NCERT Class 12 Mathematics — Vector Algebra, Exercise 10.2. All 19 questions solved.

Exercise 10.2 is the working exercise of the chapter — the one where the definitions from 10.1 turn into routine calculations you will use for the rest of the chapter and throughout Three Dimensional Geometry. Four ideas cover almost all of it.

For a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$:

• Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
• Unit vector: $\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}$
• Direction cosines: $l = \dfrac{a_1}{|\vec{a}|}$, $m = \dfrac{a_2}{|\vec{a}|}$, $n = \dfrac{a_3}{|\vec{a}|}$ — which are just the components of $\hat{a}$
• Vector joining two points: $\overrightarrow{\mathrm{PQ}} =$ (position vector of Q) $-$ (position vector of P)

Key insight. Almost every question here reduces to “find the magnitude, then divide by it”. A unit vector, a vector of specified magnitude, and the direction cosines are all the same computation with a different final scaling. Get comfortable dividing by $|\vec{a}|$ and most of this exercise disappears.

The one genuinely new tool is the section formula in questions 15 and 16, and the sign difference between its internal and external forms is where marks are lost.

Question 1

Compute the magnitude of the following vectors:

$$\vec{a} = \hat{i} + \hat{j} + \hat{k}; \quad \vec{b} = 2\hat{i} – 7\hat{j} – 3\hat{k}; \quad \vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} – \frac{1}{\sqrt{3}}\hat{k}$$

Solution. Apply $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ in each case. The squares kill every minus sign, which is why $\vec{b}$’s negative components cause no trouble.

$$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$

$$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$$

$$|\vec{c}| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$$

That $|\vec{c}| = 1$ is not a coincidence — $\vec{c}$ is already a unit vector, which is worth recognising on sight.

Question 2

Write two different vectors having same magnitude.

Solution. Two vectors are equal only if all three components match, so any two vectors with the same components in a different arrangement will do.

$$\vec{a} = 2\hat{i} + 3\hat{j} – \hat{k}, \qquad \vec{b} = 3\hat{i} – \hat{j} + 2\hat{k}$$

Both have magnitude $\sqrt{4 + 9 + 1} = \sqrt{14}$, but they point in different directions, so $\vec{a} \neq \vec{b}$.

Question 3

Write two different vectors having same direction.

Solution. Two vectors share a direction when one is a positive scalar multiple of the other. Scaling changes the length but not the way it points.

$$\vec{a} = \hat{i} + \hat{j} + \hat{k}, \qquad \vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

Both have unit vector $\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$, which is what “same direction” means precisely. Note the multiplier must be positive — $-\vec{a}$ points the opposite way.

Question 4

Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.

Solution. Two vectors are equal exactly when their corresponding components are equal — there is nothing more to it than matching coefficients.

$$x = 2, \qquad y = 3$$

Question 5

Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point $(-5, 7)$.

Solution. The vector from P to Q is always the terminal position vector minus the initial one, in that order.

$$\overrightarrow{\mathrm{PQ}} = (-5 – 2)\hat{i} + (7 – 1)\hat{j} = -7\hat{i} + 6\hat{j}$$

The scalar components are the plain numbers $-7$ and $6$. The vector components are those numbers attached to their unit vectors, $-7\hat{i}$ and $6\hat{j}$ — the two vectors which add to give $\overrightarrow{\mathrm{PQ}}$.

Question 6

Find the sum of the vectors $\vec{a} = \hat{i} – 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$.

Solution. Add component by component.

$$\hat{i}: \quad 1 – 2 + 1 = 0$$ $$\hat{j}: \quad -2 + 4 – 6 = -4$$ $$\hat{k}: \quad 1 + 5 – 7 = -1$$

The $\hat{i}$ components cancel exactly, so that term vanishes from the answer rather than appearing as $0\hat{i}$.

Question 7

Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Solution. Find the magnitude, then divide the vector by it.

$$|\vec{a}| = \sqrt{1 + 1 + 4} = \sqrt{6}$$

$$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$$

A quick check worth doing: the squares of the coefficients are $\tfrac16 + \tfrac16 + \tfrac46 = 1$, as they must be for a unit vector.

Question 8

Find the unit vector in the direction of vector $\overrightarrow{\mathrm{PQ}}$, where P and Q are the points $(1, 2, 3)$ and $(4, 5, 6)$ respectively.

Solution. Build the vector first, then normalise it.

$$\overrightarrow{\mathrm{PQ}} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$

$$|\overrightarrow{\mathrm{PQ}}| = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$$

$$\widehat{\mathrm{PQ}} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$$

The factor of $3$ cancels, which it had to — scaling a vector never changes its unit vector.

Question 9

For given vectors $\vec{a} = 2\hat{i} – \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} – \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.

