Vectors

Class 11 Physics, unit 2, book pages 22–31

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Some physical quantities need a direction as well as a size, and these are vectors. This unit shows how to break a vector into two perpendicular (rectangular) components and how to rebuild it from them, then how two vectors can be multiplied in two ways: the dot product, which gives a scalar (work, power, electric flux), and the cross product, which gives a vector (torque, angular momentum) and whose size is the area of a parallelogram.

Learning outcomes (2)

Represent a vector in 2-D as two perpendicular components.
Describe the product of two vectors (dot and cross-product) along with their properties.

Introduction to vectors

In daily life we deal with many physical quantities, such as mass, time, length, velocity, acceleration, force, work and torque. Some of them are completely described by a number and a proper unit; others also need a direction.

Scalar: A physical quantity that is completely described by a number and a proper unit only. Examples: mass, time, length.
Vector: A physical quantity that needs a direction along with a number and a unit. Examples: displacement, force, acceleration, torque, momentum.

Vectors are very useful in daily life. A signboard, for example, tells us the distances and directions of other places relative to the signboard. The distance alone is a scalar and is not enough to reach the destination: we must also know the direction from the signboard. Distance together with direction is a vector quantity called the displacement. So signboards give information about displacement from the signboard to the destination.

Scalars and vectors.
	Scalar	Vector
Described by	number and unit	number, unit and direction
Example from the signboard	distance	displacement (distance and direction)
Other examples	mass, time, length	force, acceleration, torque, momentum

Graphical representation. A vector can be completely expressed by a line with an arrow head. The length of the line represents the magnitude according to a proper scale, and the arrow head represents the direction.

The addition of vectors was studied in grade 9. In this unit we study the components of vectors and the products of vectors.

2.1Vector resolutionOften examined

Components are the effective parts of a vector in different directions. In 2-D (a plane) we take an $x$ -component and a $y$ -component. The component along the $x$ -axis is called the horizontal or $x$ -component, and the component along the $y$ -axis is called the vertical or $y$ -component. The $x$ - and $y$ -components are perpendicular to each other, so they are called rectangular components.

Rectangular components: The components of a vector which are mutually perpendicular.

Consider a vector $A$ making an angle $θ$ with the horizontal, as in Figure 2.1. Perpendiculars are drawn from the head of $A$ on the $x$ -axis and the $y$ -axis. The projection OQ of $A$ along the $x$ -axis is the $x$ -component of $A$ , written $A_{x}$ . The projection OS along the $y$ -axis is the $y$ -component, written $A_{y}$ .

Figure 2.1: Rectangular components of a vector. The accent arrows are the components

A_{x}

(along OQ) and

A_{y}

(along OS) of

A

The vector $A$ in component form is the sum of its two components:

A = A_{x} + A_{y}

2.1

The vector sum of the two rectangular components.

Here $\overset{}{^}$ and $\overset{}{^}$ represent the directions along the $+ x$ -axis and $+ y$ -axis. They are called unit vectors. Writing each component as its size times its unit vector gives

A = A_{x} \overset{}{^} + A_{y} \overset{}{^}

2.2

$A_{x}$ and $A_{y}$ : sizes of the components; $\overset{}{^}$ and $\overset{}{^}$ : unit vectors along $+ x$ and $+ y$ .

Unit vector: A vector of magnitude one. It specifies the direction of a vector and has no other job.

2.1.1Finding magnitude of rectangular componentsOften examined

To find the magnitudes of the rectangular components $A_{x}$ and $A_{y}$ of a vector $A$ we use trigonometric ratios in the right-angled triangle OPQ of Figure 2.1.

Derivation Components of a vector from its size and angle

Aim. Express $A_{x}$ and $A_{y}$ in terms of the magnitude $A$ and the angle $θ$ the vector makes with the $x$ -axis.

Assumptions.

$A$ makes the angle $θ$ with the $x$ -axis, as in Figure 2.1.
The triangle OPQ has its right angle at Q, with hypotenuse OP $= A$ , base OQ $= A_{x}$ and perpendicular QP $= A_{y}$ (QP equals OS, since OQPS is a rectangle).

In the right-angled triangle OPQ, the sine of $θ$ is the perpendicular over the hypotenuse.
$sin θ = \frac{perpendicular}{hypotenuse} = \frac{QP}{O P} = \frac{O S}{O P}$
Replace the lengths by the magnitudes: $QP = O S = A_{y}$ and $O P = A$ .
$sin θ = \frac{A _{y}}{A}$
Multiply both sides by $A$ to get the $y$ -component.
$A_{y} = A sin θ (2.3)$
In the same triangle, the cosine of $θ$ is the base over the hypotenuse.
$cos θ = \frac{base}{hypotenuse} = \frac{O Q}{O P} = \frac{A _{x}}{A}$
Multiply both sides by $A$ to get the $x$ -component.
$A_{x} = A cos θ (2.4)$

Result

A_{x} = A cos θ, A_{y} = A sin θ

Equations (2.3) and (2.4) give the magnitudes of the perpendicular components $A_{x}$ and $A_{y}$ of a given vector.

Worked example: components when the angle is given from the vertical

A kite string pulls on a boy's hand with a force of

40 N

at an angle of

3 5^{\circ}

with the vertical. Find its horizontal and vertical components.

Given

$F = 40 N$
$θ = 3 5^{\circ} from the vertical$

To find

$F_{x}$ and $F_{y}$

The angle is measured from the vertical axis, so the component next to the angle is the vertical one.

The equations $A_{x} = A cos θ$ and $A_{y} = A sin θ$ hold only when $θ$ is from the $x$ -axis. Here the angle is from the $y$ -axis, so the sine and cosine swap: the vertical component is next to the angle.
$F_{y} = F cos 3 5^{\circ}, F_{x} = F sin 3 5^{\circ}$
Horizontal component.
$F_{x} = 40 \times 0.5736 = 22.9 N$
Vertical component.
$F_{y} = 40 \times 0.8192 = 32.8 N$
Check with the magnitude formula: the components must give back $40 N$ .
$22. 9^{2} + 32. 8^{2} = 524.4 + 1075.8 = 1600.2 = 40 N$

F_{x} = 22.9 N, F_{y} = 32.8 N

2.1.2Finding a vector from its componentsOften examined

If the rectangular components $A_{x}$ and $A_{y}$ of a vector $A$ are given, its magnitude $A$ and direction $θ$ can be found. We again use the triangle OPQ of Figure 2.1.

Derivation Magnitude and direction of a vector from its components

Aim. Find the magnitude $A$ and the direction $θ$ of $A$ when $A_{x}$ and $A_{y}$ are known.

Assumptions.

The triangle OPQ of Figure 2.1 is right-angled at Q, with OP $= A$ , OQ $= A_{x}$ and QP $= A_{y}$ .

Apply Pythagoras theorem to the right-angled triangle OPQ: the square on the hypotenuse equals the sum of the squares on the other two sides.
$(O P)^{2} = (O Q)^{2} + (QP)^{2} (2.5)$
QP and OS are opposite sides of a rectangle, so $QP = O S$ . Equation (2.5) can be written with OS in place of QP.
$(O P)^{2} = (O Q)^{2} + (O S)^{2} (2.6)$
Put $O P = A$ , $O Q = A_{x}$ and $O S = A_{y}$ in Equation (2.6).
$A^{2} = (A_{x})^{2} + (A_{y})^{2}$
Take the positive square root, since a magnitude cannot be negative.
$A = (A_{x})^{2} + (A_{y})^{2} (2.7)$
For the direction, use the tangent ratio in triangle OPQ: perpendicular over base.
$tan θ = \frac{perpendicular}{base} = \frac{A _{y}}{A _{x}}$
Take the inverse tangent to get the angle.
$θ = tan^{- 1} \frac{A _{y}}{A _{x}} (2.8)$

Result

A = A_{x}^{2} + A_{y}^{2}, θ = tan^{- 1} \frac{A _{y}}{A _{x}}

Equations (2.7) and (2.8) together give the magnitude and the direction of a vector from its rectangular components.

Example 2.1: Components of a pushing force

A peddler is pushing a trolley on a horizontal road with a force of

50 N

making an angle

3 0^{\circ}

with the road. Find the horizontal and vertical components of the force.

Given

$F = 50 N$ magnitude of the force
$θ = 3 0^{\circ}$ angle of the force with the horizontal

To find

$F_{x}$
$F_{y}$

To find the $x$ -component of $F$ , use Equation (2.4).
$F_{x} = F cos θ$
Put in the values.
$F_{x} = 50 cos 3 0^{\circ} = 50 (0.866) = 43.3 N$
To find the $y$ -component, use Equation (2.3).
$F_{y} = F sin θ$
Put in the values.
$F_{y} = 50 sin 3 0^{\circ} = 50 (0.5) = 25 N$

F_{x} = 43.3 N, F_{y} = 25 N

Assignment 2.1: Fatima's trolley bag

Fatima is pulling her trolley bag while climbing up the ramp at her school gate. Find the force with which she is pulling her bag, if the

x

-component and

y

-component of her force are

12 N

and

5 N

, respectively.

Given

$F_{x} = 12 N$
$F_{y} = 5 N$

To find

$F$ (the magnitude of the force, and its direction)

The two components are perpendicular, so use Equation (2.7).
$F = F_{x}^{2} + F_{y}^{2}$
Substitute.
$F = (12)^{2} + (5)^{2} = 144 + 25 = 169$
Evaluate.
$F = 13 N$
The direction above the ramp (horizontal), from Equation (2.8).
$θ = tan^{- 1} \frac{5}{12} = 22. 6^{\circ}$

F = 13 N at 22. 6^{\circ} to the horizontal

2.2Product of vectors

When two vector quantities are multiplied, the product may be a scalar quantity or a vector quantity. The product obtained depends upon the nature of the given vectors. There are two types of products.

