(-1)=-4+0+4=0. The \hat j component of \vec v is 0, so the middle product vanishes.
The vectors are perpendicular. Option D, -2, comes from adding -4 and the +2 that is not there; the \hat k term is (-4)(-1)=+4, not -4.
2026 Annual I, Set B, MCQ 181 mark
Three vectors \vec P,\vec Q and \vec R are coplanar if:
A\vec P\times(\vec Q\times\vec R)=0
B(\vec P\cdot\vec Q)\times\vec R=0
C(\vec P\cdot\vec Q)\cdot\vec R=0
D\vec P\cdot(\vec Q\times\vec R)=0Answer
Why
Three vectors are coplanar when their scalar triple product is zero, \vec P\cdot(\vec Q\times\vec R)=0 (book 3.7.6): \vec Q\times\vec R is perpendicular to the plane of \vec Q and \vec R, and if \vec P lies in that plane it is perpendicular to \vec Q\times\vec R.
Options B and C are meaningless: \vec P\cdot\vec Q is a scalar, and a scalar has no cross or dot product with a vector. Option A is a vector (zero or not) and is a different condition.
2025 Annual I, Set B, Q.2 (i) OR4 marks
A triangular park has its three corners located at points A(2, 3, 1), B(5, 7, 2), and C(1, 4, 3). Find area of the park using cross product of vectors.
Solution
The area of a triangle with sides \overrightarrow{AB} and \overrightarrow{AC} is \dfrac12|\overrightarrow{AB}\times\overrightarrow{AC}| (book 3.6.4).
Find the two side vectors from A(2,3,1) (head minus tail).
\overrightarrow{AB}=3\hat i+4\hat j+\hat k,\qquad\overrightarrow{AC}=-\hat i+\hat j+2\hat k
A force \vec F=3\hat i-2\hat j+\hat k acts at point (1,2,3). Find the moment about the origin.
Solution
The moment of a force \vec F about a point O is \vec M=\vec r\times\vec F, where \vec r is the position vector, from O, of the point where the force acts (book 3.6.8). About the origin \vec r=\hat i+2\hat j+3\hat k.
Write the cross product as a determinant, with \vec r in row 2 and \vec F=3\hat i-2\hat j+\hat k in row 3.
Volume is the absolute value, since a volume cannot be negative.
V=|-4|=4
V=4\ \text{cubic units}
2026 Annual I, Set B, Q.2 (x) OR4 marks
Find the value of \lambda so that vectors \vec v=\hat i-\hat j+\hat k, \vec u=2\hat i+\hat j+\hat k and \vec w=-\hat i+\lambda\hat j+2\hat k are coplanar.
Solution
Three vectors are coplanar when \vec v\cdot\vec u\times\vec w=0 (book 3.7.6), that is, when their determinant is zero.
Write the scalar triple product as a determinant, one vector in each row (book 3.7.2).
Expand along the first row (the middle term has a minus sign).
\vec a\times\vec b=((-2)(-1)-4\cdot2)\hat i-(3\cdot(-1)-4\cdot1)\hat j+(3\cdot2-(-2)\cdot1)\hat k
Evaluate the brackets.
\vec a\times\vec b=-6\hat i+7\hat j+8\hat k
\vec a\times\vec b=-6\hat i+7\hat j+8\hat k
2025 Annual I, Set B, Q.3 OR (b)
If \vec a=3\hat i-2\hat j+4\hat k and \vec b=\hat i+2\hat j-\hat k, find (\vec a\times\vec b)\cdot\vec a and verify that \vec a\times\vec b\perp\vec a.
Solution
From part (a), \vec a\times\vec b=-6\hat i+7\hat j+8\hat k. Take its dot product with \vec a.
Multiply matching components and add.
(-6)(3)+(7)(-2)+(8)(4)=-18-14+32
The dot product is zero, and neither vector is zero, so \cos\theta=0: the angle between \vec a\times\vec b and \vec a is 90^\circ.
(\vec a\times\vec b)\cdot\vec a=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec a
(\vec a\times\vec b)\cdot\vec a=0,\ \text{so }\vec a\times\vec b\perp\vec a
2025 Annual I, Set B, Q.3 OR (c)
If \vec a=3\hat i-2\hat j+4\hat k and \vec b=\hat i+2\hat j-\hat k, find (\vec a\times\vec b)\cdot\vec b and verify that \vec a\times\vec b\perp\vec b.
Solution
From part (a), \vec a\times\vec b=-6\hat i+7\hat j+8\hat k. Take its dot product with \vec b.
Multiply matching components and add.
(-6)(1)+(7)(2)+(8)(-1)=-6+14-8
The dot product is zero, so the angle is 90^\circ.
