Vector Algebra: Class 12 Maths · Chapter 10

Arrows you can add, stretch and multiply

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

125 min
1 simulation
73 practice questions
30 MCQs

What you will learn in Vector Algebra

Tell scalars from vectors, and describe a vector by its magnitude, direction cosines and direction ratios
Recognise zero, unit, coinitial, collinear, equal and negative vectors
Add and subtract vectors by the triangle and parallelogram laws, and use their properties
Multiply a vector by a scalar, find unit vectors, and work with vectors in component form
Find the vector joining two points and the position vector of a point dividing a segment
Use the scalar product to find angles, test perpendicularity and compute projections
Use the vector product to find perpendicular vectors and the areas of parallelograms and triangles

Original companion text aligned to NCERT Mathematics Class 12, Chapter 10, pp. 338–376. Not affiliated with or endorsed by NCERT.

10.1 Introduction

A parcel drone takes off from a warehouse in Bengaluru. Ask how far it flew and one number answers you: 5 km. Ask where it landed and one number is no longer enough, because 5 km north-east and 5 km south-west are two different places (Fig. 10.1).

2/2Adding a direction fixes the landing point: this is a vector.

Fig. 10.1A distance alone allows a whole circle of end points; a distance with a direction picks out one.

A start point is marked with a dashed circle of radius 5 km around it and a small compass showing north and east. In the first step six possible landing points sit on the circle, each with a question mark, and a radius is labelled 5 km. In the second step those points fade and a single blue arrow runs from the start to the circle in the north-east direction, labelled 5 km NE.

Some quantities are fully described by a single real number, together with a unit: the mass of a pumpkin, the length of a cricket pitch, the noon temperature in Jaipur, the price of a bus ticket. Such quantities are called scalars (अदिश). Others need a size and a direction: the displacement of the drone, the velocity of a train, the pull on a kabaddi raider caught at the ankle. These are vectors (सदिश).

Temperature, energy, speed, volume and price are scalars; displacement, force, velocity and momentum are vectors. A quick test: if a measurement still makes sense after you add "towards the north", it is probably a vector.

In this chapter we treat vectors as objects of algebra. We learn to add them, to stretch them by numbers, and to multiply two of them in two different ways. Each rule comes with a picture, and the pictures are what make the rules easy to trust.

Quick check

1 of 2

Which of these are vector quantities?

the 22 yards between the stumpsa wind of 20 km/h blowing from the westthe 1.5 kg mass of a melona push of 40 N towards the door

Select all that apply

Show answer

Answer

(B) a wind of 20 km/h blowing from the west
(D) a push of 40 N towards the door

The wind and the push each come with a direction. A length of 22 yards and a mass of 1.5 kg are complete as single numbers, so they are scalars.

10.2 Some Basic Concepts

A vector needs a picture that carries a size and a sense together. Put an arrowhead on a straight line $l$ and travel along it is allowed one way only: we now have a directed line. Cut out the stretch from A to B and keep the arrowhead. What remains, a directed line segment, has a definite length as well as a direction, which is exactly the information a vector holds (Fig. 10.2).

Fig. 10.2A line, a directed line and a directed line segment: the last one carries both a length and a direction.

Three rows. The first row is a plain straight line labelled l. The second row is the same line with an arrowhead at its right end, showing one chosen direction. The third row shows a faint line with the piece from A to B drawn thick in blue, an arrowhead at B, A labelled initial point and B labelled terminal point, and the label vector AB above it.

Definition

Vector

A quantity that has both magnitude and direction. Geometrically, a vector is a directed line segment.

ExampleThe segment from $A$ to $B$ , written $\vec{A B}$ .

Names first. The arrow from A to B is $\vec{A B}$ ; when the end points do not matter we just call it $\vec{a}$ . A, where it starts, is the initial point (the tail), and B is the terminal point (the head). How long the arrow is, the distance AB, is its magnitude: $∣ \vec{A B} ∣$ , $∣ \vec{a} ∣$ , or simply $a$ for short.

Because it is a length, a magnitude can be 0 but never negative. An equation such as $∣ \vec{a} ∣ = - 3$ simply has no solution.

Position vector

Every point has a vector of its own. Fix the origin O of the right-handed axes from Class 11 as the tail; the arrow $\vec{O P}$ that ends at P(x, y, z) is then the position vector of P, often called $\vec{r}$ . Its length is the distance OP. By habit, points A, B, C, … get the position vectors $\vec{a}, \vec{b}, \vec{c}$ , …

Length of a position vector

∣ \vec{r} ∣ = \sqrt{x^{2} + y^{2} + z^{2}}

(10.1)

Example

How far from the origin?

A drone hovers at the point P(1, 4, 8), with distances in hundreds of metres. How far is it from the origin O?

Try it yourself first. Then reveal the solution one step at a time.

Direction cosines

Which way does $\vec{r} = \vec{O P}$ point? One complete answer is the angle it makes with each positive axis: α with the x-axis, β with the y-axis and γ with the z-axis. These direction angles lie between 0 and π. In practice we work with their cosines, $l = \cos α$ , $m = \cos β$ and $n = \cos γ$ , the direction cosines of $\vec{r}$ .

Drop a perpendicular from P to the x-axis, meeting it at A. Triangle OAP has its right angle at A, its hypotenuse is $O P = r$ and its side along the axis is $O A = x$ , so $\cos α = x / r$ (Fig. 10.3). The perpendiculars to the other two axes give two more right triangles in the same way.

2/4Triangle $O A P$ is right-angled at $A$ , so $\cos α = x / r$ .

Fig. 10.3Each direction cosine is adjacent over hypotenuse in a right triangle with hypotenuse

r

.NCERT Fig. 10.3 · p. 340

On the left, three axes meet at O and a blue arrow r runs from O to P(x, y, z), with faint guide lines to the coordinate planes. On the right, in the first step, the length formula for r is shown. In the second step the axis piece OA is highlighted with a dashed drop from P to A, and on the right the true-shape right triangle OAP appears with hypotenuse r, base x, a right angle at A, angle alpha at O and the label cos alpha equals x over r. The third and fourth steps do the same for the y-axis with B and beta, and the z-axis with C and gamma.

Direction cosines

l = \frac{x}{r}, m = \frac{y}{r}, n = \frac{z}{r}

(10.2)

Read backwards, (10.2) says $P = (l r, m r, n r)$ : the coordinates are the cosines scaled by r. Any triple obtained by scaling the cosines by some non-zero number, $a = k l$ , $b = k m$ , $c = k n$ , is called a set of direction ratios of the vector. The coordinates $x, y, z$ are the choice $k = r$ .

Squaring and adding the three cosines gives $l^{2} + m^{2} + n^{2} = (x^{2} + y^{2} + z^{2}) / r^{2} = 1$ , always. Direction ratios carry no such promise: 2, 4, 4 and 1, 2, 2 describe the same direction, and neither set has squares adding up to 1.

Example

Cosines from a point

Find the direction cosines of the position vector of P(1, 4, 8), and the angle it makes with the z-axis.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A point P is at (2, −3, 6). Find

∣ \vec{O P} ∣

Show answer

Answer 7

\sqrt{4 + 9 + 36} = \sqrt{49} = 7

10.3 Types of Vectors

A few kinds of vector turn up so often that they have names. Keep Fig. 10.4 in view as you read the list.

A zero vector, written $\vec{0}$ , starts and ends at one point, like $\vec{A A}$ . With nothing to point along, it has no particular direction, and it is convenient to let it take any direction we need.
A unit vector is any vector of length exactly 1; it records a direction and nothing else. For a non-zero $\vec{a}$ , the one pointing the same way is called $\hat{a}$ (read “a hat”).
Coinitial vectors share a tail: they all begin at one point.
Collinear vectors all lie along one line or along parallel lines. Their lengths may differ, and some may point the opposite way to others.
Equal vectors match in length and in direction; where they are drawn does not matter. We write $\vec{a} = \vec{b}$ .
Reversing a vector gives its negative: same length, opposite sense. Walking from B to A undoes walking from A to B, so $\vec{B A} = - \vec{A B}$ .

1/5Six vectors drawn on a grid.

Fig. 10.4Count grid squares to compare arrows: equal vectors make the same moves; collinear ones make proportional moves.NCERT Fig. 10.5 · p. 342

A square grid holds six arrows named p, q, r, s, t and u. In the first step all are shown. In the second step only p and q stay bright: both go 3 squares right and 1 up, so they are equal. In the third step r joins them: it goes 6 left and 2 down, along the same line direction, so p, q and r are collinear. In the fourth step s and t are bright: they start at the same grid point, so they are coinitial. In the fifth step t and u are bright: u goes 2 left and 3 down, exactly opposite to t, so u is the negative of t.

Example 1NCERT Example 1 · p. 341

Drawing a displacement

A survey drone flies 30 km in the direction 40° west of north. Show this displacement on a diagram.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 10.5The displacement of Example 1, drawn to scale.NCERT Fig. 10.4 · p. 341

Lines through a point O show north, south, east and west. A blue arrow OD leaves O between north and west, 40 degrees away from north, and is labelled 30 km. A scale bar in the corner shows the length that stands for 10 km.

Example 2NCERT Example 2 · p. 341

Scalar or vector?

Sort each measure into scalars and vectors: (i) 90 minutes, (ii) 250 mL, (iii) 3 kg, (iv) 15 m/s due east, (v) an acceleration of 9.8 m/s² straight down, (vi) ₹500, (vii) 60 °C, (viii) a 50 N upward pull.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 342

Reading types from a picture

In Fig. 10.4, which vectors are (i) equal, (ii) collinear, (iii) coinitial? Is any vector the negative of another?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

\vec{P Q}

and

\vec{R S}

have the same length and point the same way, but P and R are different points. Then the two vectors are

equalcoinitialnegatives of each othernot related at all

Show answer

Answer (A) equal

Equality needs only the same magnitude and the same direction; the starting points may differ (free vectors). They are not coinitial, because P ≠ R.

10.4 Addition of Vectors

An auto-rickshaw takes you from home H to the market M, and then from the market to school S. However much the lanes twist, your overall change of position is the straight arrow from H to S. Displacements add head to tail: $\vec{H M} + \vec{M S} = \vec{H S}$ .

The same rule adds any two vectors. To find $\vec{a} + \vec{b}$ , slide $\vec{b}$ , without turning or stretching it, until its tail sits on the head of $\vec{a}$ . The arrow from the tail of $\vec{a}$ to the head of the moved $\vec{b}$ is the sum, also called the resultant. This is the triangle law of vector addition (Fig. 10.6).

Triangle law

\vec{A B} + \vec{B C} = \vec{A C}

(10.3)

3/5 $\vec{A C} = \vec{a} + \vec{b}$ closes the triangle.