The two kinds of product of vectors.
Product	Result	Written as	Section
Scalar product (dot product)	a scalar	\vec A\cdot\vec B	2.2.1
Vector product (cross product)	a vector	\vec A\times\vec B	2.2.2

2.2.1Scalar product or dot productOften examined

Scalar product: When the product of two vector quantities gives a scalar quantity, the product is called the scalar product.

The scalar product of two vectors is represented by putting a dot (\cdot) between the symbols of the two vectors, so it is also known as the dot product. Take two vectors \vec A and \vec B making an angle \theta with each other, as in Figure 2.2. Their dot product is written \vec A\cdot\vec B and is defined as

Figure 2.2: Two vectors \vec A and \vec B that are making an angle \theta with each other.

\vec A\cdot\vec B=AB\cos\theta

2.9

A and B: magnitudes of the vectors; \theta: angle between them.

Here A and B are the magnitudes of the vectors. Thus the scalar product of two vectors is obtained by multiplying their magnitudes with the cosine of the angle between them.

Derivation Dot product as a projection

Aim. Show that the dot product is the magnitude of one vector times the component of the other vector along it.

Figure 2.3: Component of vector \vec A in the direction of vector \vec B is A\cos\theta.

Start from the definition of the dot product, Equation (2.9).
\vec A\cdot\vec B=AB\cos\theta
Group the factors: bracket A\cos\theta together.
\vec A\cdot\vec B=(A\cos\theta)\,B\qquad(2.10)
Figure 2.3 shows that a perpendicular from the head of \vec A on to the line of \vec B cuts off a length A\cos\theta. This is the component of \vec A in the direction of \vec B.
A\cos\theta=\text{component of }\vec A\text{ along }\vec B
Put this in Equation (2.10).
\vec A\cdot\vec B=(\text{component of }\vec A\text{ along }\vec B)\,B\qquad(2.11)
In the same way, grouping B\cos\theta instead shows that
\vec A\cdot\vec B=A\,(\text{component of }\vec B\text{ along }\vec A)\qquad(2.12)

Result\vec A\cdot\vec B=B\,(A\cos\theta)=A\,(B\cos\theta)

The scalar product is the product of the magnitude of either vector with the component of the other vector in the direction of the first vector.

Examples of scalar product

Work. Work W is a scalar which is the scalar product of displacement \vec d and force \vec F: W=\vec F\cdot\vec d=Fd\cos\theta.
Power. Power P is the scalar product of force \vec F and velocity \vec v: P=\vec F\cdot\vec v=Fv\cos\theta.
Electric flux. Electric flux \Phi is a scalar which is the scalar product of electric field intensity \vec E and vector area \vec A: \Phi=\vec E\cdot\vec A=EA\cos\theta.

Worked example: work done by a rope pulled at an angle

A sledge is pulled along level ground through a displacement of 12\,\mathrm{m} by a rope that pulls with a force of 80\,\mathrm{N} at 30^\circ above the horizontal. Find the work done by the force. What would the work be if the rope were pulled at 90^\circ to the ground?

Given

F = 80\,\mathrm{N}
d = 12\,\mathrm{m}
\theta = 30^\circ

To find

work W
W for \theta=90^\circ

Only the part of the force along the displacement, F\cos\theta (accent), does the work.

Work is the scalar product of force and displacement.
W=\vec F\cdot\vec d=Fd\cos\theta
Substitute the values.
W=80\times12\times\cos30^\circ=960\times0.8660
Multiply.
W=831\,\mathrm{J}
Check by the component method: the part of the force along \vec d is F\cos30^\circ=69.3\,\mathrm{N}.
W=69.3\times12=831\,\mathrm{J}
For \theta=90^\circ the force is perpendicular to the displacement and \cos90^\circ=0.
W=Fd\cos90^\circ=0

W=831\,\mathrm{J},\qquad W=0\ \text{at }90^\circ

Properties of scalar product

1) The scalar product is commutative.

\vec A\cdot\vec B=AB\cos\theta=BA\cos\theta=\vec B\cdot\vec A

Hence \vec A\cdot\vec B=\vec B\cdot\vec A. The order does not matter, because ordinary numbers A and B commute and \cos\theta is the same angle both ways.

2) The scalar product of two orthogonal (perpendicular) vectors is zero.

\vec A\cdot\vec B=AB\cos90^\circ=0

Similarly, for mutually perpendicular unit vectors:

\hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0

The dot product of a vector with the null vector is also zero.

3) The scalar product of two parallel vectors has its maximum value and is equal to the product of their magnitudes only.

\vec A\cdot\vec B=AB\cos0^\circ=AB

The scalar product of a vector with itself is equal to the square of its magnitude:

\vec A\cdot\vec A=AA\cos0^\circ=A^2

Similarly, the dot product of a unit vector with itself is unity:

\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1

4) The scalar product of two antiparallel vectors is negative.

\vec A\cdot\vec B=AB\cos180^\circ=-AB

5) Scalar product in terms of rectangular components.

Derivation Dot product from components

Aim. Write \vec A\cdot\vec B in terms of the rectangular components of the two vectors.

Write both vectors in component form, with \hat k the unit vector along +z for a third component.
\vec A\cdot\vec B=(A_x\hat\imath+A_y\hat\jmath+A_z\hat k)\cdot(B_x\hat\imath+B_y\hat\jmath+B_z\hat k)
Multiply out. This gives nine terms, such as A_xB_x\,\hat\imath\cdot\hat\imath and A_xB_y\,\hat\imath\cdot\hat\jmath.
Every term that has two different unit vectors, like \hat\imath\cdot\hat\jmath, is zero (property 2). Only the three terms with the same unit vector twice survive, and each of them has \hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1.
\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z\qquad(2.13)
From the definition, \cos\theta=\dfrac{\vec A\cdot\vec B}{AB}. Put in Equation (2.13) to get the angle in terms of components.
\cos\theta=\dfrac{A_xB_x+A_yB_y+A_zB_z}{AB}\qquad(2.14)

Result\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z

Multiply the matching components and add. Equation (2.14) then gives the angle between two vectors given in component form.

Example 2.2: Power delivered by an engine

Find the power delivered by the engine for attaining velocity (3,4)\,\mathrm{m\,s^{-1}}, while it exerts a force (8,-2)\,\mathrm{N}.

Given

\vec v = (3\hat\imath+4\hat\jmath)\,\mathrm{m\,s^{-1}} velocity
\vec F = (8\hat\imath-2\hat\jmath)\,\mathrm{N} force

To find

Power delivered P

Power is the scalar product of force and velocity.
P=\vec F\cdot\vec v
Substitute the vectors in component form.
P=(3\hat\imath+4\hat\jmath)\cdot(8\hat\imath-2\hat\jmath)
Multiply matching components and add, Equation (2.13): (3)(8)+(4)(-2).
P=24-8
Evaluate.
P=16\,\mathrm{W}

P=16\,\mathrm{W}

Assignment 2.2: Projections of two vectors

If vectors \vec A=5\hat\imath+\hat\jmath and \vec B=2\hat\imath+4\hat\jmath, then find (i) the projection of \vec A on \vec B, (ii) the projection of \vec B on \vec A.

The dot product, from Equation (2.13).
\vec A\cdot\vec B=(5)(2)+(1)(4)=14
The magnitudes of the vectors, from Equation (2.7).
A=\sqrt{25+1}=\sqrt{26}=5.10,\qquad B=\sqrt{4+16}=\sqrt{20}=4.47
(i) The component (projection) of \vec A along \vec B is A\cos\theta. From Equation (2.11), \vec A\cdot\vec B=(A\cos\theta)B, so divide the dot product by B.
A\cos\theta=\dfrac{\vec A\cdot\vec B}{B}=\dfrac{14}{\sqrt{20}}=3.13
(ii) The projection of \vec B along \vec A is B\cos\theta. From Equation (2.12), divide the dot product by A.
B\cos\theta=\dfrac{\vec A\cdot\vec B}{A}=\dfrac{14}{\sqrt{26}}=2.75

\text{(i) }3.13,\qquad\text{(ii) }2.75

2.2.2Vector product or cross productOften examined

Vector product: When the product of two vector quantities gives a vector quantity, the product is called the vector product.

The vector product of two vectors is represented by putting a cross (\times) between the symbols of the two vectors, so it is also known as the cross product. Take two vectors \vec A and \vec B making an angle \theta with each other, as in Figure 2.4. The cross product is written \vec A\times\vec B and is defined as

Figure 2.4: Cross product of two vectors \vec A and \vec B is shown by \vec A\times\vec B. The shaded plane contains \vec A and \vec B; the product points perpendicular to it.

\vec A\times\vec B=AB\sin\theta\,\hat n

2.15

A and B: magnitudes; \theta: angle between \vec A and \vec B; \hat n: unit vector perpendicular to the plane of \vec A and \vec B.

Thus the vector product is obtained by multiplying their magnitudes with the sine of the angle between them. In Equation (2.15), \hat n gives the direction of the vector product, which is always perpendicular to the plane containing \vec A and \vec B.

Right hand rule of vector product

The direction of the cross product (\hat n) can be found by the right hand rule, as in Figure 2.5.

Figure 2.5: Right hand rule. Curl the fingers of the right hand from \vec A towards \vec B and the erect thumb gives \hat n.