(\vec a\times\vec b)\cdot\vec b=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec b
(\vec a\times\vec b)\cdot\vec b=0,\ \text{so }\vec a\times\vec b\perp\vec b
2026 Annual I, Set B, Q.6 OR (a)5+3 = 5 marks
Prove that scalar triple product is invariant under cyclic permutation.
Solution
Statement
For any three vectors, \vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a=\vec c\cdot\vec a\times\vec b. Moving the vectors one place round the cycle \vec a\to\vec b\to\vec c\to\vec a leaves the scalar triple product unchanged.
Proof
Let \vec a=a_1\hat i+a_2\hat j+a_3\hat k, \vec b=b_1\hat i+b_2\hat j+b_3\hat k, \vec c=c_1\hat i+c_2\hat j+c_3\hat k. The scalar triple product is the determinant with one vector in each row (book 3.7.2).
Go from the first determinant to the second by two row interchanges: swap R_1\leftrightarrow R_2 (rows become \vec b,\vec a,\vec c), then R_2\leftrightarrow R_3 (rows become \vec b,\vec c,\vec a). Interchanging two rows changes the sign of a determinant, so two interchanges change it twice, which restores the original sign.
(-1)(-1)=+1
Hence the two determinants are equal.
\vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a
The same argument applied once more (rows \vec c,\vec a,\vec b) gives the third equality.
\vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a=\vec c\cdot\vec a\times\vec b
\vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a=\vec c\cdot\vec a\times\vec b
2026 Annual I, Set B, Q.6 OR (b)5+3 = 3 marks
Verify, using vectors \hat i,\hat j,\hat k, that the scalar triple product is invariant under cyclic permutation.
Solution
Take \vec a=\hat i, \vec b=\hat j, \vec c=\hat k and use \hat j\times\hat k=\hat i, \hat k\times\hat i=\hat j, \hat i\times\hat j=\hat k (book 3.7.3).
If dot product of two non-zero vectors is zero, then both vectors are:
Aparallel
Bequal
Cmutually perpendicularAnswer
Din opposite direction
Why
\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 with both magnitudes non-zero forces \cos\theta=0, so \theta=90^\circ: the vectors are mutually perpendicular.
Parallel vectors have \cos\theta=\pm1, so their dot product is \pm|\vec a||\vec b|, not zero. It is the cross product that is zero for parallel vectors.
2025 Annual I, Set D, MCQ 111 mark
If dot product of two vectors is zero then angle between them is:
A180^\circ
B90^\circAnswer
C45^\circ
D0^\circ
Why
\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 gives \cos\theta=0, so \theta=90^\circ.
0^\circ would make the dot product maximum (|\vec a||\vec b|), and 180^\circ makes it -|\vec a||\vec b|; it is the cross product that vanishes at 0^\circ and 180^\circ.
2026 Annual I, Set D, MCQ 121 mark
If \vec a=a_1\hat i+b_1\hat j and \vec b=a_2\hat i+b_2\hat j then \vec a\cdot\vec b is equal to:
Aa_1a_2+b_1b_2Answer
Ba_1a_2-b_1b_2
Ca_1b_1+a_2b_2
Da_2b_1-a_1b_2
Why
Multiply matching components and add (book 3.5.2): \vec a\cdot\vec b=a_1a_2+b_1b_2, because \hat i\cdot\hat i=\hat j\cdot\hat j=1 and \hat i\cdot\hat j=0.
Option D, a_2b_1-a_1b_2, has the shape of a 2\times2 determinant, which belongs to the cross product, not the dot product.
2025 Annual I, Set D, MCQ 121 mark
If \vec a\perp\vec b, then projection of \vec a on \vec b is:
A-1
B0Answer
C1
DUndefined
Why
The projection of \vec a on \vec b is \dfrac{\vec a\cdot\vec b}{|\vec b|} (book 3.5.11). If \vec a\perp\vec b then \vec a\cdot\vec b=0, so the projection is 0 (the vector has no component along \vec b).
It is not undefined: the formula only fails when |\vec b|=0, which is not the case for a perpendicular pair of non-zero vectors.
2026 Annual I, Set D, MCQ 131 mark
Two vectors \vec a and \vec b will be orthogonal, if the angle \theta between them is:
A\pi
B\dfrac{\pi}{2}Answer
C2\pi
D\dfrac{2\pi}{3}
Why
Orthogonal means perpendicular, so \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0, giving \cos\theta=0 and \theta=\dfrac{\pi}{2}.
\pi is the angle for opposite directions and 2\pi (a full turn) brings the vector back to parallel; neither is a right angle.
2025 Annual I, Set D, MCQ 131 mark
What is the volume of a parallelepiped formed by vectors \hat i+\hat j, \hat j+\hat k and \hat k+\hat i?