Solution. Add first, normalise second — normalising each vector separately and then adding gives a different, wrong answer.

$$\vec{a} + \vec{b} = (2-1)\hat{i} + (-1+1)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{k}$$

$$|\vec{a} + \vec{b}| = \sqrt{1 + 0 + 1} = \sqrt{2}$$

$$\frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$$

Question 10

Find a vector in the direction of vector $5\hat{i} – \hat{j} + 2\hat{k}$ which has magnitude 8 units.

Solution. The recipe for “direction of $\vec{v}$, magnitude $k$” is: take the unit vector, then multiply by $k$. The unit vector fixes the direction and the multiplier fixes the length.

$$|\vec{v}| = \sqrt{25 + 1 + 4} = \sqrt{30}$$

$$8\hat{v} = \frac{8}{\sqrt{30}}\left(5\hat{i} – \hat{j} + 2\hat{k}\right)$$

$$= \frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$$

Question 11

Show that the vectors $2\hat{i} – 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} – 8\hat{k}$ are collinear.

Solution. Two vectors are collinear exactly when one is a scalar multiple of the other. Compare components in order and look for a common ratio:

$$\frac{-4}{2} = -2, \qquad \frac{6}{-3} = -2, \qquad \frac{-8}{4} = -2$$

All three ratios agree, so $\vec{b} = -2\vec{a}$ and the vectors are collinear. The scalar is negative, which means they point in opposite directions along the same line — still collinear, since collinearity is about the line, not the sense.

Question 12

Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.

Solution. The direction cosines are the components divided by the magnitude — that is, the components of the unit vector.

$$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$$

$$l = \frac{1}{\sqrt{14}}, \qquad m = \frac{2}{\sqrt{14}}, \qquad n = \frac{3}{\sqrt{14}}$$

Check: $l^2 + m^2 + n^2 = \tfrac{1 + 4 + 9}{14} = 1$, which always holds and is worth using to catch arithmetic slips.

Question 13

Find the direction cosines of the vector joining the points $\mathrm{A}(1, 2, -3)$ and $\mathrm{B}(-1, -2, 1)$, directed from A to B.

Solution. “Directed from A to B” fixes the order of the subtraction — get it backwards and every direction cosine changes sign.

$$\overrightarrow{\mathrm{AB}} = (-1-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} – 4\hat{j} + 4\hat{k}$$

$$|\overrightarrow{\mathrm{AB}}| = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

$$l = \frac{-2}{6} = -\frac{1}{3}, \qquad m = \frac{-4}{6} = -\frac{2}{3}, \qquad n = \frac{4}{6} = \frac{2}{3}$$

Question 14

Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.

Solution. The angle a vector makes with an axis is read off directly from its direction cosines: $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$, where $\alpha, \beta, \gamma$ are the angles with OX, OY and OZ.

$$|\vec{a}| = \sqrt{1 + 1 + 1} = \sqrt{3}$$

$$\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}}$$

Since all three cosines are equal, all three angles are equal — each is $\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)$ — so the vector is equally inclined to the three axes.

Question 15

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} – \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio $2 : 1$ (i) internally (ii) externally.

Solution. With $\vec{p}$ and $\vec{q}$ the position vectors and ratio $m : n$, the two section formulae are

$$\text{internal: } \frac{m\vec{q} + n\vec{p}}{m + n}, \qquad \text{external: } \frac{m\vec{q} – n\vec{p}}{m – n}$$

They differ only in two signs, and the external one divides by $m – n$, which is why $m = n$ has no external point — parallel lines never meet.

Here $m = 2$, $n = 1$, $\vec{p} = \hat{i} + 2\hat{j} – \hat{k}$ and $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$.

(i) Internally

$$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} – \hat{k})}{2 + 1}$$

$$= \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} – \hat{k})}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3}$$

(ii) Externally

$$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) – 1(\hat{i} + 2\hat{j} – \hat{k})}{2 – 1}$$

$$= (-2\hat{i} + 2\hat{j} + 2\hat{k}) – (\hat{i} + 2\hat{j} – \hat{k}) = -3\hat{i} + 0\hat{j} + 3\hat{k}$$

Question 16

Find the position vector of the mid point of the vector joining the points $\mathrm{P}(2, 3, 4)$ and $\mathrm{Q}(4, 1, -2)$.

Solution. The midpoint is the internal section formula with ratio $1 : 1$, which simplifies to the average of the two position vectors.

$$\vec{r} = \frac{\vec{p} + \vec{q}}{2} = \frac{(2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k}}{2}$$

$$= \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$$

Question 17

Show that the points A, B and C with position vectors $\vec{a} = 3\hat{i} – 4\hat{j} – 4\hat{k}$, $\vec{b} = 2\hat{i} – \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} – 3\hat{j} – 5\hat{k}$ respectively, form the vertices of a right angled triangle.