Examples of vector product

Torque. Torque \vec\tau is the cross product of force \vec F and position vector \vec r: \vec\tau=\vec r\times\vec F=rF\sin\theta\,\hat n.
Angular momentum. Angular momentum \vec L is the cross product of position vector \vec r and linear momentum \vec p: \vec L=\vec r\times\vec p=rp\sin\theta\,\hat n.

Properties of vector product

1) The vector product is not commutative but anti-commutative. Changing the order of the vectors reverses the direction of the vector product, as in Figure 2.6.

Figure 2.6: By changing the order of vectors, the direction of vector product gets reversed.

\vec A\times\vec B\ne\vec B\times\vec A\qquad\qquad \vec A\times\vec B=-\,\vec B\times\vec A

The magnitude of \vec A\times\vec B is the same as that of \vec B\times\vec A. By the right hand rule, \vec A\times\vec B points upward (along \hat n) and \vec B\times\vec A points downward (along -\hat n).

2) The vector product of two parallel or anti-parallel vectors is a null vector.

Null vector: A vector with zero magnitude and arbitrary direction.

\vec A\times\vec B=AB\sin0^\circ\,\hat n=\vec 0\qquad\text{and}\qquad \vec A\times\vec B=AB\sin180^\circ\,\hat n=\vec 0

The vector product of a vector with itself is also the null vector:

\vec A\times\vec A=AA\sin0^\circ\,\hat n=\vec 0

Similarly, the vector product of a unit vector with itself is the null vector:

\hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0

3) The vector product of two perpendicular vectors has its maximum value and is equal to the product of their magnitudes only.

\vec A\times\vec B=AB\sin90^\circ\,\hat n=AB\,\hat n

Similarly, for mutually perpendicular unit vectors:

\hat\imath\times\hat\jmath=\hat k,\qquad \hat\jmath\times\hat k=\hat\imath,\qquad \hat k\times\hat\imath=\hat\jmath

4) Vector product in terms of rectangular components.

Derivation Cross product from components

Aim. Write \vec A\times\vec B in terms of the rectangular components of the two vectors.

Write both vectors in component form.
\vec A\times\vec B=(A_x\hat\imath+A_y\hat\jmath+A_z\hat k)\times(B_x\hat\imath+B_y\hat\jmath+B_z\hat k)
Multiply out the nine terms. The three terms with the same unit vector twice (\hat\imath\times\hat\imath and so on) are null vectors and vanish. Use the unit vector results above for the other six.
\begin{aligned}\vec A\times\vec B&=A_xB_y\,\hat k-A_xB_z\,\hat\jmath-A_yB_x\,\hat k\\&\quad+A_yB_z\,\hat\imath+A_zB_x\,\hat\jmath-A_zB_y\,\hat\imath\end{aligned}
Collect the terms along each unit vector.
\vec A\times\vec B=(A_yB_z-A_zB_y)\,\hat\imath+(A_zB_x-A_xB_z)\,\hat\jmath+(A_xB_y-A_yB_x)\,\hat k\qquad(2.17)
This is exactly what we get by expanding the determinant whose rows are the unit vectors, the components of \vec A and the components of \vec B.
\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}\qquad(2.16)

Result\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}

The determinant is the easy way to remember and evaluate the cross product. The order of the rows matters: swapping the rows of \vec A and \vec B reverses the sign, as anti-commutativity requires.

Physical significance of vector product

The vector product gives us the area of a plane. For example, if two vectors \vec A and \vec B represent adjacent sides of a plane figure, as in Figure 2.7, then the magnitude of \vec A\times\vec B gives the magnitude of that area (the figure may be a rectangle or a parallelogram).

Figure 2.7: Magnitude of \vec A\times\vec B gives area of parallelogram.

\text{Area of parallelogram}=|\vec A\times\vec B|=AB\sin\theta

The unit vector \hat n gives the direction of that area, which is normal to the plane and can be found by the right hand rule, as in Figure 2.5.

Worked example: area of a parallelogram as the angle between the sides changes

Two adjacent sides of a parallelogram are 8.0\,\mathrm{cm} and 5.0\,\mathrm{cm} long and the angle between them is 40^\circ. Find its area. For what angle is the area greatest, and what happens when the angle is 140^\circ?

The magnitude of the cross product of the two sides is the area.
\text{Area}=|\vec A\times\vec B|=AB\sin\theta
Substitute.
\text{Area}=8.0\times5.0\times\sin40^\circ=40\times0.6428=25.7\,\mathrm{cm^{2}}
The area is greatest when \sin\theta=1, that is \theta=90^\circ (a rectangle).
\text{Area}_{max}=40\,\mathrm{cm^{2}}
At 140^\circ use \sin140^\circ=\sin40^\circ.
\text{Area}=40\sin140^\circ=25.7\,\mathrm{cm^{2}}
At 0^\circ or 180^\circ the sides lie along one line, the parallelogram collapses to a line and the area is zero (the cross product of parallel vectors is a null vector).

Area =40\sin\theta for sides of 8.0\,\mathrm{cm} and 5.0\,\mathrm{cm}. It is greatest at 90^\circ and equal at 40^\circ and 140^\circ.

\text{Area}=25.7\,\mathrm{cm^{2}};\ \text{maximum }40\,\mathrm{cm^{2}}\text{ at }90^\circ

Example 2.3: Area of a parallelogram

Find the area of the parallelogram whose adjacent sides are given by the vectors \vec A=(\hat\imath+6\hat\jmath+2\hat k)\,\mathrm{m} and \vec B=(7\hat\imath+\hat\jmath+5\hat k)\,\mathrm{m}.

Given

\vec A = (\hat\imath+6\hat\jmath+2\hat k)\,\mathrm{m}
\vec B = (7\hat\imath+\hat\jmath+5\hat k)\,\mathrm{m}

To find

\vec A\times\vec B

The vector product gives the area of the parallelogram. Write it as a determinant, Equation (2.16).
\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\1&6&2\\7&1&5\end{vmatrix}
Expand along the first row, with the signs +,-,+.
\vec A\times\vec B=\hat\imath\begin{vmatrix}6&2\\1&5\end{vmatrix}-\hat\jmath\begin{vmatrix}1&2\\7&5\end{vmatrix}+\hat k\begin{vmatrix}1&6\\7&1\end{vmatrix}
Evaluate each 2\times2 determinant.
\vec A\times\vec B=\hat\imath(30-2)-\hat\jmath(5-14)+\hat k(1-42)
Simplify.
\vec A\times\vec B=(28\hat\imath+9\hat\jmath-41\hat k)\,\mathrm{m^2}

\vec A\times\vec B=(28\hat\imath+9\hat\jmath-41\hat k)\,\mathrm{m^2}

Assignment 2.3: Magnitude of a cross product

Two vectors \vec A and \vec B having magnitudes 3.2 unit and 5.2 unit respectively, making an angle of 60^\circ with each other. What is the magnitude of their cross product?

Given

A = 3.2\ \text{unit}
B = 5.2\ \text{unit}
\theta = 60^\circ

To find

|\vec A\times\vec B|

The magnitude of the cross product is, from Equation (2.15), AB\sin\theta.
|\vec A\times\vec B|=AB\sin\theta
Substitute.
|\vec A\times\vec B|=(3.2)(5.2)\sin60^\circ=(16.64)(0.866)
Evaluate.
|\vec A\times\vec B|=14.4\ \text{unit}

|\vec A\times\vec B|=14.4\ \text{unit}

Worked example: cross product of two vectors by the determinant, with a check

Find \vec A\times\vec B for \vec A=3\hat\imath+2\hat\jmath-\hat k and \vec B=\hat\imath-4\hat\jmath+2\hat k. Check the answer, and write down \vec B\times\vec A.

Write the cross product as a determinant.
\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\3&2&-1\\1&-4&2\end{vmatrix}
Expand along the first row with the signs +,-,+.
=\hat\imath\begin{vmatrix}2&-1\\-4&2\end{vmatrix}-\hat\jmath\begin{vmatrix}3&-1\\1&2\end{vmatrix}+\hat k\begin{vmatrix}3&2\\1&-4\end{vmatrix}
Evaluate the three 2\times2 determinants.
=\hat\imath(4-4)-\hat\jmath(6+1)+\hat k(-12-2)
Simplify.
\vec A\times\vec B=-7\hat\jmath-14\hat k
Check 1: the cross product is perpendicular to both vectors, so both dot products must be zero.
\begin{aligned}\vec A\cdot(\vec A\times\vec B)&=0+2(-7)+(-1)(-14)=0\\ \vec B\cdot(\vec A\times\vec B)&=0+(-4)(-7)+2(-14)=0\end{aligned}
Reversing the order reverses the direction (anti-commutative).
\vec B\times\vec A=-(\vec A\times\vec B)=7\hat\jmath+14\hat k

\vec A\times\vec B=-7\hat\jmath-14\hat k,\qquad|\vec A\times\vec B|=\sqrt{245}=15.7

Formula sheet

Vector in component form

\vec A=A_x\hat\imath+A_y\hat\jmath

A_x,A_y: rectangular components; \hat\imath,\hat\jmath: unit vectors along +x, +y