Fig. 10.6Head to tail: the third side of the triangle is the sum.NCERT Fig. 10.8 · p. 343

Step one shows a blue vector a from A to B and, separately, an orange vector b on the right. In step two b slides, without turning, until its tail is at B; its head is now C. In step three a purple arrow from A to C is labelled a plus b. In step four the arrow from C back to A is dashed and the label says AB plus BC plus CA equals the zero vector. In step five the purple sum is gone; a dashed red arrow minus b runs from B down to D, and a purple arrow from A to D is labelled a minus b.

Going once round a triangle brings you back to the start, so the three sides taken in order add up to the zero vector: $\vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$ . Subtraction is addition of the negative: $\vec{a} - \vec{b} = \vec{a} + (- \vec{b})$ .

The parallelogram law

A delivery drone is set to fly due north at 40 km/h while a crosswind carries it east at 30 km/h. The two velocities act at the same point at the same time, so we draw them as coinitial arrows. The drone actually moves along the diagonal of the parallelogram they span: at 50 km/h, about 37° east of north.

LawParallelogram law of vector addition

Draw two vectors from one point and complete the parallelogram they span. The diagonal that starts at that point is their sum.

The two laws agree. In parallelogram OACB, with $\vec{O A} = \vec{a}$ and $\vec{O B} = \vec{b}$ , the side AC is equal and parallel to OB, so $\vec{A C} = \vec{b}$ . The triangle law in triangle OAC then gives $\vec{O C} = \vec{a} + \vec{b}$ (Fig. 10.7).

3/4The diagonal $\vec{O C}$ is $\vec{a} + \vec{b}$ .

Fig. 10.7The diagonal through the common tail is the sum.NCERT Fig. 10.9 · p. 344

From a point O a blue vector a runs to A and an orange vector b runs to B. In step two dashed lines through A and B complete a shaded parallelogram with fourth corner C. In step three a purple diagonal from O to C is labelled a plus b. In step four the dashed sides become arrows: b from A to C and a from B to C, showing that both routes, a then b and b then a, end at C.

InteractiveVector addition lab

A square grid from minus 5 to 5 on both axes with two vectors from the origin: a in blue, starting at 3i + j, and b in orange, starting at i + 2j. Their tips are round handles that can be dragged to whole-number points, or set with four number boxes. Four buttons choose the view: triangle law (b redrawn from the head of a and the purple sum), parallelogram (the shaded parallelogram and its diagonal), a minus b (minus b drawn from the head of a and the purple difference), and lambda a (a slider for lambda from minus 3 to 3 and the purple multiple). A status panel gives the result in components with the lengths. Nothing moves until the learner acts.

Fig. 10.8Drag the tips of

\vec{a}

and

\vec{b}

(or type components). Switch between the triangle law, the parallelogram,

\vec{a} - \vec{b}

and

λ \vec{a}

, and read the components and lengths.

Properties of vector addition

Property 1Commutative property

For any two vectors, $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ .

Property 2Associative property

For any three vectors, $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$ .

1/2 $\vec{P R} + \vec{R S} = \vec{P S}$ .

Fig. 10.9However the three vectors are grouped, the sum is

\vec{P S}

.NCERT Fig. 10.11 · p. 345

Four points P, Q, R, S are joined head to tail by a blue vector a from P to Q, an orange vector b from Q to R and a green vector c from R to S, with a purple arrow from P to S labelled a plus b plus c. In the first step a dashed red diagonal from P to R is labelled a plus b, the first grouping. In the second step the dashed diagonal is from Q to S, labelled b plus c, the second grouping. Both groupings reach S.

Because the grouping does not matter, we simply write $\vec{a} + \vec{b} + \vec{c}$ . Adding the zero vector changes nothing, $\vec{a} + \vec{0} = \vec{a}$ , so $\vec{0}$ is the additive identity for vector addition.

Quick check

1 of 4

In triangle XYZ,

\vec{X Y} + \vec{Y Z}

equals

\vec{Z X}

\vec{X Z}

\vec{Y X}

\vec{0}

Show answer

Answer (B) $\vec{X Z}$

Head to tail: start at X, go to Y, then to Z. The net move is from X to Z.

\vec{Z X}

has the right length but the wrong direction.

10.5 Multiplication of a Vector by a Scalar

Doubling a displacement means going twice as far the same way; negating it means going back. In general, for a real number λ, the vector $λ \vec{a}$ is $\vec{a}$ stretched by the factor |λ| and turned round when λ is negative (Fig. 10.10). Either way it stays parallel to $\vec{a}$ : a vector and its multiples are always collinear.

Length of a scalar multiple

∣ λ \vec{a} ∣ = ∣ λ ∣ ∣ \vec{a} ∣

(10.4)

4/4 $\hat{a} = \frac{1}{2.5} \vec{a}$ has length 1.

Fig. 10.10Multiplying by λ stretches by |λ|, and reverses the direction when λ < 0.NCERT Fig. 10.12 · p. 346

All arrows start from one dashed vertical line, on a background ruled in units. In step one a blue arrow a points right with length 2.5. In step two a second arrow, 2a, points right with length 5. In step three a red arrow, minus 1.5 a, points left with length 3.75. In step four a short purple arrow, a hat, points right with length 1. A column on the right lists the lengths 2.5, 5, 3.75 and 1.

Three factors deserve names. λ = −1 reverses $\vec{a}$ into $- \vec{a}$ , which cancels it, $\vec{a} + (- \vec{a}) = \vec{0}$ ; so $- \vec{a}$ is the additive inverse of $\vec{a}$ . λ = 0 squashes every vector to $\vec{0}$ , and no factor can make $\vec{0}$ any longer: $k \vec{0} = \vec{0}$ .

The third is $λ = 1 / ∣ \vec{a} ∣$ , for $\vec{a} \neq \vec{0}$ . It rescales $\vec{a}$ to length exactly 1 without turning it, which gives $\hat{a}$ , the unit vector pointing the same way as $\vec{a}$ :

Unit vector along a

\hat{a} = \frac{1}{∣ \vec{a} ∣} \vec{a}

(10.5)

10.5.1 Components of a vector

Place a unit vector along each positive axis: $\hat{i}$ , $\hat{j}$ and $\hat{k}$ run from O to (1, 0, 0), (0, 1, 0) and (0, 0, 1). Every vector in space can be built from these three.

For a point P(x, y, z), let A be (x, 0, 0) and Q be (x, y, 0), the point of the xy-plane directly below or above P. The path O → A → Q → P makes three moves parallel to the axes: $x \hat{i}$ , then $y \hat{j}$ , then $z \hat{k}$ (Fig. 10.11). By the triangle law the position vector of P is their sum.

Component form

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

(10.6)

4/4Third move: $\vec{Q P} = z \hat{k}$ ; right angle at Q.

Fig. 10.11Three moves parallel to the axes add up to

\vec{r}

; Pythagoras twice gives its length.NCERT Fig. 10.14 · p. 347

Three axes meet at O. Step one shows short unit arrows i, j and k along the x-, y- and z-axes. Step two adds a blue arrow x i from O to A on the x-axis. Step three adds an orange arrow y j from A to Q, parallel to the y-axis, a dashed line OQ, a right-angle mark at A and the formula OQ squared equals x squared plus y squared. Step four adds a green arrow z k from Q up to P, a purple arrow r from O to P, a right-angle mark at Q and the formula OP squared equals OQ squared plus z squared.

Written like this, a vector is in component form. The three numbers x, y, z, its scalar components, say how far to go along each axis; the three moves themselves, $x \hat{i}$ , $y \hat{j}$ and $z \hat{k}$ , are its vector components (some books call them rectangular components).

Its length comes from Pythagoras, used twice. Triangle OAQ has its right angle at A, so $O Q^{2} = x^{2} + y^{2}$ . Triangle OQP has its right angle at Q, so $O P^{2} = O Q^{2} + z^{2} = x^{2} + y^{2} + z^{2}$ , which is (10.1) again.

In components, the algebra of vectors becomes arithmetic on matching entries. Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ .

Sum: add matching components, so $\vec{a} + \vec{b}$ has components $a_{1} + b_{1}$ , $a_{2} + b_{2}$ and $a_{3} + b_{3}$ .
Difference: subtract matching components: $a_{1} - b_{1}$ , $a_{2} - b_{2}$ and $a_{3} - b_{3}$ .
Equality: $\vec{a} = \vec{b}$ exactly when $a_{1} = b_{1}$ , $a_{2} = b_{2}$ and $a_{3} = b_{3}$ .
Scalar multiple: $λ \vec{a}$ has components $λ a_{1}$ , $λ a_{2}$ and $λ a_{3}$ .

PropertyDistributive laws

For vectors $\vec{a}, \vec{b}$ and scalars $k, m$ : (i) $k \vec{a} + m \vec{a} = (k + m) \vec{a}$ ; (ii) $k (m \vec{a}) = (k m) \vec{a}$ ; (iii) $k (\vec{a} + \vec{b}) = k \vec{a} + k \vec{b}$ .

Example 4NCERT Example 4 · p. 349

Equal vectors

Find p, q and r so that $2 p \hat{i} + (q - 3) \hat{j} + 5 \hat{k}$ and $6 \hat{i} + 2 \hat{j} + (r + 1) \hat{k}$ are equal.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 350

Same length, different vectors

Let $\vec{a} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k}$ and $\vec{b} = - 8 \hat{i} + \hat{j} + 4 \hat{k}$ . Is $∣ \vec{a} ∣ = ∣ \vec{b} ∣$ ? Are the vectors equal?

Try it yourself first. Then reveal the solution one step at a time.

Example 6NCERT Example 6 · p. 350

A unit vector

Which unit vector points the same way as $\vec{a} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ ?

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 350

A vector of given length

Find a vector of magnitude 12 in the direction of $2 \hat{i} + \hat{j} - 2 \hat{k}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 350

Unit vector along a sum

Add $\vec{a} = 5 \hat{i} + \hat{j} - 3 \hat{k}$ and $\vec{b} = - 3 \hat{i} + \hat{j} + 4 \hat{k}$ , then give the vector of length 1 that points along their sum.

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 351

Ratios and cosines

Write direction ratios of $\vec{v} = 2 \hat{i} - 6 \hat{j} + 3 \hat{k}$ and find its direction cosines.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Compared with

\vec{a}

, the vector

- 3 \vec{a}

3 times as long, same direction3 times as long, opposite directiona third as long, opposite directionas long, opposite direction

Show answer

Answer (B) 3 times as long, opposite direction

∣ - 3 \vec{a} ∣ = 3 ∣ \vec{a} ∣

, and the negative factor reverses the direction. A third as long would be

- \frac{1}{3} \vec{a}

10.5.2 Vector joining two points

Let $P_{1} (x_{1}, y_{1}, z_{1})$ and $P_{2} (x_{2}, y_{2}, z_{2})$ be two points. In triangle $O P_{1} P_{2}$ the triangle law says $\vec{O P_{1}} + \vec{P_{1} P_{2}} = \vec{O P_{2}}$ (Fig. 10.12). So the vector from $P_{1}$ to $P_{2}$ is "position vector of the head minus position vector of the tail".