A0
B1
C\sqrt2
D2Answer
Why
Volume is |\vec a\cdot\vec b\times\vec c| (book 3.7.5). Write the determinant with the three vectors as rows.
\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}
Expand along the first row.
=1(1\cdot1-1\cdot0)-1(0\cdot1-1\cdot1)+0=1+1=2
Option A, 0, would mean the vectors are coplanar; they are not, since the determinant is non-zero. Option B, 1, is the volume of the unit cube, whereas these three edges span a solid of volume 2.
2026 Annual I, Set D, Q.2 (i) OR4 marks
A force \vec F=2\hat i+3\hat j+\hat k moves an object through displacement \vec d=\hat i+2\hat j+2\hat k. Find the work done.
Solution
Work done is the dot product of force and displacement, W=\vec F\cdot\vec d (book 3.5.13).
Multiply matching components and add.
W=(2)(1)+(3)(2)+(1)(2)=2+6+2
Add.
W=10
W=10\ \text{units of work}
2025 Annual I, Set D, Q.2 (i) OR (a)
A mechanic applies a force \vec F=3\hat i+4\hat j+5\hat k at point A(14,-3,6) to turn a bolt with pivot at P(2,-2,1). Find the moment arm \vec r.
Solution
The moment arm is the position vector from the pivot P to the point of application A (head minus tail).
Subtract P(2,-2,1) from A(14,-3,6).
\vec r=\overrightarrow{PA}=(14-2)\hat i+(-3+2)\hat j+(6-1)\hat k=12\hat i-\hat j+5\hat k
\vec r=12\hat i-\hat j+5\hat k
2025 Annual I, Set D, Q.2 (i) OR (b)
A mechanic applies a force \vec F=3\hat i+4\hat j+5\hat k at point A(14,-3,6) to turn a bolt with pivot at P(2,-2,1). Find the moment vector \vec M about the bolt.
Solution
The moment about the pivot is \vec M=\vec r\times\vec F (book 3.6.8), with \vec r=12\hat i-\hat j+5\hat k from part (a) and \vec F=3\hat i+4\hat j+5\hat k.
Write the cross product as a determinant, \vec r in row 2 and \vec F in row 3.
A force \vec F=5\hat i+3\hat j+4\hat k is applied to move an object from A(2,3,1) to B(5,7,4). Find work done by the force.
Solution
Displacement from A(2,3,1) to B(5,7,4) (head minus tail).
\vec d=\overrightarrow{AB}=3\hat i+4\hat j+3\hat k
Work done is W=\vec F\cdot\vec d (book 3.5.13), with \vec F=5\hat i+3\hat j+4\hat k.
W=(5)(3)+(3)(4)+(4)(3)=15+12+12
Add.
W=39
W=39\ \text{units of work}
2025 Annual I, Set D, Q.2 (viii) OR (a)
A boat is sailing north at 10 km/h but is also being pushed east by ocean currents at 5 km/h. Find the actual speed (V) of the boat.
Solution
The two motions act at right angles, so the velocity vectors \vec V_N (10 km/h north) and \vec V_E (5 km/h east) are the legs of a right triangle and the actual velocity \vec V is its hypotenuse (resultant, by the triangle law of vector addition).
Velocity triangle: V_N=10, V_E=5, resultant V and the angle \theta from north.
A boat is sailing north at 10 km/h but is also being pushed east by ocean currents at 5 km/h. Find the direction angle \theta of the boat relative to due north.
Solution
In the velocity triangle of part (a), \theta is the angle at the start between the north leg V_N and the resultant V. The side opposite \theta is V_E and the side adjacent is V_N.
Expand along the first row (the middle term has a minus sign).
\vec a\times\vec b=((-2)(-1)-4\cdot2)\hat i-(3\cdot(-1)-4\cdot1)\hat j+(3\cdot2-(-2)\cdot1)\hat k
Evaluate the brackets.
\vec a\times\vec b=-6\hat i+7\hat j+8\hat k
\vec a\times\vec b=-6\hat i+7\hat j+8\hat k
2026 Annual I, Set D, Q.3 OR (b)
If \vec a=3\hat i-2\hat j+4\hat k and \vec b=\hat i+2\hat j-\hat k, find (\vec a\times\vec b)\cdot\vec a and verify that \vec a\times\vec b\perp\vec a.
Solution
From part (a), \vec a\times\vec b=-6\hat i+7\hat j+8\hat k. Take its dot product with \vec a.
Multiply matching components and add.
(-6)(3)+(7)(-2)+(8)(4)=-18-14+32
The dot product is zero, and neither vector is zero, so \cos\theta=0: the angle between \vec a\times\vec b and \vec a is 90^\circ.