Unit: same as \vec A

Rectangular components

A_x=A\cos\theta,\quad A_y=A\sin\theta

A: magnitude; \theta: angle with the x-axis

Unit: same as A

Magnitude from components

A=\sqrt{A_x^{2}+A_y^{2}}

A_x,A_y: rectangular components

Unit: same as the components

Direction from components

\theta=\tan^{-1}\dfrac{A_y}{A_x}

\theta: angle of the vector with the x-axis

Units: degree or radian

Position vector

\vec r=a\hat\imath+b\hat\jmath,\quad r=\sqrt{a^2+b^2}

location of point P(a,b) with respect to the origin

Unit: \mathrm{m}

Scalar (dot) product

\vec A\cdot\vec B=AB\cos\theta

\theta: angle between \vec A and \vec B

Unit: a scalar

Dot product in components

\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z

components of \vec A and \vec B

Unit: a scalar

Angle from components

\cos\theta=\dfrac{A_xB_x+A_yB_y+A_zB_z}{AB}

components and magnitudes of \vec A and \vec B

Unit: none

Work, power, electric flux

W=\vec F\cdot\vec d,\quad P=\vec F\cdot\vec v,\quad \Phi=\vec E\cdot\vec A

examples of scalar product

\mathrm{J},\ \mathrm{W}, flux unit

Unit vector dot products

\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1,\quad \hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0

unit vectors along x,y,z

Unit: none

Vector (cross) product

\vec A\times\vec B=AB\sin\theta\,\hat n

\hat n: unit vector perpendicular to the plane of \vec A and \vec B (right hand rule)

Unit: a vector

Cross product as determinant

\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}

components of \vec A and \vec B

Unit: a vector

Unit vector cross products

\hat\imath\times\hat\jmath=\hat k,\ \hat\jmath\times\hat k=\hat\imath,\ \hat k\times\hat\imath=\hat\jmath,\ \hat\imath\times\hat\imath=\vec0

unit vectors along x,y,z

Unit: none

Anti-commutative property

\vec A\times\vec B=-\vec B\times\vec A

order of the vectors reverses the direction

Unit: none

Torque and angular momentum

\vec\tau=\vec r\times\vec F,\quad \vec L=\vec r\times\vec p

\vec r: position vector; \vec F: force; \vec p: linear momentum

Units: \mathrm{N\,m}, \mathrm{kg\,m^2\,s^{-1}}

Area of parallelogram

\text{Area}=|\vec A\times\vec B|

\vec A,\vec B: adjacent sides

Unit: \mathrm{m^2}

Key terms

Scalar: A physical quantity completely described by a number and a proper unit only, such as mass, time and length.
Vector: A physical quantity that needs a direction along with a magnitude and a unit, such as displacement, force, acceleration, torque and momentum.
Rectangular components: The components of a vector which are mutually perpendicular, the x-component A_x=A\cos\theta and the y-component A_y=A\sin\theta.
Unit vector: A vector of magnitude one. It specifies the direction of any vector: \hat\imath,\hat\jmath,\hat k along +x,+y,+z.
Position vector: A vector which specifies the location of a point P(a,b) with respect to the origin, \vec r=a\hat\imath+b\hat\jmath, of magnitude \sqrt{a^2+b^2}.
Null vector: A vector with zero magnitude and arbitrary direction.
Scalar product (dot product): The product of two vectors that gives a scalar, \vec A\cdot\vec B=AB\cos\theta: the magnitude of one vector times the component of the other along it.
Vector product (cross product): The product of two vectors that gives a vector, \vec A\times\vec B=AB\sin\theta\,\hat n, perpendicular to the plane of the two vectors; its magnitude is the area of the parallelogram on them.
Right hand rule: Rotate the fingers of the right hand from the first vector to the second through the smaller angle; the erect thumb gives the direction of the vector product.

Multiple choice questionsp. 30

The book prints no answer key for these six questions. Each key below was worked out from the book's formulas, and the reason shows why the tempting option fails.

The number of perpendicular components of a force (in 2-D) are:

A1
B2Answer
C3
D4

Why

A force in 2-D resolves into exactly two rectangular components, F_x=F\cos\theta along the x-axis and F_y=F\sin\theta along the y-axis, and these two are perpendicular to each other (equations 2.3 and 2.4). Option A describes only one component, and a plane has no third perpendicular direction for options C and D.

\hat{j}\times\hat{i}=

A0
B1
C\hat{k}
D-\hat{k}Answer

Why

The cross product is anticommutative, so reversing the order reverses the direction. From the unit vector rules \hat i\times\hat j=\hat k, we get \hat j\times\hat i=-\hat i\times\hat j=-\hat k.

Option C is \hat i\times\hat j, the product in the opposite order. Option A would be correct only for parallel vectors, and \hat j and \hat i are perpendicular, so their cross product is not zero. Option B is a scalar, but a cross product is a vector.

A force of 10\,\mathrm{N} is making an angle of 30^\circ with the horizontal. Its x-component will be:

A4\,\mathrm{N}
B5\,\mathrm{N}
C7\,\mathrm{N}
D8.7\,\mathrm{N}Answer

Why

The x-component is the one along the horizontal, so F_x=F\cos\theta (equation 2.4): F_x=10\cos30^\circ=10\times0.866\approx8.7\,\mathrm{N}.

Option B is F\sin30^\circ, which is the vertical (y) component, so it is the trap for students who swap sine and cosine.

If two forces of magnitude 3\,\mathrm{N} and 4\,\mathrm{N} are acting at right angle to each other then their resultant force will be:

A7\,\mathrm{N}
B5\,\mathrm{N}Answer
C1\,\mathrm{N}
DNull vector

Why

The forces are perpendicular, so they act as the two rectangular components of the resultant. By Pythagoras (equation 2.7): R=\sqrt{3^2+4^2}=\sqrt{25}=5\,\mathrm{N}.

Option A is the plain sum, which is correct only for forces in the same direction. Option C is the difference, which is correct only for opposite directions. Option D needs equal and opposite forces.

Angle between two vectors A and B can be easily determined by:

Adot productAnswer
Bcross product
Chead to tail rule
Dright hand rule

Why

The dot product gives \cos\theta=\dfrac{\mathbf A\cdot\mathbf B}{AB} (equation 2.14). Taking the inverse cosine gives \theta directly from the two magnitudes and the scalar product.

The cross product also contains the angle (through \sin\theta), but it gives a vector and the dot product is the standard method. The head to tail rule and the right hand rule are used to find a resultant or a direction, not an angle.

62025 Annual I, Set D, MCQ 2

For which angle the equation |\mathbf A\cdot\mathbf B|=|\mathbf A\times\mathbf B| is correct?

A30^\circ
B45^\circAnswer
C60^\circ
D90^\circ

Why

At 90^\circ the dot product is zero while the cross product is AB, so option D fails. At 30^\circ and 60^\circ the sine and cosine differ.

Short questionsp. 30–31

2.1

If the cross product of two vectors vanishes, what will you say about their orientation?

Solution

The cross product is \mathbf A\times\mathbf B=AB\sin\theta\,\hat n (equation 2.15). If it is zero and neither vector is a null vector, then AB\neq0, so \sin\theta=0.

So \theta=0^\circ or 180^\circ. The vectors lie along the same line, so they are parallel or antiparallel (pointing the same way or opposite ways).

\theta=0^\circ\ \text{or}\ 180^\circ

2.2

Find the dot product of unit vectors with each other at (a) 0^\circ and (b) 90^\circ.

Solution

Two unit vectors have magnitude 1, so from equation 2.9 their dot product is \hat u\cdot\hat v=1\times1\times\cos\theta=\cos\theta, where \theta is the angle between them.

(a) Unit vectors at 0^\circ.

Put \theta=0^\circ in \cos\theta. The vectors are parallel, as \hat i\cdot\hat i is.
\hat u\cdot\hat v=\cos0^\circ=1

(b) Unit vectors at 90^\circ.

Put \theta=90^\circ. The vectors are perpendicular, as \hat i\cdot\hat j is.
\hat u\cdot\hat v=\cos90^\circ=0

2.3

Show that scalar product obeys commutative property.

Solution

Let \theta be the angle between A and B. The angle is the same whichever vector is written first, so the definition gives two forms of the same number:

Write the dot product with A first, using equation 2.9.
\mathbf A\cdot\mathbf B=AB\cos\theta
Write it with B first. The magnitudes and the angle are unchanged.
\mathbf B\cdot\mathbf A=BA\cos\theta
Multiplication of real numbers is commutative, so AB=BA. Therefore the two results are equal.
\mathbf A\cdot\mathbf B=\mathbf B\cdot\mathbf A

The same result follows from the component form (equation 2.13), because A_xB_x=B_xA_x and the same holds for each term.

\mathbf A\cdot\mathbf B=\mathbf B\cdot\mathbf A

2.4

Solve by using the properties of dot and cross product: (a) \hat i\cdot(\hat j\times\hat k) (b) \hat j\times(\hat j\times\hat k)?

Solution

First use the unit vector rule for the cross product in the bracket: \hat j\times\hat k=\hat i (the cyclic order \hat i\to\hat j\to\hat k\to\hat i gives \hat j\times\hat k=\hat i, \hat k\times\hat i=\hat j and \hat i\times\hat j=\hat k).

(a) \hat i\cdot(\hat j\times\hat k)

Replace the bracket by \hat i.
\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i
The dot product of a unit vector with itself is 1.
\hat i\cdot\hat i=1

(b) \hat j\times(\hat j\times\hat k)

Replace the inner cross product by \hat i. The result is a vector, so this is a cross product of two vectors.
\hat j\times(\hat j\times\hat k)=\hat j\times\hat i
Reverse the order of the cross product. It changes the sign, so \hat j\times\hat i=-\hat k.
\hat j\times\hat i=-(\hat i\times\hat j)=-\hat k

-\hat k

2.5

If both the dot product and the cross product of two vectors are zero. What would you conclude about the individual vectors?

Solution

Take the two conditions one at a time, with both vectors non-zero:

\mathbf A\cdot\mathbf B=0 with AB\neq0 needs \cos\theta=0, so the vectors are perpendicular.
\mathbf A\times\mathbf B=0 with AB\neq0 needs \sin\theta=0, so the vectors are parallel or antiparallel.
A vector cannot be both perpendicular and parallel to another non-zero vector, so the two conditions cannot hold together for two non-zero vectors.