Fig. 10.12

\vec{P_{1} P_{2}} = \vec{O P_{2}} - \vec{O P_{1}}

: head minus tail.NCERT Fig. 10.15 · p. 351

From the origin O a blue arrow goes to P1 and an orange arrow goes to P2. A purple arrow joins P1 to P2. The picture shows OP1 plus P1P2 equals OP2, so P1P2 is OP2 minus OP1.

Vector joining two points

\begin{aligned} \vec{P_{1} P_{2}} & = (x_{2} - x_{1}) \hat{i} \\ + (y_{2} - y_{1}) \hat{j} \\ + (z_{2} - z_{1}) \hat{k} \end{aligned}

(10.7)

Its magnitude is the distance $P_{1} P_{2}$ :

\begin{aligned} ∣ \vec{P_{1} P_{2}} ∣^{2} & = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \\ + (z_{2} - z_{1})^{2} \end{aligned}

(10.8)

Example 10NCERT Example 10 · p. 352

Head minus tail

Find the vector from A(4, −1, 2) to B(1, 3, −10), and its length.

Try it yourself first. Then reveal the solution one step at a time.

10.5.3 Section formula

Let P and Q have position vectors $\vec{a}$ and $\vec{b}$ , and let R be a point on the line PQ. If R lies between P and Q with $\frac{P R}{R Q} = \frac{m}{n}$ (m, n > 0), we say R divides PQ internally in the ratio m : n. If R lies on the line outside the segment with $\frac{P R}{Q R} = \frac{m}{n}$ , it divides PQ externally in that ratio (Fig. 10.13).

1/2R between P and Q, with $P R : R Q = m : n$ .

Fig. 10.13Internal division puts R between P and Q; external division puts R on the line outside the segment.NCERT Fig. 10.16 · p. 352

From the origin O, a blue vector a goes to P and an orange vector b goes to Q, and P and Q are joined. In the first step R lies between P and Q, two thirds of the way from P; braces above the segment mark PR as m and RQ as n, and a purple vector r goes from O to R. In the second step R lies beyond Q on the extended line; a brace above marks PR as m and a brace below marks QR as n, and the purple vector r goes from O to this new R.

DerivationPosition vector of the dividing point

Let R divide PQ internally in the ratio m : n and let its position vector be $\vec{r}$ . By head minus tail, $\vec{P R} = \vec{r} - \vec{a}$ and $\vec{R Q} = \vec{b} - \vec{r}$ .

These two vectors both point from P towards Q, and their lengths are in the ratio m : n. So n copies of $\vec{P R}$ match m copies of $\vec{R Q}$ :

\begin{aligned} n (\vec{r} - \vec{a}) \\ = m (\vec{b} - \vec{r}) \end{aligned}

Collect the $\vec{r}$ terms: $(m + n) \vec{r} = m \vec{b} + n \vec{a}$ . Dividing by $m + n$ gives the section formula for internal division.

Internal division

\vec{r} = \frac{m \vec{b} + n \vec{a}}{m + n}

(10.9)

For external division (m ≠ n), $\vec{P R}$ and $\vec{Q R}$ point the same way and $n \vec{P R} = m \vec{Q R}$ . Solving $n (\vec{r} - \vec{a}) = m (\vec{r} - \vec{b})$ in the same way gives the second formula.

External division

\vec{r} = \frac{m \vec{b} - n \vec{a}}{m - n}

(10.10)

For the midpoint, put m = n in (10.9):

Midpoint

\vec{r} = \frac{\vec{a} + \vec{b}}{2}

(10.11)

SimulationWhere does R divide PQ?

A coordinate plane from minus 10 to 10 on both axes with two fixed points, P at (−4, −5) and Q at (2, 3), and a draggable point R that starts at the midpoint (−1, −1). The distances PR and RQ are drawn and measured as R moves, so the learner can check the ratio in which R divides PQ, internally or externally.

Fig. 10.14P(−4, −5) and Q(2, 3) are fixed, 10 units apart. Drag R and compare the distances PR and RQ.

Try this

Use (10.9) to predict the point that divides PQ internally in the ratio 2 : 3. Drag R there. Do PR and RQ read 4 and 6?
Drag R to (5, 7). It lies outside the segment. In what ratio does it divide PQ externally?
Predict the point dividing PQ externally in the ratio 1 : 3, then drag R to check.

Example 11NCERT Example 11 · p. 353

Dividing both ways

Points P and Q have position vectors $\vec{a} = 2 \hat{i} + \hat{j} - 3 \hat{k}$ and $\vec{b} = 7 \hat{i} - 4 \hat{j} + 2 \hat{k}$ . Find the position vector of the point dividing PQ in the ratio 2 : 3, (i) internally, (ii) externally.

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 353

A right-angled triangle

Show that the triangle with corners A(2, 0, 1), B(3, 2, 3) and C(5, 3, 1) has a right angle.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Find the magnitude of