(\vec a\times\vec b)\cdot\vec a=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec a
(\vec a\times\vec b)\cdot\vec a=0,\ \text{so }\vec a\times\vec b\perp\vec a
2026 Annual I, Set D, Q.3 OR (c)
If \vec a=3\hat i-2\hat j+4\hat k and \vec b=\hat i+2\hat j-\hat k, find (\vec a\times\vec b)\cdot\vec b and verify that \vec a\times\vec b\perp\vec b.
Solution
From part (a), \vec a\times\vec b=-6\hat i+7\hat j+8\hat k. Take its dot product with \vec b.
Multiply matching components and add.
(-6)(1)+(7)(2)+(8)(-1)=-6+14-8
The dot product is zero, so the angle is 90^\circ.
(\vec a\times\vec b)\cdot\vec b=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec b
(\vec a\times\vec b)\cdot\vec b=0,\ \text{so }\vec a\times\vec b\perp\vec b
2025 Annual I, Set D, Q.3 OR (a)
If \vec a=5\hat i-3\hat j+\hat k and \vec b=-2\hat i+6\hat j+4\hat k, find \vec a\times\vec b.
Solution
Write the cross product as a determinant (book 3.6.3), the components of the first vector in row 2 and of the second in row 3.
Expand along the first row (the middle term has a minus sign).
\vec a\times\vec b=((-3)(4)-1\cdot6)\hat i-(5\cdot4-1\cdot(-2))\hat j+(5\cdot6-(-3)(-2))\hat k
Evaluate the brackets.
\vec a\times\vec b=-18\hat i-22\hat j+24\hat k
\vec a\times\vec b=-18\hat i-22\hat j+24\hat k
2025 Annual I, Set D, Q.3 OR (b)
If \vec a=5\hat i-3\hat j+\hat k and \vec b=-2\hat i+6\hat j+4\hat k, find (\vec a\times\vec b)\cdot\vec a and verify that \vec a\times\vec b\perp\vec a.
Solution
From part (a), \vec a\times\vec b=-18\hat i-22\hat j+24\hat k. Take its dot product with \vec a.
Multiply matching components and add.
(-18)(5)+(-22)(-3)+(24)(1)=-90+66+24
The dot product is zero, and neither vector is zero, so \cos\theta=0: the angle between \vec a\times\vec b and \vec a is 90^\circ.
(\vec a\times\vec b)\cdot\vec a=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec a
(\vec a\times\vec b)\cdot\vec a=0,\ \text{so }\vec a\times\vec b\perp\vec a
2025 Annual I, Set D, Q.3 OR (c)
If \vec a=5\hat i-3\hat j+\hat k and \vec b=-2\hat i+6\hat j+4\hat k, find (\vec a\times\vec b)\cdot\vec b and verify that \vec a\times\vec b\perp\vec b.
Solution
From part (a), \vec a\times\vec b=-18\hat i-22\hat j+24\hat k. Take its dot product with \vec b.
Multiply matching components and add.
(-18)(-2)+(-22)(6)+(24)(4)=36-132+96
The dot product is zero, so the angle is 90^\circ.
(\vec a\times\vec b)\cdot\vec b=0\;\Longrightarrow\;\vec a\times\vec b\perp\vec b
(\vec a\times\vec b)\cdot\vec b=0,\ \text{so }\vec a\times\vec b\perp\vec b
Exam focus
Find \lambda (or p) so that three given vectors are coplanar, using \vec a\cdot\vec b\times\vec c=0. (short)
Asked in every paper: 2026 Set B Q.2 (x) OR, 2025 Set B Q.2 (iii) (quadratic in p) and 2025 Set D Q.2 (iii); 2026 Set B MCQ 18 tests the condition. Book Example 27 and Exercise 3.4 Q.3 are the same type.
Find \vec a\times\vec b, then show it is perpendicular to \vec a and to \vec b by dot products. (long)
The 8-mark Section C OR question in 2026 Set D (Q.3 OR) and in 2025 Sets B and D (Q.3 OR), parts (a), (b), (c) each time with different vectors. Practise Exercise 3.3 Q.2.
Moment of a force \vec M=\vec r\times\vec F about a point. (short)
2026 Set B Q.2 (vi) OR (about the origin) and 2025 Set D Q.2 (i) OR (moment arm and moment vector about a pivot). Exercise 3.3 Q.14 to Q.17 and Examples 20 and 21.
Dot product facts: zero dot product means perpendicular, angle between vectors, components and projections. (mcq)
Eight MCQs: 2026 Set B MCQ 16 and 17, Set D MCQ 11 to 13, 2025 Set B MCQ 11 and Set D MCQ 11 and 12; and 2026 Set D Q.2 (v) OR (angle, result 90^\circ).