Hence at least one of the two vectors must be a null vector (zero magnitude). Both products vanish automatically if either vector is zero.

\text{At least one of the vectors is a null vector.}

2.6

What are rectangular components of a vector? How they can be found?

Solution

The rectangular components of a vector are its parts along two mutually perpendicular directions, the x-axis and the y-axis in 2-D. Their vector sum gives the original vector: \mathbf A=A_x+A_y=A_x\hat\imath+A_y\hat\jmath (equations 2.1 and 2.2).

If the magnitude A and the angle \theta with the x-axis are known, the components are found from the right-angled triangle they make:

A_x=A\cos\theta\qquad A_y=A\sin\theta

If the components are known, the magnitude and direction follow from Pythagoras and the tangent ratio: A=\sqrt{A_x^2+A_y^2} and \theta=\tan^{-1}\dfrac{A_y}{A_x}.

A_x=A\cos\theta,\quad A_y=A\sin\theta

2.7

Give two examples for each of the scalar and vector product.

Solution

Scalar product (dot product):

Work: W=\mathbf F\cdot\mathbf d=Fd\cos\theta, the scalar product of force and displacement.
Power: P=\mathbf F\cdot\mathbf v=Fv\cos\theta, the scalar product of force and velocity.

Vector product (cross product):

Torque: \boldsymbol\tau=\mathbf r\times\mathbf F=rF\sin\theta\,\hat n, the cross product of position vector and force.
Angular momentum: \mathbf L=\mathbf r\times\mathbf p=rp\sin\theta\,\hat n, the cross product of position vector and linear momentum.

2.82026 Annual I, Set B, Q.2 (iii)

Show that: \mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z.

Solution

Write each vector in terms of its unit vectors, then expand the dot product term by term:

Write the two vectors in component form.
\mathbf A=A_x\hat\imath+A_y\hat\jmath+A_z\hat k,\qquad \mathbf B=B_x\hat\imath+B_y\hat\jmath+B_z\hat k
Multiply out the brackets with the distributive law. This gives nine products.
\mathbf A\cdot\mathbf B=A_xB_x\,\hat\imath\cdot\hat\imath+A_xB_y\,\hat\imath\cdot\hat\jmath+\cdots+A_zB_z\,\hat k\cdot\hat k
For unit vectors, \hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1, because the angle is 0^\circ.
\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1
The mixed products are zero, because the unit vectors are mutually perpendicular (\cos90^\circ=0).
\hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0
Keep only the three non-zero terms.
\mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z

\mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z

2.9

What units are associated with the unit vectors \hat i, \hat j, and \hat k?

Solution

A unit vector has magnitude one by definition. It only fixes a direction, so its magnitude is a pure number, |\hat i|=1 with no unit.

\text{None: unit vectors are dimensionless.}

Comprehensive questionsp. 31

2.1

Explain the resolution of a vector in 2-D as two rectangular components.

Solution

Resolution of a vector

Resolving a vector means splitting it into two components, which together have the same effect as the original vector. In two dimensions the components are taken along the x-axis and the y-axis. These two directions are perpendicular, so the parts are called rectangular components. The component along the x-axis is A_x and the component along the y-axis is A_y. The unit vectors \hat\imath and \hat\jmath show the directions of the axes and each has magnitude one.

Derivation Components of a vector A at angle \theta

Aim. Write the vector A as the sum of its rectangular components and find their magnitudes in terms of A and \theta.

Assumptions.

The vector A makes angle \theta with the positive x-axis, measured anticlockwise.
The tail of A is at the origin O, and its head is at P.

Figure 2.1: Rectangular components of a vector A. OQ is the x-component and OS the y-component.

Draw A from the origin O to its head P, at angle \theta to the positive x-axis. Drop perpendiculars from P to the axes, meeting the x-axis at Q and the y-axis at S. The figure OQPS is a rectangle, so QP=OS. The triangle OPQ has a right angle at Q, and OQ=A_x, QP=A_y.
\overrightarrow{OP}=\overrightarrow{OQ}+\overrightarrow{QP}
Replace the sides by the components, with the unit vectors for direction. This is equation 2.2.
\mathbf A=A_x\,\hat\imath+A_y\,\hat\jmath
In the right angled triangle OPQ, \cos\theta is the base over the hypotenuse, so OQ is the adjacent side to \theta.
\cos\theta=\dfrac{OQ}{OP}=\dfrac{A_x}{A}
Make A_x the subject. This is equation 2.4.
A_x=A\cos\theta
\sin\theta is the opposite side over the hypotenuse, and the opposite side is QP=A_y.
\sin\theta=\dfrac{A_y}{A}
Make A_y the subject. This is equation 2.3.
A_y=A\sin\theta

ResultA_x=A\cos\theta,\qquad A_y=A\sin\theta

The components are the projections of A on the two axes. For 0^\circ<\theta<90^\circ both are positive. For an angle above 90^\circ, \cos\theta is negative, so A_x points along -\hat\imath.

Magnitude and direction from the components

Derivation Finding A and \theta from A_x and A_y

Aim. Find the magnitude and direction of A when its components are known.

In triangle OPQ, Pythagoras gives (OP)^2=(OQ)^2+(QP)^2.
A^2=A_x^2+A_y^2
Take the positive square root, since a magnitude is never negative. This is equation 2.7.
A=\sqrt{A_x^2+A_y^2}
Use the tangent ratio, perpendicular over base, in the same triangle.
\tan\theta=\dfrac{QP}{OQ}=\dfrac{A_y}{A_x}
Invert the tangent to get the angle. This is equation 2.8.
\theta=\tan^{-1}\dfrac{A_y}{A_x}

ResultA=\sqrt{A_x^2+A_y^2},\qquad \theta=\tan^{-1}\dfrac{A_y}{A_x}

A calculator's inverse tangent gives angles only between -90^\circ and 90^\circ. Check the signs of A_x and A_y to place \theta in the correct quadrant.

Example (book Example 2.1): a force of 50\,\mathrm{N} at 30^\circ to the road has F_x=50\cos30^\circ=43.3\,\mathrm{N} and F_y=50\sin30^\circ=25\,\mathrm{N}.

2.22026 Annual I, Set D, Q.2 (i) OR

Describe the dot product of two vectors along with their properties.

Solution

Scalar (dot) product

When the product of two vector quantities gives a scalar, the product is called the scalar product, or dot product, written \mathbf A\cdot\mathbf B. The dot product of A and B is the product of their magnitudes and the cosine of the angle \theta between them, taken tail to tail:

\mathbf A\cdot\mathbf B=AB\cos\theta\qquad(2.9)

Derivation Dot product as a component times a magnitude

Aim. Show that the dot product can be written as the magnitude of one vector times the component of the other along it.

Assumptions.

\theta is the angle between the tails of A and B, with 0^\circ\le\theta\le180^\circ.

Figure 2.3: Component of vector A in the direction of vector B, which is A\cos\theta.

Draw A and B tail to tail with angle \theta between them. From the head of A drop a perpendicular on the line of B. The part of A lying along B is the component A\cos\theta.
\text{component of }\mathbf A\text{ along }\mathbf B=A\cos\theta
Multiply that component by the magnitude of B, then use equation 2.9.
\mathbf A\cdot\mathbf B=(A\cos\theta)\,B=AB\cos\theta
Repeat with the component of B along A, which is B\cos\theta. The angle is the same, so the result is the same.
\mathbf A\cdot\mathbf B=A\,(B\cos\theta)

Result\mathbf A\cdot\mathbf B=AB\cos\theta=(A\cos\theta)B=A(B\cos\theta)

Only the part of A along B contributes to the dot product. This is why work W=\mathbf F\cdot\mathbf d counts only the force component along the displacement.

Examples of the scalar product

Work: W=\mathbf F\cdot\mathbf d=Fd\cos\theta.
Power: P=\mathbf F\cdot\mathbf v=Fv\cos\theta.
Electric flux: \Phi=\mathbf E\cdot\mathbf A=EA\cos\theta, where \mathbf A is the vector area.

Properties of the dot product

Commutative: \mathbf A\cdot\mathbf B=\mathbf B\cdot\mathbf A, because AB\cos\theta=BA\cos\theta.
Perpendicular vectors: \mathbf A\cdot\mathbf B=AB\cos90^\circ=0. So \hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0. The dot product with a null vector is also zero.
Parallel vectors: \mathbf A\cdot\mathbf B=AB\cos0^\circ=AB, the maximum value. In particular \mathbf A\cdot\mathbf A=A^2 and \hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1.
Antiparallel vectors: \mathbf A\cdot\mathbf B=AB\cos180^\circ=-AB, the minimum value.
Components: \mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z (equation 2.13), and so \cos\theta=\dfrac{A_xB_x+A_yB_y+A_zB_z}{AB} (equation 2.14).

Derivation Dot product in rectangular components

Aim. Derive equation 2.13 from the unit vector rules.

Write A and B in component form and multiply them term by term.
\mathbf A\cdot\mathbf B=(A_x\hat\imath+A_y\hat\jmath+A_z\hat k)\cdot(B_x\hat\imath+B_y\hat\jmath+B_z\hat k)
Each term is a number times a dot product of unit vectors. Unit vectors dotted with themselves give 1, and mutually perpendicular unit vectors give 0.
\mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z

Result\mathbf A\cdot\mathbf B=A_xB_x+A_yB_y+A_zB_z

2.32026 Annual I, Set B, Q.2 (i) OR2025 Annual I, Set B, Q.32025 Annual I, Set D, Q.2 (xiv) OR

Describe the cross product of two vectors along with their properties.