- 4 \hat{i} + 4 \hat{j} + 2 \hat{k}

Show answer

Answer 6

\sqrt{16 + 16 + 4} = \sqrt{36} = 6

Practice · 19 questions

Practice 10.2: questions with answers

1
Quick answerEasyNCERT Exercise 10.2, Q1
Find the magnitude of each vector.
1. $\vec{p} = 2 \hat{i} - \hat{j} + 2 \hat{k}$
  Show answer
  Answer 3
  $\sqrt{4 + 1 + 4} = 3$ .
2. $\vec{q} = 6 \hat{i} - 2 \hat{j} + 3 \hat{k}$
  Show answer
  Answer 7
  $\sqrt{36 + 4 + 9} = 7$ .
3. $\vec{s} = \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} - \frac{1}{\sqrt{2}} \hat{k}$
  Show answer
  Answer 1
  $\sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = 1$ : a unit vector.
Short answerEasyNCERT Exercise 10.2, Q2
Give an example of two unequal vectors whose lengths are equal.
Show solution
AnswerFor example $3 \hat{i} + 4 \hat{j}$ and $5 \hat{k}$ .
$∣ 3 \hat{i} + 4 \hat{j} ∣ = \sqrt{9 + 16} = 5$ and $∣ 5 \hat{k} ∣ = 5$ . The vectors point in different directions, so they are different vectors with the same magnitude.
Short answerEasyNCERT Exercise 10.2, Q3
Give an example of two unequal vectors that point in exactly the same direction.
Show solution
AnswerFor example $2 \hat{i} - \hat{j} + \hat{k}$ and $6 \hat{i} - 3 \hat{j} + 3 \hat{k}$ .
The second is 3 times the first. Multiplying by a positive scalar keeps the direction and changes the length (from $\sqrt{6}$ to $3 \sqrt{6}$ ), so the vectors differ.
4
Quick answerEasyNCERT Exercise 10.2, Q4
Find x and y so that $(x + 2) \hat{i} - 4 \hat{j} = 5 \hat{i} + 2 y \hat{j}$ .
1. x = ?
  Show answer
  Answer 3
  $x + 2 = 5$ .
2. y = ?
  Show answer
  Answer -2
  $2 y = - 4$ .
Short answerEasyNCERT Exercise 10.2, Q5
A vector starts at (−3, 4) and ends at (5, −2). Find its scalar and vector components.
Show solution
AnswerScalar: 8 and −6. Vector: $8 \hat{i}$ and $- 6 \hat{j}$ .
The vector is $(5 + 3) \hat{i} + (- 2 - 4) \hat{j} = 8 \hat{i} - 6 \hat{j}$ .
Its scalar components are 8 and −6; its vector components are $8 \hat{i}$ and $- 6 \hat{j}$ .
Short answerEasyNCERT Exercise 10.2, Q6
Find the sum of $\vec{a} = 3 \hat{i} + \hat{j} - 4 \hat{k}$ , $\vec{b} = - \hat{i} + 5 \hat{j} + 2 \hat{k}$ and $\vec{c} = - 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .
Show solution
Answer $3 \hat{j} + 4 \hat{k}$
$\hat{i}$ : $3 - 1 - 2 = 0$ . $\hat{j}$ : $1 + 5 - 3 = 3$ . $\hat{k}$ : $- 4 + 2 + 6 = 4$ .
So the sum is $3 \hat{j} + 4 \hat{k}$ , a vector of length 5.
Short answerEasyNCERT Exercise 10.2, Q7
Which unit vector points the same way as $\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ ?
Show solution
Answer $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$
$∣ \vec{a} ∣ = \sqrt{4 + 4 + 1} = 3$ , so $\hat{a} = \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}$ .
Short answerEasyNCERT Exercise 10.2, Q8
P and Q are the points (−1, 0, 2) and (1, 6, −1). Give the unit vector along $\vec{P Q}$ .
Show solution
Answer $\frac{1}{7} (2 \hat{i} + 6 \hat{j} - 3 \hat{k})$
$\vec{P Q} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $∣ \vec{P Q} ∣ = \sqrt{4 + 36 + 9} = 7$ .
The unit vector is $\frac{2}{7} \hat{i} + \frac{6}{7} \hat{j} - \frac{3}{7} \hat{k}$ .
Short answerMediumNCERT Exercise 10.2, Q9
For $\vec{a} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ and $\vec{b} = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ , give the unit vector along $\vec{a} + \vec{b}$ .
Show solution
Answer $\frac{1}{\sqrt{6}} (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{a} + \vec{b} = 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$ , of length $\sqrt{24} = 2 \sqrt{6}$ .
Dividing: $\frac{1}{2 \sqrt{6}} (2 \hat{i} + 4 \hat{j} + 2 \hat{k}) = \frac{1}{\sqrt{6}} (\hat{i} + 2 \hat{j} + \hat{k})$ .
Short answerEasyNCERT Exercise 10.2, Q10
Find a vector of magnitude 18 in the direction of $4 \hat{i} + 4 \hat{j} - 7 \hat{k}$ .
Show solution
Answer $8 \hat{i} + 8 \hat{j} - 14 \hat{k}$
$∣ 4 \hat{i} + 4 \hat{j} - 7 \hat{k} ∣ = \sqrt{16 + 16 + 49} = 9$ .
The required vector is $\frac{18}{9} (4 \hat{i} + 4 \hat{j} - 7 \hat{k}) = 8 \hat{i} + 8 \hat{j} - 14 \hat{k}$ .
Quick answerEasyNCERT Exercise 10.2, Q11
The vectors $- 3 \hat{i} + \hat{j} + 5 \hat{k}$ and $6 \hat{i} - 2 \hat{j} - 10 \hat{k}$ are collinear: the second is λ times the first. Find λ.
Show answer
Answer -2
$\frac{6}{- 3} = \frac{- 2}{1} = \frac{- 10}{5} = - 2$ . All three ratios agree, which is exactly the collinear test.
Short answerEasyNCERT Exercise 10.2, Q12
Find the direction cosines of $\hat{i} - 2 \hat{j} + 2 \hat{k}$ .
Show solution
Answer $\frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$
The length is $\sqrt{1 + 4 + 4} = 3$ , so $l, m, n = \frac{1}{3}, - \frac{2}{3}, \frac{2}{3}$ .
Short answerMediumNCERT Exercise 10.2, Q13
Find the direction cosines of the vector from A(3, −1, 4) to B(−3, 2, 2).
Show solution
Answer $- \frac{6}{7}, \frac{3}{7}, - \frac{2}{7}$
$\vec{A B} = - 6 \hat{i} + 3 \hat{j} - 2 \hat{k}$ , of length $\sqrt{36 + 9 + 4} = 7$ .
So $l, m, n = - \frac{6}{7}, \frac{3}{7}, - \frac{2}{7}$ .
14
Short answerMediumNCERT Exercise 10.2, Q14
Angles with the axes.
1. Show that $2 \hat{i} + 2 \hat{j} + 2 \hat{k}$ makes equal angles with OX, OY and OZ, and find that angle.
  Show solution
  Answer $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$
  The length is $\sqrt{12} = 2 \sqrt{3}$ , so each direction cosine is $\frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}$ .
  Equal cosines give equal angles, each $\cos^{- 1} \frac{1}{\sqrt{3}} \approx {54.7}^{\circ}$ .
2. Does $\hat{i} - \hat{j} + \hat{k}$ make equal angles with the positive directions of the three axes?
  Show solution
  AnswerNo
  Its direction cosines are $\frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ . The angle with OY is obtuse (about ${125.3}^{\circ}$ ), while the other two are about ${54.7}^{\circ}$ .
15
Short answerMediumNCERT Exercise 10.2, Q15
P and Q have position vectors $3 \hat{i} - \hat{j} + 2 \hat{k}$ and $- \hat{i} + 3 \hat{j} + 6 \hat{k}$ . Find the position vector of the point R that divides PQ in the ratio 3 : 1, (a) internally, (b) externally.
1. internally
  Show solution
  Answer $2 \hat{j} + 5 \hat{k}$
  $\vec{r} = \frac{3 \vec{q} + \vec{p}}{4} = \frac{1}{4} (0 \hat{i} + 8 \hat{j} + 20 \hat{k}) = 2 \hat{j} + 5 \hat{k}$ .
2. externally
  Show solution
  Answer $- 3 \hat{i} + 5 \hat{j} + 8 \hat{k}$
  $\vec{r} = \frac{3 \vec{q} - \vec{p}}{2} = \frac{1}{2} (- 6 \hat{i} + 10 \hat{j} + 16 \hat{k}) = - 3 \hat{i} + 5 \hat{j} + 8 \hat{k}$ .
Short answerEasyNCERT Exercise 10.2, Q16
Find the position vector of the midpoint of the segment joining P(−3, 5, 1) and Q(7, −1, 4).
Show solution
Answer $2 \hat{i} + 2 \hat{j} + \frac{5}{2} \hat{k}$
$\frac{1}{2} [(- 3 + 7) \hat{i} + (5 - 1) \hat{j} + (1 + 4) \hat{k}] = 2 \hat{i} + 2 \hat{j} + \frac{5}{2} \hat{k}$ .
ProofMediumNCERT Exercise 10.2, Q17
A, B, C have position vectors $4 \hat{i} + \hat{k}$ , $5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $9 \hat{i} + 2 \hat{j} + \hat{k}$ . Show that ABC is a right-angled triangle.
Show solution
AnswerRight-angled at B.
$\vec{A B} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , $\vec{B C} = 4 \hat{i} - 2 \hat{k}$ and $\vec{A C} = 5 \hat{i} + 2 \hat{j}$ .
$A B^{2} = 9$ , $B C^{2} = 20$ and $A C^{2} = 29$ .
Since $9 + 20 = 29$ , the angle at B (opposite AC) is a right angle.
Quick answerEasyNCERT Exercise 10.2, Q18
In triangle PQR, which statement is false?
$\vec{P Q} + \vec{Q R} = \vec{P R}$ $\vec{Q R} + \vec{R P} + \vec{P Q} = \vec{0}$ $\vec{P Q} - \vec{R Q} = \vec{P R}$ $\vec{P R} + \vec{R Q} = \vec{Q P}$
Show answer
Answer (D) $\vec{P R} + \vec{R Q} = \vec{Q P}$
P→R→Q ends at Q, so $\vec{P R} + \vec{R Q} = \vec{P Q}$ , not $\vec{Q P}$ . C is true because $- \vec{R Q} = \vec{Q R}$ .
19
Quick answerMediumNCERT Exercise 10.2, Q19
$\vec{a}$ and $\vec{b}$ are non-zero collinear vectors. Which statements must be true?
$\vec{b} = λ \vec{a}$ for some scalar λ $∣ \vec{a} ∣ = ∣ \vec{b} ∣$ their components are proportionalthey point the same way
Select all that apply
Show answer
Answer
(A) $\vec{b} = λ \vec{a}$ for some scalar λ
(C) their components are proportional
Collinearity gives $\vec{b} = λ \vec{a}$ , so the components are proportional. Nothing fixes λ: it may be negative (opposite directions) or have $∣ λ ∣ \neq 1$ (different lengths).

10.6 Product of Two Vectors

The work a force does on a moving cart depends only on the part of the force along the motion. The turning effect of a spanner depends on the part of the push across the handle. These two needs lead to two different ways of multiplying vectors: the scalar (or dot) product, whose value is a number, and the vector (or cross) product, whose value is a vector.

10.6.1 Scalar (or dot) product of two vectors

Definition

Scalar (dot) product

$\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ$ , where $0 \leq θ \leq π$ is the angle between the non-zero vectors $\vec{a}$ and $\vec{b}$ . When one of them is $\vec{0}$ there is no angle to speak of, and the product is defined to be 0.

Scalar product

\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ

(10.12)

The angle θ is measured between the two vectors drawn from a common point, so it never exceeds π. The magnitudes are positive, so the sign of $\vec{a} \cdot \vec{b}$ is the sign of cos θ (Fig. 10.15).

1/3cos θ > 0, so $\vec{a} \cdot \vec{b} > 0$ .

Fig. 10.15Acute:

\vec{a} \cdot \vec{b} > 0

. Right angle:

\vec{a} \cdot \vec{b} = 0

. Obtuse:

\vec{a} \cdot \vec{b} < 0

.NCERT Fig. 10.19 · p. 356

A blue vector a points right from O along a line that is dashed behind O. An orange vector b starts at O. In the first step b makes an acute angle with a; its shadow on the line of a lies along a and is shaded green with a plus sign, and the label says a dot b is greater than 0. In the second step b is at a right angle, marked with a square corner, there is no shadow, and the label says a dot b equals 0. In the third step the angle is obtuse; the shadow falls behind O, is hatched red with a minus sign, and the label says a dot b is less than 0.

Some consequences of the definition, worth knowing by heart:

$\vec{a} \cdot \vec{b}$ is a real number, not a vector.
For non-zero vectors, $\vec{a} \cdot \vec{b} = 0$ exactly when $\vec{a}$ and $\vec{b}$ are perpendicular.
If θ = 0, $\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣$ . In particular $\vec{a} \cdot \vec{a} = ∣ \vec{a} ∣^{2}$ .
If θ = π, $\vec{a} \cdot \vec{b} = - ∣ \vec{a} ∣ ∣ \vec{b} ∣$ . In particular $\vec{a} \cdot (- \vec{a}) = - ∣ \vec{a} ∣^{2}$ .
For the axis unit vectors, $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ , while $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ .
The order does not matter: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ , since the angle and its cosine are the same either way.

Rearranging the definition gives the angle between two non-zero vectors:

Angle between two vectors

\cos θ = \frac{\vec{a} \cdot \vec{b}}{∣ \vec{a} ∣ ∣ \vec{b} ∣}

(10.13)

PropertyTwo rules for the scalar product

(1) Distributivity: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$ . (2) Scalars move freely: $(λ \vec{a}) \cdot \vec{b} = λ (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (λ \vec{b})$ .

DerivationThe scalar product from components

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ . Expanding $\vec{a} \cdot \vec{b}$ with the two rules gives nine terms, each a number times a product such as $\hat{i} \cdot \hat{j}$ .

The six mixed products such as $\hat{i} \cdot \hat{j}$ are 0, and the three squares such as $\hat{i} \cdot \hat{i}$ are 1. Only three terms survive.

Scalar product in components

\begin{aligned} \vec{a} \cdot \vec{b} = a_{1} b_{1} \\ + a_{2} b_{2} + a_{3} b_{3} \end{aligned}

(10.14)

InteractiveVector product lab

A square grid from minus 5 to 5 on both axes with two vectors from the origin: a in blue, starting at 4i + j, and b in orange, starting at i + 3j. Their tips can be dragged to whole-number points or set with number boxes. Three buttons choose the view. Dot product shows the angle arc and a panel with a dot b worked out from components, cos theta, the angle in degrees and whether it is acute, right or obtuse. Projection draws the line of b, drops a dashed perpendicular from the tip of a and shows the projection vector in purple with its signed length. Cross product shades the parallelogram on a and b, shows the turn from a to b, and gives a cross b as a multiple of k, its size as the area, and whether it points out of or into the screen. Nothing moves until the learner acts.

Fig. 10.16Drag the tips of

\vec{a}

and

\vec{b}

. Dot product shows

\vec{a} \cdot \vec{b}

and the angle; Projection drops

\vec{a}

onto the line of

\vec{b}

; Cross product (for 10.6.3) shades the parallelogram whose area is

∣ \vec{a} \times \vec{b} ∣

10.6.2 Projection of a vector on a line

Hold a pencil $\vec{A B}$ above a ruler lying along a directed line l, and let light fall straight down onto the ruler. The shadow of the pencil is its projection on l. With θ the angle between $\vec{a} = \vec{A B}$ and l, the shadow is $∣ \vec{a} ∣ ∣ \cos θ ∣$ long. Whether it runs with l or against it is decided by the sign of cos θ (Fig. 10.17).