Solution. Without the dot product (which arrives in the next section), the route is Pythagoras: find the three side lengths and check whether the squares of two of them add to the square of the third.

$$\overrightarrow{\mathrm{AB}} = \vec{b} – \vec{a} = -\hat{i} + 3\hat{j} + 5\hat{k}, \qquad |\overrightarrow{\mathrm{AB}}|^2 = 1 + 9 + 25 = 35$$

$$\overrightarrow{\mathrm{BC}} = \vec{c} – \vec{b} = -\hat{i} – 2\hat{j} – 6\hat{k}, \qquad |\overrightarrow{\mathrm{BC}}|^2 = 1 + 4 + 36 = 41$$

$$\overrightarrow{\mathrm{CA}} = \vec{a} – \vec{c} = 2\hat{i} – \hat{j} + \hat{k}, \qquad |\overrightarrow{\mathrm{CA}}|^2 = 4 + 1 + 1 = 6$$

Now test the three combinations. The two smaller squares are $35$ and $6$:

$$|\overrightarrow{\mathrm{AB}}|^2 + |\overrightarrow{\mathrm{CA}}|^2 = 35 + 6 = 41 = |\overrightarrow{\mathrm{BC}}|^2$$

The converse of Pythagoras’ theorem then gives a right angle, and it sits at A — the vertex shared by the two shorter sides AB and CA, opposite the longest side BC.

Working with the squares throughout is deliberate: the square roots would be irrational and would obscure the exact relationship.

Question 18

In triangle ABC, which of the following is not true:

A note on the printing. In the current reprint of the textbook, options (B) and (C) are both set as $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{AC}} = \vec{0}$, which is a typesetting error — two identical options cannot have one correct answer between them. Option (C) is meant to read $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{CA}} = \vec{0}$, and that is how it is written above. The answer key gives (C).

Solution. Everything follows from the triangle law: travelling A to B and then B to C is the same as travelling A to C, so

$$\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}$$

and from reversing a vector, $\overrightarrow{\mathrm{CA}} = -\overrightarrow{\mathrm{AC}}$.

(A) is true. Going right round the triangle returns you to the start: $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} + \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}} – \overrightarrow{\mathrm{AC}} = \vec{0}$.

(B) is true. It is the triangle law rearranged: $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{AC}} = \overrightarrow{\mathrm{AC}} – \overrightarrow{\mathrm{AC}} = \vec{0}$.

(C) is not true. Substituting gives

$$\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} – \overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}} – (-\overrightarrow{\mathrm{AC}}) = 2\overrightarrow{\mathrm{AC}}$$

which is $\vec{0}$ only if A and C coincide — impossible in a genuine triangle.

(D) is true. Since $-\overrightarrow{\mathrm{CB}} = \overrightarrow{\mathrm{BC}}$, option (D) is option (A) in disguise.

Question 19

If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:

Solution. Note that this question asks which statements are incorrect, and that more than one option qualifies — a departure from the usual single-answer format.

(A) is correct. This is the definition of collinearity: $\vec{b} = \lambda\vec{a}$ for some scalar $\lambda$.

(B) is incorrect. It forces $\lambda = \pm 1$, so the two vectors would have to have equal magnitude. But $\hat{i}$ and $5\hat{i}$ are collinear with different magnitudes.

(C) is incorrect. It states the exact opposite of the truth. If $\vec{b} = \lambda\vec{a}$ then every component of $\vec{b}$ is $\lambda$ times the corresponding component of $\vec{a}$, so the components are proportional.

(D) is incorrect. It is too restrictive in both halves. Collinear vectors may point in opposite directions when $\lambda < 0$ (as in question 11, where $\lambda = -2$), and they may have the same magnitude when $\lambda = \pm 1$.

Common mistakes

• Subtracting the position vectors the wrong way round. $\overrightarrow{\mathrm{PQ}} = \vec{q} – \vec{p}$, terminal minus initial. In question 13 the reversed order flips the sign of all three direction cosines, and the answer is then wrong even though the magnitude is right.
• Normalising before adding. In question 9 the unit vector of $\vec{a} + \vec{b}$ is not $\hat{a} + \hat{b}$. Combine the vectors first, then divide by the magnitude of the result.
• Confusing scalar components with vector components. Question 5 asks for both: $-7$ and $6$ are the scalar components, $-7\hat{i}$ and $6\hat{j}$ the vector ones. Giving the same thing twice loses a mark.
• Mixing up the two section formulae. Internal divides by $m + n$, external by $m – n$, and the external form subtracts the second term. Question 15 asks for both precisely so the difference is exercised.
• Thinking collinear means “same direction”. A negative scalar multiple is still collinear, as questions 11 and 19(D) both show.
• Assuming a multiple-choice question has exactly one answer. Question 19 has three, and stopping at the first incorrect statement found costs two marks.
• Taking square roots too early in question 17. Comparing $\sqrt{35}$, $\sqrt{41}$ and $\sqrt{6}$ as decimals invites rounding error. Work with the squared lengths, where the relationship is exact.

Practise next

• Exercise 10.3 — the scalar (dot) product, which turns question 17’s Pythagoras argument into a single line and gives a direct way to find angles between vectors.
• Exercise 10.4 — the vector (cross) product, areas and perpendicularity.