Solution

Vector (cross) product

When the product of two vector quantities gives a vector, the product is called the vector product, or cross product, written \mathbf A\times\mathbf B. It is defined as

\mathbf A\times\mathbf B=AB\sin\theta\,\hat n\qquad(2.15)

where A and B are the magnitudes, \theta is the angle between them and \hat n is the unit vector that gives the direction. The direction of \hat n is perpendicular to the plane containing A and B. It is found with the right hand rule: rotate the fingers of the right hand from the first vector to the second through the smaller angle, and the erect thumb gives the direction of the product.

Figure 2.4: Cross product of A and B, written A × B. It points perpendicular to the plane of A and B (drawn upward, as in the book).

Examples of the vector product

Torque: \boldsymbol\tau=\mathbf r\times\mathbf F=rF\sin\theta\,\hat n, the cross product of position vector and force.
Angular momentum: \mathbf L=\mathbf r\times\mathbf p=rp\sin\theta\,\hat n, the cross product of position vector and linear momentum.

Properties of the cross product

Anticommutative: \mathbf B\times\mathbf A=-\mathbf A\times\mathbf B. The magnitude is the same, but the direction reverses (Figure 2.6 of the book).
Parallel or antiparallel vectors: \mathbf A\times\mathbf B=AB\sin0^\circ\,\hat n=0 and AB\sin180^\circ\,\hat n=0. So \mathbf A\times\mathbf A=\mathbf 0 and \hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\mathbf 0.
Perpendicular vectors: \mathbf A\times\mathbf B=AB\sin90^\circ\,\hat n=AB\,\hat n, the maximum magnitude.
Unit vectors: \hat\imath\times\hat\jmath=\hat k, \hat\jmath\times\hat k=\hat\imath and \hat k\times\hat\imath=\hat\jmath (the cyclic order).
Components: the product in rectangular components is given by a determinant (equations 2.16 and 2.17).

Derivation Cross product in rectangular components

Aim. Derive equation 2.17 from the unit vector rules.

Write A and B in component form and multiply term by term using the distributive law.
\mathbf A\times\mathbf B=(A_x\hat\imath+A_y\hat\jmath+A_z\hat k)\times(B_x\hat\imath+B_y\hat\jmath+B_z\hat k)
Terms with the same unit vector on both sides are zero. Keep the six terms with different unit vectors.
=A_xB_y\,\hat\imath\times\hat\jmath+A_xB_z\,\hat\imath\times\hat k+A_yB_x\,\hat\jmath\times\hat\imath+A_yB_z\,\hat\jmath\times\hat k+A_zB_x\,\hat k\times\hat\imath+A_zB_y\,\hat k\times\hat\jmath
Use \hat\imath\times\hat\jmath=\hat k, \hat\jmath\times\hat k=\hat\imath, \hat k\times\hat\imath=\hat\jmath, and reverse the order where needed: \hat\jmath\times\hat\imath=-\hat k, \hat k\times\hat\jmath=-\hat\imath, \hat\imath\times\hat k=-\hat\jmath.
=A_xB_y\hat k-A_xB_z\hat\jmath-A_yB_x\hat k+A_yB_z\hat\imath+A_zB_x\hat\jmath-A_zB_y\hat\imath
Collect the coefficients of \hat\imath, \hat\jmath and \hat k.
=(A_yB_z-A_zB_y)\hat\imath+(A_zB_x-A_xB_z)\hat\jmath+(A_xB_y-A_yB_x)\hat k
Write the same result as a determinant, which is easier to remember. The first row holds the unit vectors, the other rows hold the components.
\mathbf A\times\mathbf B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}

Result\mathbf A\times\mathbf B=(A_yB_z-A_zB_y)\,\hat\imath+(A_zB_x-A_xB_z)\,\hat\jmath+(A_xB_y-A_yB_x)\,\hat k

2.4

What geometric interpretation does the cross product have? Explain with the help of a diagram.

Solution

Geometric meaning

The magnitude of the cross product gives the area of a parallelogram. If A and B are two adjacent sides of a parallelogram, then |\mathbf A\times\mathbf B| is the area of that parallelogram. Its direction, given by the right hand rule, is normal to the plane of the parallelogram. The rectangle or parallelogram may have any shape, so the rule applies to any two adjacent sides.

Derivation Area of the parallelogram

Aim. Show that the area of the parallelogram with adjacent sides A and B equals |\mathbf A\times\mathbf B|.

Assumptions.

A and B are adjacent sides of the parallelogram, with \theta the angle between them.

Figure 2.7: The magnitude of A × B gives the area of the parallelogram with sides A and B.

Take the side A as the base of the parallelogram. Its length is A. The height is the perpendicular distance from the tip of B to the line of A.
\text{base}=A
In the right angled triangle formed by B and the height, the height is the opposite side to \theta in a triangle with hypotenuse B.
h=B\sin\theta
Area is base times height.
\text{Area}=A\,(B\sin\theta)=AB\sin\theta
Compare with the magnitude of the cross product, from equation 2.15.
|\mathbf A\times\mathbf B|=AB\sin\theta=\text{Area}

Result\text{Area of parallelogram}=|\mathbf A\times\mathbf B|

If A and B are parallel the parallelogram collapses to a line with zero area, which agrees with \mathbf A\times\mathbf B=\mathbf 0.

Check with book Example 2.3: for \mathbf A=\hat\imath+6\hat\jmath+2\hat k and \mathbf B=7\hat\imath+\hat\jmath+5\hat k (in metres), \mathbf A\times\mathbf B=28\hat\imath+9\hat\jmath-41\hat k\,\mathrm{m^2}. Its magnitude is \sqrt{28^2+9^2+41^2}=\sqrt{2546}\approx50.5\,\mathrm{m^2}, which is the area of the parallelogram.

Numerical problemsp. 31

2.1

If the magnitude of cross product between two vectors is \sqrt{3} times the dot product, find angle between them.

Solution

Given

|\mathbf A\times\mathbf B| = \sqrt3\,(\mathbf A\cdot\mathbf B)

To find

\theta

Write both products in terms of the magnitudes and the angle, using equations 2.9 and 2.15.
AB\sin\theta=\sqrt3\,AB\cos\theta
Cancel AB and divide both sides by \cos\theta, which is not zero because the dot product is not zero.
\tan\theta=\sqrt3
Take the inverse tangent.
\theta=\tan^{-1}\sqrt3=60^\circ

\theta=60^\circ

Book answer. 60^\circ

2.2

A force is acting on a body making an angle of 30^\circ with the horizontal. The horizontal component of the force is 20\,\mathrm{N}. Find the force.

Solution

Given

\theta = 30^\circ
F_x = 20\,\mathrm{N}

To find

The horizontal component is F_x=F\cos\theta (equation 2.4), so the force is F=\dfrac{F_x}{\cos\theta}.
F=\dfrac{F_x}{\cos\theta}
Substitute the values.
F=\dfrac{20}{\cos30^\circ}=\dfrac{20}{0.866}
Evaluate.
F=23.1\,\mathrm{N}

F\approx23.1\,\mathrm{N}

Book answer. 23.1\,\mathrm{N}

2.3

A vector F having magnitude 5.5\,\mathrm{N} makes 10^\circ with x-axis and a vector r with magnitude 4.3\,\mathrm{m} makes 80^\circ with x-axis. What is the magnitude of their dot and cross products?

Solution

Given

F = 5.5\,\mathrm{N} at 10^\circ to the x-axis
r = 4.3\,\mathrm{m} at 80^\circ to the x-axis

To find

|\mathbf F\cdot\mathbf r|
|\mathbf F\times\mathbf r|

Both vectors make angles with the same x-axis, so the angle between them is the difference of the two angles: \theta=80^\circ-10^\circ=70^\circ.

(i) Magnitude of the dot product

Use the scalar product (equation 2.9).
\mathbf F\cdot\mathbf r=Fr\cos\theta
Substitute the values. Fr=5.5\times4.3=23.65\,\mathrm{N\,m}.
\mathbf F\cdot\mathbf r=23.65\times\cos70^\circ=23.65\times0.342
Evaluate.
\mathbf F\cdot\mathbf r=8.1\,\mathrm{N\,m}

8.1\,\mathrm{N\,m}

(ii) Magnitude of the cross product

Use the magnitude of the vector product (equation 2.15).
|\mathbf F\times\mathbf r|=Fr\sin\theta
Substitute the values.
|\mathbf F\times\mathbf r|=23.65\times\sin70^\circ=23.65\times0.940
Evaluate.
|\mathbf F\times\mathbf r|=22.2\,\mathrm{N\,m}

22.2\,\mathrm{N\,m}

Book answer. 8.1\,\mathrm{N\,m} and 22.2\,\mathrm{N\,m}

2.42026 Annual I, Set D, Q.2 (iii)

The magnitude of dot and cross product of two vectors are 6\sqrt{3} and 6, respectively. Find the angle between the vectors.

Solution

Given

\mathbf A\cdot\mathbf B = 6\sqrt3
|\mathbf A\times\mathbf B| = 6

To find

\theta

Write the two products in magnitude form.
AB\cos\theta=6\sqrt3,\qquad AB\sin\theta=6
Divide the second equation by the first. AB cancels.
\tan\theta=\dfrac{AB\sin\theta}{AB\cos\theta}=\dfrac{6}{6\sqrt3}=\dfrac{1}{\sqrt3}
Take the inverse tangent.
\theta=\tan^{-1}\dfrac{1}{\sqrt3}=30^\circ
Check: squaring and adding the two equations gives (AB)^2=108+36=144, so AB=12, which is consistent with positive magnitudes.
(AB)^2=(6\sqrt3)^2+6^2=144

\theta=30^\circ

Book answer. \theta=30^\circ

2.5

A force of 200\,\mathrm{N} is making an angle of 30^\circ with the x-axis. Find the horizontal and vertical components of this force.