Definition

Projection on a line

Slide $\vec{a} = \vec{A B}$ so that A lies on the directed line l, and drop a perpendicular from B to l, meeting it at C. The vector $\vec{A C}$ is the projection vector of $\vec{a}$ on l, and the signed number $∣ \vec{a} ∣ \cos θ$ is the projection of $\vec{a}$ on l.

ExampleA vector of length 10 at $60^{\circ}$ to l has projection $10 \cos 60^{\circ} = 5$ .

3/3 $\vec{A C}$ is the projection vector; its signed length is $∣ \vec{a} ∣ \cos θ$ .

Fig. 10.17The projection of

\vec{a}

on the line of

\vec{b}

∣ \vec{a} ∣ \cos θ = \vec{a} \cdot \hat{b}

.NCERT Fig. 10.20 · p. 357

A directed line l runs left to right. From a point A on it, an orange vector b lies along the line and a blue vector a = AB rises at angle theta. In step two a dashed perpendicular drops from B to C on the line, with a right-angle mark. In step three a purple arrow just below the line runs from A to C: the projection vector, labelled a-length times cos theta, which equals a dot b-hat.

If $\hat{p}$ is the unit vector along l, the projection of $\vec{a}$ on l is $\vec{a} \cdot \hat{p}$ , because $∣ \hat{p} ∣ = 1$ .
The projection of $\vec{a}$ on another vector $\vec{b}$ uses $\hat{p} = \hat{b}$ , which gives (10.15). The projection vector is that number times $\hat{b}$ .
When θ = 0 or θ = π, $\vec{a}$ lies along l and is its own projection vector; the projection is $∣ \vec{a} ∣$ or $- ∣ \vec{a} ∣$ .
When θ = π/2, the projection vector is $\vec{0}$ .

Projection of a on b

\vec{a} \cdot \hat{b} = \frac{\vec{a} \cdot \vec{b}}{∣ \vec{b} ∣}

(10.15)

Example 13NCERT Example 13 · p. 358

Angle from the dot product

Vectors $\vec{a}$ and $\vec{b}$ have magnitudes 4 and 3, and $\vec{a} \cdot \vec{b} = - 6$ . Find the angle between them.

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 359

Angle from components

Find the angle between $\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = 6 \hat{i} - 3 \hat{j} + 2 \hat{k}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 15NCERT Example 15 · p. 359

Diagonals of a rhombus

For $\vec{a} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 2 \hat{k}$ , show that the diagonals $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ of the parallelogram on $\vec{a}$ and $\vec{b}$ are perpendicular.

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 359

A projection

Find the projection of $\vec{a} = 4 \hat{i} - \hat{j} + 3 \hat{k}$ on $\vec{b} = 2 \hat{i} + 2 \hat{j} - \hat{k}$ , and the projection vector.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 359

Length of a sum

If $∣ \vec{a} ∣ = 3$ , $∣ \vec{b} ∣ = 5$ and $\vec{a} \cdot \vec{b} = 7$ , find $∣ \vec{a} + \vec{b} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 360

An unknown length

If $∣ \vec{a} ∣ = 2$ and $(\vec{x} + 2 \vec{a}) \cdot (\vec{x} - 2 \vec{a}) = 9$ , find $∣ \vec{x} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 19NCERT Example 19 · p. 360

Cauchy–Schwarz inequality

Show that $∣ \vec{a} \cdot \vec{b} ∣ \leq ∣ \vec{a} ∣ ∣ \vec{b} ∣$ for any two vectors.

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 360

Triangle inequality

Show that $∣ \vec{a} + \vec{b} ∣ \leq ∣ \vec{a} ∣ + ∣ \vec{b} ∣$ for any two vectors.

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 361

Three points on a line

Show that A(1, −2, 3), B(3, 1, 2) and C(7, 7, 0) are collinear.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Find

(2 \hat{i} - \hat{j} + 3 \hat{k}) \cdot (\hat{i} + 4 \hat{j} + \hat{k})

Show answer

Answer 1

2 \cdot 1 + (- 1) \cdot 4 + 3 \cdot 1 = 2 - 4 + 3 = 1

Practice · 18 questions

Practice 10.3: questions with answers

Short answerEasyNCERT Exercise 10.3, Q1
Find the angle between two vectors of magnitudes 2 and $\sqrt{2}$ whose scalar product is −2.
Show solution
Answer $\frac{3 π}{4}$
$\cos θ = \frac{- 2}{2 \sqrt{2}} = - \frac{1}{\sqrt{2}}$ , so $θ = \frac{3 π}{4}$ .
Short answerEasyNCERT Exercise 10.3, Q2
Find the angle between $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} - \hat{j} + 2 \hat{k}$ .
Show solution
Answer $\cos^{- 1} (- \frac{1}{6}) \approx {99.6}^{\circ}$
Dot product: $2 - 1 - 2 = - 1$ . Both lengths are $\sqrt{6}$ .
$\cos θ = \frac{- 1}{6}$ , so $θ = \cos^{- 1} (- \frac{1}{6}) \approx {99.6}^{\circ}$ .
Quick answerEasyNCERT Exercise 10.3, Q3
Find the projection of $3 \hat{i} + \hat{j}$ on $\hat{i} - \hat{j}$ .
Show answer
Answer 1.4142
$\frac{3 - 1}{\sqrt{2}} = \sqrt{2} \approx 1.414$ .
Short answerEasyNCERT Exercise 10.3, Q4
Find the projection of $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ on $\hat{i} + 2 \hat{j} + 2 \hat{k}$ .
Show solution
Answer $\frac{8}{3}$
$\frac{2 - 6 + 12}{3} = \frac{8}{3}$ .
ProofMediumNCERT Exercise 10.3, Q5
Show that $\frac{1}{9} (\hat{i} + 4 \hat{j} + 8 \hat{k})$ , $\frac{1}{9} (4 \hat{i} + 7 \hat{j} - 4 \hat{k})$ and $\frac{1}{9} (8 \hat{i} - 4 \hat{j} + \hat{k})$ are unit vectors, and that each is perpendicular to the other two.
Show solution
AnswerAll three have length 1 and all three dot products are 0.
Lengths: $\frac{1}{9} \sqrt{1 + 16 + 64} = \frac{1}{9} \sqrt{16 + 49 + 16} = \frac{1}{9} \sqrt{64 + 16 + 1} = 1$ .
Dot products, ignoring the factor $\frac{1}{81}$ : $4 + 28 - 32 = 0$ , $32 - 28 - 4 = 0$ and $8 - 16 + 8 = 0$ .
Short answerMediumNCERT Exercise 10.3, Q6
If $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 27$ and $∣ \vec{a} ∣ = 2 ∣ \vec{b} ∣$ , find $∣ \vec{a} ∣$ and $∣ \vec{b} ∣$ .
Show solution
Answer $∣ \vec{a} ∣ = 6$ , $∣ \vec{b} ∣ = 3$
$∣ \vec{a} ∣^{2} - ∣ \vec{b} ∣^{2} = 27$ and $∣ \vec{a} ∣^{2} = 4 ∣ \vec{b} ∣^{2}$ , so $3 ∣ \vec{b} ∣^{2} = 27$ .
Hence $∣ \vec{b} ∣ = 3$ and $∣ \vec{a} ∣ = 6$ .
Short answerMediumNCERT Exercise 10.3, Q7
Expand $(2 \vec{a} + 3 \vec{b}) \cdot (4 \vec{a} - \vec{b})$ . Then evaluate it when $∣ \vec{a} ∣ = 1$ , $∣ \vec{b} ∣ = 2$ and $\vec{a} \cdot \vec{b} = 1$ .
Show solution
Answer $8 ∣ \vec{a} ∣^{2} + 10 \vec{a} \cdot \vec{b} - 3 ∣ \vec{b} ∣^{2} = 6$
$8 \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + 12 \vec{b} \cdot \vec{a} - 3 \vec{b} \cdot \vec{b} = 8 ∣ \vec{a} ∣^{2} + 10 \vec{a} \cdot \vec{b} - 3 ∣ \vec{b} ∣^{2}$ .
With the given values: $8 + 10 - 12 = 6$ .
Quick answerEasyNCERT Exercise 10.3, Q8
Two vectors have equal magnitudes, the angle between them is $120^{\circ}$ , and their scalar product is −8. Find their common magnitude.
Show answer
Answer 4
$∣ \vec{a} ∣^{2} \cdot (- \frac{1}{2}) = - 8$ , so $∣ \vec{a} ∣^{2} = 16$ and $∣ \vec{a} ∣ = 4$ .
Quick answerMediumNCERT Exercise 10.3, Q9
$\vec{a}$ is a unit vector and $(2 \vec{x} - \vec{a}) \cdot (2 \vec{x} + \vec{a}) = 15$ . Find $∣ \vec{x} ∣$ .
Show answer
Answer 2
$4 ∣ \vec{x} ∣^{2} - ∣ \vec{a} ∣^{2} = 15$ , so $4 ∣ \vec{x} ∣^{2} = 16$ and $∣ \vec{x} ∣ = 2$ .
Quick answerMediumNCERT Exercise 10.3, Q10
Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + 3 \hat{j} + \hat{k}$ . For which λ is $\vec{a} + λ \vec{b}$ perpendicular to $\vec{c}$ ?
Show answer
Answer 0.5
$\vec{a} + λ \vec{b} = (1 + 2 λ) \hat{i} + (λ - 2) \hat{j} + (3 - λ) \hat{k}$ . Dotting with $\vec{c}$ : $(1 + 2 λ) + 3 (λ - 2) + (3 - λ) = 4 λ - 2 = 0$ , so $λ = \frac{1}{2}$ .
ProofMediumNCERT Exercise 10.3, Q11
For non-zero vectors $\vec{a}$ and $\vec{b}$ , show that $∣ \vec{b} ∣^{2} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ is perpendicular to $\vec{b}$ .
Show solution
AnswerIts dot product with $\vec{b}$ is 0.
$(∣ \vec{b} ∣^{2} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot \vec{b} = ∣ \vec{b} ∣^{2} (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{b}) ∣ \vec{b} ∣^{2} = 0$ .
So the vector is perpendicular to $\vec{b}$ (or is $\vec{0}$ , which happens when $\vec{a}$ and $\vec{b}$ are collinear). Dividing by $∣ \vec{b} ∣^{2}$ shows it is $\vec{a}$ minus its projection vector on $\vec{b}$ .
Short answerEasyNCERT Exercise 10.3, Q12
Suppose $\vec{u} \cdot \vec{u} = 0$ . What is $\vec{u}$ ? What does $\vec{u} \cdot \vec{v} = 0$ then tell you about $\vec{v}$ ?
Show solution
Answer $\vec{u} = \vec{0}$ ; nothing about $\vec{v}$ .
$∣ \vec{u} ∣^{2} = 0$ forces $\vec{u} = \vec{0}$ .
Then $\vec{u} \cdot \vec{v} = 0$ for every $\vec{v}$ , so $\vec{v}$ can be any vector.
Quick answerMediumNCERT Exercise 10.3, Q13
$\vec{a}$ , $\vec{b}$ , $\vec{c}$ each have magnitude 2 and $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . Find $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ .
Show answer
Answer -6
$0 = ∣ \vec{a} + \vec{b} + \vec{c} ∣^{2} = 4 + 4 + 4 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ , so the sum is $- 6$ . (Each pair is at $120^{\circ}$ .)
Short answerEasyNCERT Exercise 10.3, Q14
Find two vectors, neither of them $\vec{0}$ , whose scalar product is 0. What does your example say about the claim that $\vec{a} \cdot \vec{b} = 0$ forces one of the vectors to be $\vec{0}$ ?
Show solution
Answer $\vec{a} = 2 \hat{i} - \hat{j}$ , $\vec{b} = \hat{i} + 2 \hat{j}$
$(2 \hat{i} - \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = 2 - 2 = 0$ , yet neither vector is zero: they are perpendicular.
Short answerMediumNCERT Exercise 10.3, Q15
A triangle has vertices A(3, 1, −1), B(1, 2, 1) and C(2, 4, 3). Find ∠ABC, the angle between $\vec{B A}$ and $\vec{B C}$ .
Show solution
Answer $\cos^{- 1} (- \frac{4}{9}) \approx {116.4}^{\circ}$
$\vec{B A} = 2 \hat{i} - \hat{j} - 2 \hat{k}$ and $\vec{B C} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , each of length 3.
Their dot product is $2 - 2 - 4 = - 4$ , so $\cos ∠ A B C = - \frac{4}{9}$ and $∠ A B C \approx {116.4}^{\circ}$ .
ProofEasyNCERT Exercise 10.3, Q16
Show that A(−1, 4, 2), B(0, 2, 4) and C(2, −2, 8) are collinear.
Show solution
Answer $∣ A B ∣ + ∣ B C ∣ = 3 + 6 = 9 = ∣ A C ∣$ .
$\vec{A B} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ (length 3), $\vec{B C} = 2 \hat{i} - 4 \hat{j} + 4 \hat{k}$ (length 6), $\vec{A C} = 3 \hat{i} - 6 \hat{j} + 6 \hat{k}$ (length 9).
Since 3 + 6 = 9, B lies on segment AC, so the points are collinear.
ProofMediumNCERT Exercise 10.3, Q17
Show that A(2, −1, 3), B(3, 1, 1) and C(4, 1, 6) form a right-angled triangle, using a scalar product.
Show solution
Answer $\vec{A B} \cdot \vec{A C} = 0$ : right angle at A.
$\vec{A B} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\vec{A C} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ .
$\vec{A B} \cdot \vec{A C} = 2 + 4 - 6 = 0$ , so the angle at A is a right angle. (Check: $A B^{2} + A C^{2} = 9 + 17 = 26 = B C^{2}$ .)
Quick answerEasyNCERT Exercise 10.3, Q18
$\vec{a}$ has magnitude 5. For which λ is $λ \vec{a}$ a unit vector?
λ = 5λ = ±1/5λ = 1/25λ = −5
Show answer
Answer (B) λ = ±1/5
$∣ λ ∣ \cdot 5 = 1$ gives $∣ λ ∣ = \frac{1}{5}$ ; both signs work, since a negative λ only reverses the direction.

10.6.3 Vector (or cross) product of two vectors

Recall what makes the coordinate axes right-handed. Curl the fingers of your right hand from the positive x-axis towards the positive y-axis, and your thumb points along the positive z-axis. Equally, a right-handed screw turned from x towards y moves along z.

Definition

Vector (cross) product

For non-zero vectors $\vec{a}$ , $\vec{b}$ with angle θ between them ( $0 \leq θ \leq π$ ), $\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ . Here $\hat{n}$ is the unit vector perpendicular to both for which $\vec{a}$ , $\vec{b}$ , $\hat{n}$ form a right-handed system. If either vector is $\vec{0}$ , we set $\vec{a} \times \vec{b} = \vec{0}$ .

Vector product

\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}

(10.16)

"Right-handed system" means: turning $\vec{a}$ towards $\vec{b}$ through θ drives a right-handed screw along $\hat{n}$ (Fig. 10.18).

2/3Turning $\vec{a}$ towards $\vec{b}$ : $\vec{a} \times \vec{b}$ points up the normal.

Fig. 10.18The turn from the first vector to the second decides which way

\hat{n}

points.NCERT Fig. 10.23 · p. 363

A tilted, shaded plane holds a blue vector a and an orange vector b starting at a common point O, with the angle theta between them. In step two a curved arrow turns from a towards b and a purple arrow n-hat rises straight up from O, labelled a cross b. In step three the curved arrow turns the other way, from b towards a, and a dashed red arrow points straight down from O through the plane, labelled b cross a.

$\vec{a} \times \vec{b}$ is a vector, perpendicular to both $\vec{a}$ and $\vec{b}$ .
For non-zero vectors, $\vec{a} \times \vec{b} = \vec{0}$ exactly when $\vec{a}$ and $\vec{b}$ are parallel (θ = 0 or π). In particular $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{a} \times (- \vec{a}) = \vec{0}$ .
If θ = π/2, then $∣ \vec{a} \times \vec{b} ∣ = ∣ \vec{a} ∣ ∣ \vec{b} ∣$ .
For the axis vectors, $\hat{i} \times \hat{j} = \hat{k}$ , and going on round the cycle, $\hat{j} \times \hat{k} = \hat{i}$ and then $\hat{k} \times \hat{i} = \hat{j}$ . Each of $\hat{i} \times \hat{i}$ , $\hat{j} \times \hat{j}$ and $\hat{k} \times \hat{k}$ is $\vec{0}$ .
The angle can be read off the length: $\sin θ = \frac{∣ \vec{a} \times \vec{b} ∣}{∣ \vec{a} ∣ ∣ \vec{b} ∣}$ .
Order matters: $\vec{b} \times \vec{a} = - (\vec{a} \times \vec{b})$ . Turning $\vec{b}$ towards $\vec{a}$ is the opposite turn, so the screw moves the other way.
So $\hat{j} \times \hat{i} = - \hat{k}$ , $\hat{k} \times \hat{j} = - \hat{i}$ and $\hat{i} \times \hat{k} = - \hat{j}$ (Fig. 10.19).

Fig. 10.19Going round the cycle

\hat{i} \to \hat{j} \to \hat{k}

gives +; going backwards gives −.NCERT Fig. 10.24 · p. 364

Three circles labelled i, j and k sit around a ring, with purple arrows going clockwise from i to j, j to k and k to i, and the words clockwise plus in the middle. On the right the forward products are listed: i cross j equals k, j cross k equals i, k cross i equals j. On the left the reversed products are listed with minus signs: j cross i equals minus k, k cross j equals minus i, i cross k equals minus j.

Areas from the cross product

The size $∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ$ has a picture. Lay $\vec{a}$ along the base of a parallelogram with $\vec{b}$ as its slanting side. The height is $∣ \vec{b} ∣ \sin θ$ , so the area is base times height (Fig. 10.20).

Area of a parallelogram

area = ∣ \vec{a} \times \vec{b} ∣

(10.17)

A diagonal cuts the parallelogram into two congruent triangles. So the triangle with sides $\vec{a}$ and $\vec{b}$ drawn from one vertex has half that area.

Area of a triangle

area = \frac{1}{2} ∣ \vec{a} \times \vec{b} ∣

(10.18)

2/3Its height is $∣ \vec{b} ∣ \sin θ$ , so its area is $∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ$ .

Fig. 10.20The area is base

∣ \vec{a} ∣

times height

∣ \vec{b} ∣ \sin θ

, which equals

∣ \vec{a} \times \vec{b} ∣

for the parallelogram and half of that for the triangle.NCERT Fig. 10.27 · p. 365

A shaded parallelogram OACB has a blue base vector a from O to A and an orange side vector b from O to B, with angle theta at O. In step two a dashed height drops from B to E on the base, with a right-angle mark, labelled length of b times sine theta. In step three the diagonal AB is drawn and the triangle OAB is shaded more darkly, labelled one half of the length of a cross b.

Computing the cross product

Property 3Distributivity of the vector product

For vectors $\vec{a}, \vec{b}, \vec{c}$ and a scalar λ: (i) $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ ; (ii) $λ (\vec{a} \times \vec{b}) = (λ \vec{a}) \times \vec{b} = \vec{a} \times (λ \vec{b})$ .

DerivationThe cross product from components

Expand $(a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k})$ term by term. The three terms with a repeated unit vector vanish. Each of the other six is plus or minus one of $\hat{i}, \hat{j}, \hat{k}$ , read off the cycle in Fig. 10.19:

\begin{aligned} \vec{a} \times \vec{b} \\ = a_{1} b_{2} \hat{k} - a_{1} b_{3} \hat{j} \\ - a_{2} b_{1} \hat{k} + a_{2} b_{3} \hat{i} \\ + a_{3} b_{1} \hat{j} - a_{3} b_{2} \hat{i} \end{aligned}

Grouping the terms by unit vector gives a pattern that is easiest to remember as a determinant, expanded along its first row:

Cross product in components

\begin{aligned} \vec{a} \times \vec{b} \\ = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array} ∣ \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} \\ - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \end{aligned}

(10.19)

The product lab (Fig. 10.16) has a Cross product mode for vectors in the xy-plane. There $\vec{a} \times \vec{b} = (a_{1} b_{2} - a_{2} b_{1}) \hat{k}$ : it points out of the screen when the turn from $\vec{a}$ to $\vec{b}$ is anticlockwise, and its length is the shaded area.

Example 22NCERT Example 22 · p. 366

Length of a cross product

Find $∣ \vec{a} \times \vec{b} ∣$ for $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3 \hat{j} - 2 \hat{k}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 23NCERT Example 23 · p. 367

A unit normal

Find a unit vector perpendicular to both $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 24NCERT Example 24 · p. 367

Area of a triangle

A triangle has corners A(1, 1, −1), B(3, 0, 1) and C(2, 3, −1). How large is its area?

Try it yourself first. Then reveal the solution one step at a time.