Solution

Given

F = 200\,\mathrm{N}
\theta = 30^\circ

To find

Horizontal component (equation 2.4).
F_x=F\cos\theta=200\cos30^\circ=200\times0.866
Vertical component (equation 2.3).
F_y=F\sin\theta=200\sin30^\circ=200\times0.5
Evaluate both.
F_x=173.2\,\mathrm{N},\qquad F_y=100\,\mathrm{N}

F_x=173.2\,\mathrm{N},\qquad F_y=100\,\mathrm{N}

Book answer. 173.2\,\mathrm{N} and 100\,\mathrm{N}

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 31 mark

\hat{j}.(\hat{i}\times\hat{k}) =

A0
B-1Answer
C1
D2

Why

By the unit vector rules \hat k\times\hat\imath=\hat\jmath, so reversing the order gives \hat\imath\times\hat k=-\hat\jmath. Then \hat\jmath\cdot(\hat\imath\times\hat k)=\hat\jmath\cdot(-\hat\jmath)=-(\hat\jmath\cdot\hat\jmath)=-1. So B is correct.

Option C (1) comes from taking \hat\imath\times\hat k=+\hat\jmath, which is true only for the cyclic order \hat\imath\to\hat\jmath\to\hat k\to\hat\imath. Option A (0) would hold if the bracket were perpendicular to \hat\jmath, but \hat\imath\times\hat k lies along \hat\jmath.

Note. Option C is not clearly printed in the paper and is read here as 1; the answer B (-1) does not depend on it.

2026 Annual I, Set B, Q.2 (i) OR1+1+1 = 3 marks

Describe any three properties of cross product of two vectors.

Solution

The cross product of two vectors is \vec A\times\vec B=AB\sin\theta\,\hat n, where \theta is the angle between them and \hat n is a unit vector perpendicular to their plane (right hand rule). Three properties (1 mark each):

It is anticommutative. \vec A\times\vec B=-\,\vec B\times\vec A. The magnitude AB\sin\theta is the same, but reversing the order reverses the direction. For example \hat\imath\times\hat\jmath=\hat k but \hat\jmath\times\hat\imath=-\hat k.
It is zero for parallel or antiparallel vectors. For \theta=0^\circ or 180^\circ, \sin\theta=0, so \vec A\times\vec B=\vec 0. In particular \vec A\times\vec A=\vec 0 and \hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0.
It is largest for perpendicular vectors. For \theta=90^\circ, \sin\theta=1, so |\vec A\times\vec B|=AB. For the unit vectors, \hat\imath\times\hat\jmath=\hat k, \hat\jmath\times\hat k=\hat\imath and \hat k\times\hat\imath=\hat\jmath.

Other properties are that the product is a vector (not a scalar) and that, in components, \vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}.

2026 Annual I, Set B, Q.2 (iii)3 marks

Show that: \vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z

Solution

Write each vector in terms of its rectangular components along the unit vectors \hat\imath, \hat\jmath, \hat k, and multiply them with the dot product (1 mark).

Component form (1 mark).
\vec A=A_x\hat\imath+A_y\hat\jmath+A_z\hat k,\qquad \vec B=B_x\hat\imath+B_y\hat\jmath+B_z\hat k
Multiply term by term using the distributive law.
\vec A\cdot\vec B=A_xB_x(\hat\imath\cdot\hat\imath)+A_xB_y(\hat\imath\cdot\hat\jmath)+A_xB_z(\hat\imath\cdot\hat k)+A_yB_x(\hat\jmath\cdot\hat\imath)+A_yB_y(\hat\jmath\cdot\hat\jmath)+A_yB_z(\hat\jmath\cdot\hat k)+A_zB_x(\hat k\cdot\hat\imath)+A_zB_y(\hat k\cdot\hat\jmath)+A_zB_z(\hat k\cdot\hat k)
Use the unit vector rules (1 mark). A unit vector dotted with itself gives \cos0^\circ=1, and two different (perpendicular) unit vectors give \cos90^\circ=0. Only three terms survive.
\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1,\qquad \hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0
Result.
\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z

\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z

2025 Annual I, Set B, Q.2 (v) OR3 marks

Verify that: (\vec{A}\cdot\vec{B})^2 + (\vec{A}\times\vec{B})^2 = (AB)^2

Solution

Let \theta be the angle between \vec A and \vec B. Square the magnitude of each product.

Dot product (1 mark). \vec A\cdot\vec B=AB\cos\theta, a scalar. Square it.
(\vec A\cdot\vec B)^{2}=A^{2}B^{2}\cos^{2}\theta
Cross product (1 mark). \vec A\times\vec B=AB\sin\theta\,\hat n has magnitude AB\sin\theta and \hat n\cdot\hat n=1. Square it.
(\vec A\times\vec B)^{2}=A^{2}B^{2}\sin^{2}\theta
Add (1 mark). Take A^{2}B^{2} common and use \sin^{2}\theta+\cos^{2}\theta=1.
(\vec A\cdot\vec B)^{2}+(\vec A\times\vec B)^{2}=A^{2}B^{2}\left(\cos^{2}\theta+\sin^{2}\theta\right)=A^{2}B^{2}=(AB)^{2}

Hence the relation is verified. It holds for every angle, because the dot product picks out the part of one vector along the other and the cross product the part perpendicular to it.

(\vec A\cdot\vec B)^{2}+(\vec A\times\vec B)^{2}=(AB)^{2}

2026 Annual I, Set B, Q.2 (viii) OR3 marks

Find the area of the parallelogram whose adjacent sides are given by the vectors: \vec{A} = (\hat{i}+6\hat{j}+2\hat{k})\,m and \vec{B} = (7\hat{i}+\hat{j}+5\hat{k})\,m.

Solution

The area of the parallelogram whose adjacent sides are the vectors \vec A and \vec B is the magnitude of their cross product, \text{Area}=|\vec A\times\vec B|=AB\sin\theta (base A times height B\sin\theta) (1 mark).

Given

\vec A = (\hat\imath+6\hat\jmath+2\hat k)\,\mathrm m
\vec B = (7\hat\imath+\hat\jmath+5\hat k)\,\mathrm m

To find

area of the parallelogram

Cross product by the determinant (1 mark).
\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\1&6&2\\7&1&5\end{vmatrix}=(6\cdot5-2\cdot1)\hat\imath-(1\cdot5-2\cdot7)\hat\jmath+(1\cdot1-6\cdot7)\hat k
Simplify.
\vec A\times\vec B=28\hat\imath+9\hat\jmath-41\hat k\ \ \mathrm{m^{2}}
Magnitude gives the area (1 mark).
\text{Area}=\sqrt{28^{2}+9^{2}+(-41)^{2}}=\sqrt{784+81+1681}=\sqrt{2546}=50.5\,\mathrm{m^{2}}

\text{Area}=50.5\,\mathrm{m^{2}}

2025 Annual I, Set B, Q.33+4 = 7 marks

State and explain vector product with an example. Also show that \vec{A}\times\vec{B} \neq \vec{B}\times\vec{A}.

Solution

Vector product: statement and explanation (3 marks)

When the product of two vectors gives a vector, it is called the vector product or cross product, written \vec A\times\vec B. For two vectors of magnitudes A and B with angle \theta between them it is defined as

\vec A\times\vec B=AB\sin\theta\,\hat n

The magnitude is the product of the magnitudes and the sine of the angle between them. The unit vector \hat n gives the direction: it is perpendicular to the plane containing \vec A and \vec B, and it is found by the right hand rule. Curl the fingers of the right hand from the first vector towards the second through the smaller angle; the erect thumb gives the direction of \hat n.

Example: torque. The torque of a force \vec F acting at position vector \vec r is the cross product \vec\tau=\vec r\times\vec F=rF\sin\theta\,\hat n. Its magnitude is largest when \vec F is perpendicular to \vec r (\sin90^\circ=1) and zero when \vec F is along \vec r. Another example is angular momentum, \vec L=\vec r\times\vec p.

Showing that A × B is not equal to B × A (4 marks)

Compare the two products in magnitude and in direction.

Magnitudes. The angle between the vectors is the same whichever is taken first, so the magnitudes are equal.
|\vec A\times\vec B|=AB\sin\theta=|\vec B\times\vec A|
Directions. By the right hand rule, rotating from \vec A to \vec B gives a thumb pointing up, +\hat n. Rotating from \vec B to \vec A gives a thumb pointing down, -\hat n (Figure 2.6).
\vec A\times\vec B=AB\sin\theta\,\hat n\qquad \vec B\times\vec A=AB\sin\theta\,(-\hat n)
Conclusion. The two products have the same magnitude but opposite directions, so they are not equal. The cross product is anticommutative.
\vec A\times\vec B=-\,\vec B\times\vec A\;\ne\;\vec B\times\vec A

Figure 2.6: By changing the order of the vectors, the direction of the vector product is reversed. A × B points up and B × A points down (right hand rule), with the same magnitude AB\sin\theta.

Check with unit vectors. \hat\imath\times\hat\jmath=\hat k, but reversing the order gives \hat\jmath\times\hat\imath=-\hat k. The two products are equal only when each is the null vector, that is when \vec A and \vec B are parallel or antiparallel (\sin\theta=0).