Example 25NCERT Example 25 · p. 368

Area of a parallelogram

Find the area of the parallelogram with adjacent sides $\vec{a} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 12 questions

Practice 10.4: questions with answers

Short answerEasyNCERT Exercise 10.4, Q1
Find $∣ \vec{a} \times \vec{b} ∣$ for $\vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$ .
Show solution
Answer $5 \sqrt{3}$
$\vec{a} \times \vec{b} = (6 - 1) \hat{i} - (4 + 1) \hat{j} + (- 2 - 3) \hat{k} = 5 \hat{i} - 5 \hat{j} - 5 \hat{k}$ .
Its length is $\sqrt{75} = 5 \sqrt{3}$ .
Short answerMediumNCERT Exercise 10.4, Q2
For $\vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ , find a unit vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ .
Show solution
Answer $\pm \frac{1}{\sqrt{42}} (4 \hat{i} - 5 \hat{j} - \hat{k})$
$\vec{a} + \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{a} - \vec{b} = \hat{i} + 4 \hat{k}$ .
Their cross product is $(8 - 0) \hat{i} - (12 - 2) \hat{j} + (0 - 2) \hat{k} = 8 \hat{i} - 10 \hat{j} - 2 \hat{k}$ , of length $\sqrt{168} = 2 \sqrt{42}$ .
So the unit vectors are $\pm \frac{1}{\sqrt{42}} (4 \hat{i} - 5 \hat{j} - \hat{k})$ . (Shortcut: $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = - 2 (\vec{a} \times \vec{b})$ .)
Short answerMediumNCERT Exercise 10.4, Q3
A unit vector $\vec{a}$ makes angle $\frac{2 π}{3}$ with $\hat{i}$ , angle $\frac{π}{3}$ with $\hat{j}$ and an acute angle θ with $\hat{k}$ . Find θ and the components of $\vec{a}$ .
Show solution
Answer $θ = \frac{π}{4}$ ; $\vec{a} = - \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$
$\cos^{2} θ = 1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}$ . As θ is acute, $\cos θ = \frac{1}{\sqrt{2}}$ and $θ = \frac{π}{4}$ .
A unit vector equals its direction cosines in component form, so $\vec{a} = - \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ .
ProofMediumNCERT Exercise 10.4, Q4
Show that $(\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) = - 5 (\vec{a} \times \vec{b})$ .
Show solution
AnswerExpand and simplify.
$(\vec{a} + 2 \vec{b}) \times (2 \vec{a} - \vec{b}) = 2 \vec{a} \times \vec{a} - \vec{a} \times \vec{b} + 4 \vec{b} \times \vec{a} - 2 \vec{b} \times \vec{b}$ .
The first and last terms are $\vec{0}$ , and $4 \vec{b} \times \vec{a} = - 4 \vec{a} \times \vec{b}$ . The total is $- 5 (\vec{a} \times \vec{b})$ .
Short answerMediumNCERT Exercise 10.4, Q5
Find λ and μ if $(3 \hat{i} - 6 \hat{j} + 9 \hat{k}) \times (2 \hat{i} + λ \hat{j} + μ \hat{k}) = \vec{0}$ .
Show solution
Answer $λ = - 4$ , $μ = 6$
The vectors must be parallel, so their components are proportional: $\frac{2}{3} = \frac{λ}{- 6} = \frac{μ}{9}$ .
Hence $λ = - 4$ and $μ = 6$ . (Indeed the second vector is $\frac{2}{3}$ of the first.)
Short answerEasyNCERT Exercise 10.4, Q6
Can two non-zero vectors have a zero scalar product and a zero vector product at the same time? What must be true when $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$ hold together?
Show solution
Answer $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$
If both were non-zero, the first condition would make them perpendicular (cos θ = 0) and the second parallel (sin θ = 0). No angle has both sine and cosine zero.
So at least one of the vectors is the zero vector.
Short answerMediumNCERT Exercise 10.4, Q7
Check the distributive law $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ for $\vec{a} = \hat{i} + 2 \hat{j}$ , $\vec{b} = 3 \hat{j} - \hat{k}$ and $\vec{c} = 2 \hat{i} + \hat{k}$ . Then explain why the same working proves it for all vectors.
Show solution
AnswerBoth sides equal $- \hat{k}$ .
$\vec{b} + \vec{c} = 2 \hat{i} + 3 \hat{j}$ , so $\vec{a} \times (\vec{b} + \vec{c}) = (1 \cdot 3 - 2 \cdot 2) \hat{k} = - \hat{k}$ .
$\vec{a} \times \vec{b} = - 2 \hat{i} + \hat{j} + 3 \hat{k}$ and $\vec{a} \times \vec{c} = 2 \hat{i} - \hat{j} - 4 \hat{k}$ ; their sum is $- \hat{k}$ .
In general, each component of (10.19) is linear in $b_{1}, b_{2}, b_{3}$ : for example $a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2}) = (a_{2} b_{3} - a_{3} b_{2}) + (a_{2} c_{3} - a_{3} c_{2})$ . So the law holds component by component.
Short answerEasyNCERT Exercise 10.4, Q8
Is it true that $\vec{a} \times \vec{b} = \vec{0}$ only when one of the vectors is $\vec{0}$ ? Settle it with an example.
Show solution
AnswerNo: e.g. $\hat{i} - 2 \hat{j} + 3 \hat{k}$ and $- 2 \hat{i} + 4 \hat{j} - 6 \hat{k}$ .
The second vector is $- 2$ times the first, so they are parallel and their cross product is $\vec{0}$ , although neither is $\vec{0}$ .
Short answerMediumNCERT Exercise 10.4, Q9
What is the area of triangle PQR, where P is (2, −1, 1), Q is (3, 1, 0) and R is (1, 0, 3)?
Show solution
Answer $\frac{\sqrt{35}}{2} \approx 2.96$
$\vec{P Q} = \hat{i} + 2 \hat{j} - \hat{k}$ and $\vec{P R} = - \hat{i} + \hat{j} + 2 \hat{k}$ .
$\vec{P Q} \times \vec{P R} = (4 + 1) \hat{i} - (2 - 1) \hat{j} + (1 + 2) \hat{k} = 5 \hat{i} - \hat{j} + 3 \hat{k}$ , of length $\sqrt{35}$ .
The area is $\frac{\sqrt{35}}{2}$ square units.
Short answerEasyNCERT Exercise 10.4, Q10
Find the area of the parallelogram with adjacent sides $3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\hat{i} - 3 \hat{j} + 4 \hat{k}$ .
Show solution
Answer $10 \sqrt{3} \approx 17.3$
The cross product is $(4 - 6) \hat{i} - (12 + 2) \hat{j} + (- 9 - 1) \hat{k} = - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$ .
Its length is $\sqrt{4 + 196 + 100} = \sqrt{300} = 10 \sqrt{3}$ square units.
Quick answerMediumNCERT Exercise 10.4, Q11
$∣ \vec{a} ∣ = 4$ and $∣ \vec{b} ∣ = \frac{1}{2}$ . If $\vec{a} \times \vec{b}$ is a unit vector and the angle θ between $\vec{a}$ and $\vec{b}$ is acute, then θ is
$\frac{π}{3}$ $\frac{π}{6}$ $\frac{π}{2}$ $\frac{π}{4}$
Show answer
Answer (B) $\frac{π}{6}$
$4 \cdot \frac{1}{2} \sin θ = 1$ gives $\sin θ = \frac{1}{2}$ , and the acute solution is $\frac{π}{6}$ .
Quick answerMediumNCERT Exercise 10.4, Q12
A rectangle has vertices A(−2, 1, 3), B(2, 1, 3), C(2, −1, 3) and D(−2, −1, 3). Its area is
164 $2 \sqrt{5}$ 8
Show answer
Answer (D) 8
$\vec{A B} = 4 \hat{i}$ and $\vec{A D} = - 2 \hat{j}$ , so $\vec{A B} \times \vec{A D} = - 8 \hat{k}$ and the area is 8. (Sides 4 and 2.)

Miscellaneous Examples

Example 26NCERT Example 26 · p. 369

Every unit vector in a plane

A robot cleaner on a flat floor (take it as the XY-plane) can face any direction. Write a formula for the unit vector along its heading, and show that the formula covers every possible heading.

Try it yourself first. Then reveal the solution one step at a time.

1/3 $\vec{r} = \cos θ \hat{i} + \sin θ \hat{j}$ .

Fig. 10.21Unit vectors in a plane have their heads on the unit circle.NCERT Fig. 10.28 · p. 369

Axes cross at O with a dashed circle of radius 1. In step one a purple unit vector from O to P makes angle theta with the x-axis; a blue arrow cos theta i along the x-axis and an orange arrow sin theta j upwards show its components, and P is labelled cos theta, sin theta. In step two eight blue unit vectors point in different directions, each ending on the circle. In step three the circle is drawn solid and labelled x squared plus y squared equals 1: the heads of all unit vectors in the plane.

Example 27NCERT Example 27 · p. 370

Angle that reveals collinearity

A, B, C, D have position vectors $2 \hat{i} + \hat{j} - \hat{k}$ , $4 \hat{i} + 2 \hat{k}$ , $3 \hat{j} + \hat{k}$ and $- 4 \hat{i} + 5 \hat{j} - 5 \hat{k}$ . Find the angle between $\vec{A B}$ and $\vec{C D}$ , and deduce that they are collinear.

Try it yourself first. Then reveal the solution one step at a time.

Example 28NCERT Example 28 · p. 370

Each perpendicular to the others’ sum

Vectors $\vec{a}, \vec{b}, \vec{c}$ have magnitudes 2, 3 and 6, and each of them is at right angles to the sum of the remaining two vectors. Find $∣ \vec{a} + \vec{b} + \vec{c} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 29NCERT Example 29 · p. 371

A sum of dot products

Vectors with $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ have $∣ \vec{a} ∣ = 5$ , $∣ \vec{b} ∣ = 12$ and $∣ \vec{c} ∣ = 13$ . Evaluate $μ = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 30NCERT Example 30 · p. 371

Splitting a vector along and across another

With $\vec{α} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{β} = 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , write $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$ where ${\vec{β}}_{1}$ is parallel to $\vec{α}$ and ${\vec{β}}_{2}$ is perpendicular to $\vec{α}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which is the unit vector in the XY-plane making 60° with the positive x-axis?

\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}

\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}

\hat{i} + \sqrt{3} \hat{j}

\frac{1}{2} \hat{i} + \frac{1}{2} \hat{j}

Show answer

Answer (A) $\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$

\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}

. Option B is the unit vector at 30°; C points the right way but has length 2; D is not a unit vector.

From historyNCERT Historical Note · p. 375

From the parallelogram of forces to vector analysis

Long before the word vector existed, engineers combined two pushes or pulls with the parallelogram rule. Simon Stevin, a Flemish engineer, used it for forces in his 1586 book on statics, De Beghinselen der Weeghconst ("The Principles of the Art of Weighing").

Around 1800, Caspar Wessel and Jean-Robert Argand pictured a complex number a + ib as an arrow in the plane, so adding complex numbers became adding arrows. William Rowan Hamilton wanted the same for arrows in space. In 1843 he invented quaternions, numbers of the form a + bi + cj + dk, and he gave us the word vector, Latin for "carrier".