2025 Annual I, Set D, MCQ 21 mark

If \vec{A}\cdot\vec{B} = |\vec{A}\times\vec{B}| then angle '\theta' between vectors \vec{A} and \vec{B} is:

A0
B\pi/4Answer
C\pi/2
D\pi

Why

Put \vec A\cdot\vec B=AB\cos\theta and |\vec A\times\vec B|=AB\sin\theta in the given condition: AB\cos\theta=AB\sin\theta. Cancel AB to get \tan\theta=1, so \theta=\pi/4 (45^\circ). So B is correct.

At \theta=0 or \pi the cross product is zero while the dot product is \pm AB, and at \theta=\pi/2 the dot product is zero while the cross product is AB. So A, C and D cannot satisfy the condition.

2026 Annual I, Set D, Q.2 (i) OR1+1+1 = 3 marks

Describe any three properties of dot product of two vectors.

Solution

The dot product of two vectors is \vec A\cdot\vec B=AB\cos\theta, where \theta is the angle between them. It is a scalar. Three properties (1 mark each):

It is commutative. \vec A\cdot\vec B=\vec B\cdot\vec A, because AB\cos\theta=BA\cos\theta.
It is zero for perpendicular vectors. For \theta=90^\circ, \cos\theta=0, so \vec A\cdot\vec B=0. Hence \hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0.
It is largest for parallel vectors. For \theta=0^\circ, \cos\theta=1, so \vec A\cdot\vec B=AB. In particular \vec A\cdot\vec A=A^{2} and \hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1. For antiparallel vectors (\theta=180^\circ) it is -AB, the smallest value.

In components, \vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z. Work W=\vec F\cdot\vec d and power P=\vec F\cdot\vec v are examples of the dot product.

2026 Annual I, Set D, Q.2 (iii)3 marks

The magnitudes of dot and cross product of two vectors are 6\sqrt{3} and 6 respectively. Find the angle between the vectors.

Solution

Given

\vec A\cdot\vec B = 6\sqrt3
|\vec A\times\vec B| = 6

To find

angle \theta between the vectors

Write the two products (1 mark). The dot product is AB\cos\theta and the magnitude of the cross product is AB\sin\theta.
AB\cos\theta=6\sqrt3,\qquad AB\sin\theta=6
Divide the second by the first (1 mark). AB cancels.
\tan\theta=\dfrac{AB\sin\theta}{AB\cos\theta}=\dfrac{6}{6\sqrt3}=\dfrac{1}{\sqrt3}
Find the angle (1 mark).
\theta=\tan^{-1}\!\left(\dfrac{1}{\sqrt3}\right)=30^\circ

Check: AB=\sqrt{(6\sqrt3)^{2}+6^{2}}=\sqrt{144}=12, and 12\cos30^\circ=6\sqrt3, 12\sin30^\circ=6.

\theta=30^\circ

2025 Annual I, Set D, Q.2 (iii)3 marks

Under what condition the rectangular (perpendicular) components of a vector have same magnitude?

Solution

A vector \vec A at angle \theta to the x-axis has rectangular components A_x=A\cos\theta and A_y=A\sin\theta (1 mark).

Put the two magnitudes equal (1 mark).
A\cos\theta=A\sin\theta
Cancel A and divide by \cos\theta.
\tan\theta=1\quad\Rightarrow\quad\theta=45^\circ
Value of each component (1 mark).
A_x=A_y=A\cos45^\circ=\dfrac{A}{\sqrt2}=0.707\,A

So the rectangular components of a vector have the same magnitude when the vector makes an angle of 45^\circ with the x-axis (and so also 45^\circ with the y-axis). The same holds in the other quadrants, at 135^\circ, 225^\circ and 315^\circ, where the two components differ only in sign.

\theta=45^\circ,\quad A_x=A_y=\dfrac{A}{\sqrt2}

2026 Annual I, Set D, Q.2 (iv) OR3 marks

Find the power delivered by the engine for attaining velocity (3,4)\,\mathrm{ms^{-1}} while it exerts a force (8,-2)\,\mathrm{N}.

Solution

Given

\vec v = (3,4)\,\mathrm{m\,s^{-1}}
\vec F = (8,-2)\,\mathrm{N}

To find

power P delivered by the engine

Formula (1 mark). Power is the scalar product of force and velocity.
P=\vec F\cdot\vec v
Write the vectors in components (1 mark). The scalar product of two vectors is the sum of the products of their matching components.
\vec F=8\hat i-2\hat j,\qquad \vec v=3\hat i+4\hat j
Multiply and add (1 mark). Since \hat i\cdot\hat i=\hat j\cdot\hat j=1 and \hat i\cdot\hat j=0:
P=(8)(3)+(-2)(4)=24-8=16\,\mathrm{W}

Power is a scalar, so the answer is a single number with the unit watt (1\,\mathrm{W}=1\,\mathrm{N\,m\,s^{-1}}). The negative y-component of the force reduces the power.

P=16\,\mathrm{W}

2025 Annual I, Set D, Q.2 (viii) OR3 marks

Find the power delivered by an engine for attaining velocity = (4\hat{i}+5\hat{j})\,\mathrm{m\,s^{-1}} while it exerts a force = (8\hat{i}-2\hat{j})\,\mathrm{N}.

Solution

Given

\vec v = (4\hat\imath+5\hat\jmath)\,\mathrm{m\,s^{-1}}
\vec F = (8\hat\imath-2\hat\jmath)\,\mathrm{N}

To find

Power P

Formula (1 mark). Power is the scalar product of force and velocity.
P=\vec F\cdot\vec v
Substitution (1 mark). Multiply matching components and add (the dot product in components).
P=(8\hat\imath-2\hat\jmath)\cdot(4\hat\imath+5\hat\jmath)=(8)(4)+(-2)(5)
Answer with unit (1 mark).
P=32-10=22\,\mathrm{W}

P=22\,\mathrm{W}

2025 Annual I, Set D, Q.2 (xiv) OR3 marks

State the characteristics of vector product.

Solution

The vector product is \vec A\times\vec B=AB\sin\theta\,\hat n. Its characteristics are:

It is a vector. Its magnitude is AB\sin\theta and its direction (\hat n) is perpendicular to the plane of \vec A and \vec B, given by the right hand rule.
It is anticommutative, not commutative. \vec A\times\vec B=-\vec B\times\vec A: the magnitude is the same but the direction is reversed.
It is a null vector for parallel or antiparallel vectors. For \theta=0^\circ or 180^\circ, \sin\theta=0. So \vec A\times\vec A=\vec 0 and \hat\imath\times\hat\imath=\hat\jmath\times\hat\jmath=\hat k\times\hat k=\vec 0.
It is maximum for perpendicular vectors. For \theta=90^\circ, \vec A\times\vec B=AB\,\hat n. For the unit vectors, \hat\imath\times\hat\jmath=\hat k, \hat\jmath\times\hat k=\hat\imath and \hat k\times\hat\imath=\hat\jmath; reversing the order gives a minus sign.
In components it is the determinant of the unit vectors, the components of \vec A and those of \vec B. Its magnitude equals the area of the parallelogram with \vec A and \vec B as adjacent sides.

Exam focus

Describe the vector (cross) product: definition, right hand rule, examples, and its properties (anticommutative, zero for parallel vectors, maximum for perpendicular ones). (long)
2025 Set B Q.3 (7 marks, with a proof that \vec A\times\vec B\ne\vec B\times\vec A) and Set D Q.2 (xiv) OR; 2026 Set B Q.2 (i) OR (any three properties, 3 marks). Comprehensive question 2.3.
Go to answer
Describe the dot product of two vectors with its properties and its form in components. (long)
2026 Set D Q.2 (i) OR (any three properties of the dot product, 3 marks); 2025 Set D Q.2 (viii) OR (power P=\vec F\cdot\vec v=22\,\mathrm W); 2026 Set D Q.2 (iv) OR (power =16\,\mathrm W from \vec F=(8,-2) and \vec v=(3,4), the same as Example 2.2). Comprehensive question 2.2 and short questions 2.2, 2.3.
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Show that \vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z using the dot products of the unit vectors. (short)
2026 Set B Q.2 (iii), 3 marks. Short question 2.8 asks the same proof.
Go to answer
Find the angle between two vectors from their dot and cross products, e.g. 6\sqrt3 and 6 gives 30^\circ. (numerical)
2026 Set D Q.2 (iii) (identical to book numerical 2.4) and 2025 Set D MCQ 2 (|\vec A\cdot\vec B|=|\vec A\times\vec B| gives 45^\circ). Book exercise MCQ 6 and numerical 2.1 are the same type.
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Area of a parallelogram from the cross product of two adjacent sides, using the determinant. (numerical)
2026 Set B Q.2 (viii) OR, 3 marks (\vec A=\hat\imath+6\hat\jmath+2\hat k, \vec B=7\hat\imath+\hat\jmath+5\hat k). The geometric meaning is comprehensive question 2.4 and Example 2.3.
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Verify identities that combine the dot and cross products, e.g. (\vec A\cdot\vec B)^2+(\vec A\times\vec B)^2=(AB)^2. (short)
2025 Set B Q.2 (v) OR, 3 marks. The same idea (comparing AB\cos\theta with AB\sin\theta) solves the angle questions above.
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Products of unit vectors, e.g. \hat\jmath\cdot(\hat\imath\times\hat k)=-1 and \hat\jmath\times\hat\imath=-\hat k. (mcq)
2025 Set B MCQ 3 tested the sign of \hat\imath\times\hat k; book exercise MCQ 2 and short question 2.4 are the same type.
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Rectangular components of a vector: when are they equal, and how to find them from A and \theta. (short)
2025 Set D Q.2 (iii), 3 marks (equal components at 45^\circ). Short question 2.6, MCQ 3 and numericals 2.2, 2.3, 2.5 use A_x=A\cos\theta, A_y=A\sin\theta. Not asked in 2026.
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