In the 1880s Josiah Willard Gibbs in the United States and Oliver Heaviside in England, working separately, pulled the quaternion product apart into the two products of this chapter, the dot and the cross product. Gibbs printed his Elements of Vector Analysis for his students in 1881 and 1884, and their vector analysis is the language of physics and engineering today.

Recap

Vector Algebra: key points and summary

A scalar is one number; a vector has a magnitude and a direction (Section 10.1).
The position vector of P(x, y, z) is $\vec{O P} = x \hat{i} + y \hat{j} + z \hat{k}$ , of length $\sqrt{x^{2} + y^{2} + z^{2}}$ ((10.6)).
Direction cosines are $l = \frac{x}{r}$ , $m = \frac{y}{r}$ , $n = \frac{z}{r}$ with $l^{2} + m^{2} + n^{2} = 1$ ; any proportional triple is a set of direction ratios ((10.2)).
Zero, unit, coinitial, collinear, equal and negative vectors; vectors are free to slide (Section 10.3).
Vectors add head to tail (triangle law) or as the diagonal of a parallelogram; the sides of a triangle in order add to $\vec{0}$ (Section 10.4).
Addition is commutative and associative, with $\vec{0}$ as identity (Properties of vector addition).
$λ \vec{a}$ has length $∣ λ ∣ ∣ \vec{a} ∣$ and reverses direction when λ < 0; $\hat{a} = \vec{a} / ∣ \vec{a} ∣$ (Section 10.5).
In components, add, subtract, compare and scale entry by entry; collinear vectors have proportional components (Section 10.5.1).
The vector from $P_{1}$ to $P_{2}$ is head minus tail ((10.7)).
Section formula: $\frac{m \vec{b} + n \vec{a}}{m + n}$ internally, $\frac{m \vec{b} - n \vec{a}}{m - n}$ externally (Derivation).
$\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ ; zero for perpendicular vectors ((10.14)).
The projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{∣ \vec{b} ∣}$ ((10.15)).
$\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ , the determinant (10.19); its length is the parallelogram area and half of it the triangle area ((10.19)).

Miscellaneous Exercise on Chapter 10

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 10.1, Q1	Practice 10.1 · Q1
NCERT Exercise 10.1, Q2	Practice 10.1 · Q2
NCERT Exercise 10.1, Q3	Practice 10.1 · Q3
NCERT Exercise 10.1, Q4	Practice 10.1 · Q4
NCERT Exercise 10.1, Q5	Practice 10.1 · Q5
NCERT Exercise 10.2, Q1	Practice 10.2 · Q1
NCERT Exercise 10.2, Q2	Practice 10.2 · Q2
NCERT Exercise 10.2, Q3	Practice 10.2 · Q3
NCERT Exercise 10.2, Q4	Practice 10.2 · Q4
NCERT Exercise 10.2, Q5	Practice 10.2 · Q5
NCERT Exercise 10.2, Q6	Practice 10.2 · Q6
NCERT Exercise 10.2, Q7	Practice 10.2 · Q7
NCERT Exercise 10.2, Q8	Practice 10.2 · Q8
NCERT Exercise 10.2, Q9	Practice 10.2 · Q9
NCERT Exercise 10.2, Q10	Practice 10.2 · Q10
NCERT Exercise 10.2, Q11	Practice 10.2 · Q11
NCERT Exercise 10.2, Q12	Practice 10.2 · Q12
NCERT Exercise 10.2, Q13	Practice 10.2 · Q13
NCERT Exercise 10.2, Q14	Practice 10.2 · Q14
NCERT Exercise 10.2, Q15	Practice 10.2 · Q15
NCERT Exercise 10.2, Q16	Practice 10.2 · Q16
NCERT Exercise 10.2, Q17	Practice 10.2 · Q17
NCERT Exercise 10.2, Q18	Practice 10.2 · Q18
NCERT Exercise 10.2, Q19	Practice 10.2 · Q19
NCERT Exercise 10.3, Q1	Practice 10.3 · Q1
NCERT Exercise 10.3, Q2	Practice 10.3 · Q2
NCERT Exercise 10.3, Q3	Practice 10.3 · Q3
NCERT Exercise 10.3, Q4	Practice 10.3 · Q4
NCERT Exercise 10.3, Q5	Practice 10.3 · Q5
NCERT Exercise 10.3, Q6	Practice 10.3 · Q6
NCERT Exercise 10.3, Q7	Practice 10.3 · Q7
NCERT Exercise 10.3, Q8	Practice 10.3 · Q8
NCERT Exercise 10.3, Q9	Practice 10.3 · Q9
NCERT Exercise 10.3, Q10	Practice 10.3 · Q10
NCERT Exercise 10.3, Q11	Practice 10.3 · Q11
NCERT Exercise 10.3, Q12	Practice 10.3 · Q12
NCERT Exercise 10.3, Q13	Practice 10.3 · Q13
NCERT Exercise 10.3, Q14	Practice 10.3 · Q14
NCERT Exercise 10.3, Q15	Practice 10.3 · Q15
NCERT Exercise 10.3, Q16	Practice 10.3 · Q16
NCERT Exercise 10.3, Q17	Practice 10.3 · Q17
NCERT Exercise 10.3, Q18	Practice 10.3 · Q18
NCERT Exercise 10.4, Q1	Practice 10.4 · Q1
NCERT Exercise 10.4, Q2	Practice 10.4 · Q2
NCERT Exercise 10.4, Q3	Practice 10.4 · Q3
NCERT Exercise 10.4, Q4	Practice 10.4 · Q4
NCERT Exercise 10.4, Q5	Practice 10.4 · Q5
NCERT Exercise 10.4, Q6	Practice 10.4 · Q6
NCERT Exercise 10.4, Q7	Practice 10.4 · Q7
NCERT Exercise 10.4, Q8	Practice 10.4 · Q8
NCERT Exercise 10.4, Q9	Practice 10.4 · Q9
NCERT Exercise 10.4, Q10	Practice 10.4 · Q10
NCERT Exercise 10.4, Q11	Practice 10.4 · Q11
NCERT Exercise 10.4, Q12	Practice 10.4 · Q12
NCERT Miscellaneous Exercise 10, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise 10, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise 10, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise 10, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise 10, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise 10, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise 10, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise 10, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise 10, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise 10, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise 10, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise 10, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise 10, Q13	Miscellaneous exercise · Q13
NCERT Miscellaneous Exercise 10, Q14	Miscellaneous exercise · Q14
NCERT Miscellaneous Exercise 10, Q15	Miscellaneous exercise · Q15
NCERT Miscellaneous Exercise 10, Q16	Miscellaneous exercise · Q16
NCERT Miscellaneous Exercise 10, Q17	Miscellaneous exercise · Q17
NCERT Miscellaneous Exercise 10, Q18	Miscellaneous exercise · Q18
NCERT Miscellaneous Exercise 10, Q19	Miscellaneous exercise · Q19

Quick check: MCQs on Vector Algebra

10.1 Introduction2 questions
Direction cosines3 questions
10.3 Types of Vectors3 questions
Properties of vector addition4 questions
10.5.1 Components of a vector3 questions
10.5.3 Section formula4 questions
10.6.2 Projection of a vector on a line4 questions
Computing the cross product4 questions
Miscellaneous Examples3 questions

Glossary · 16

Key terms (English–Hindi)

coinitial vectorsसह-आदिम सदिशsah-aadim sadish: Vectors with the same initial point.First seen in the chapter →
collinear vectorsसंरेख सदिशsanrekh sadish: Vectors parallel to the same line; one is a scalar multiple of the other.First seen in the chapter →

direction cosinesदिक्-कोसाइनdik-cosine: The cosines $l, m, n$ of the angles a vector makes with the positive axes; $l^{2} + m^{2} + n^{2} = 1$ .First seen in the chapter →
direction ratiosदिक्-अनुपातdik-anupaat: Any three numbers proportional to the direction cosines of a vector.First seen in the chapter →

equal vectorsसमान सदिशsamaan sadish: Vectors with the same magnitude and direction, wherever they start.First seen in the chapter →

magnitudeपरिमाणparimaan: The length of a vector, written $∣ \vec{a} ∣$ ; never negative.First seen in the chapter →

position vectorस्थिति सदिशsthiti sadish: The vector $\vec{O P}$ from the origin to a point P.First seen in the chapter →
projectionप्रक्षेपprakshep: The signed length $∣ \vec{a} ∣ \cos θ$ of the shadow of $\vec{a}$ on a directed line.

resultantपरिणामीparinaami: The sum of two or more vectors.First seen in the chapter →

scalarअदिशadish: A quantity described completely by one real number, such as mass or temperature.First seen in the chapter →
scalar componentsअदिश घटकadish ghatak: The numbers $x, y, z$ in $x \hat{i} + y \hat{j} + z \hat{k}$ .First seen in the chapter →
scalar productअदिश गुणनफलadish gunanphal: $\vec{a} \cdot \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \cos θ$ , a number.

unit vectorमात्रक सदिशmaatrak sadish: A vector of magnitude 1; $\hat{a} = \vec{a} / ∣ \vec{a} ∣$ .First seen in the chapter →

vectorसदिशsadish: A quantity with both magnitude and direction, drawn as a directed line segment.First seen in the chapter →
vector productसदिश गुणनफलsadish gunanphal: $\vec{a} \times \vec{b} = ∣ \vec{a} ∣ ∣ \vec{b} ∣ \sin θ \hat{n}$ , a vector perpendicular to both.

zero vectorशून्य सदिशshunya sadish: The vector $\vec{0}$ of length 0, whose initial and terminal points coincide.First seen in the chapter →

Vector Algebra: Class 12 Maths · Chapter 10

What you will learn in Vector Algebra

10.1 Introduction

10.2 Some Basic Concepts

Position vector

Direction cosines

10.3 Types of Vectors

Practice 10.1: questions with answers

10.4 Addition of Vectors

The parallelogram law

Properties of vector addition

10.5 Multiplication of a Vector by a Scalar

10.5.1 Components of a vector

10.5.2 Vector joining two points

10.5.3 Section formula

Practice 10.2: questions with answers

10.6 Product of Two Vectors

10.6.1 Scalar (or dot) product of two vectors

10.6.2 Projection of a vector on a line

Practice 10.3: questions with answers

10.6.3 Vector (or cross) product of two vectors

Areas from the cross product

Computing the cross product

Practice 10.4: questions with answers

Miscellaneous Examples

Vector Algebra: key points and summary

Miscellaneous Exercise on Chapter 10

Practice questions with answers (NCERT “Miscellaneous exercise”)

Find your NCERT exercise question

Quick check: MCQs on Vector Algebra

Key terms (English–Hindi)

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