NCERT Class 12 Mathematics — Vector Algebra, Miscellaneous Exercise on Chapter 10. All 19 questions solved.

The Miscellaneous Exercise draws on the whole chapter at once, so a question may need the section formula from Exercise 10.2, the dot product from 10.3 and the cross product from 10.4 in the same working. Nothing new is introduced; what is tested is whether you can tell which tool a question is asking for.

A quick guide to that choice, since it is most of the difficulty here:

• Asking for an angle, a projection, or perpendicularity → dot product.
• Asking for an area, or a vector perpendicular to two others → cross product.
• Asking for a point dividing a line → section formula.
• Asking for collinearity → either show one vector is a multiple of another, or show the three distances add up.

Key insight. When a question asks for “a vector with property X and magnitude/condition Y”, the reliable method is two-step: first find any vector with property X, then scale it by whatever λ makes condition Y hold. Questions 6, 7 and 12 are all this pattern, and question 12 is much harder if you try to do both at once.

Question 1

Write down a unit vector in XY-plane, making an angle of $30°$ with the positive direction of $x$-axis.

Solution. A unit vector in the XY-plane at angle $\theta$ to the $x$-axis is $\cos\theta\,\hat{i} + \sin\theta\,\hat{j}$ — the components are the cosine and sine, which is what makes it a unit vector.

$$\vec{r} = \cos 30°\,\hat{i} + \sin 30°\,\hat{j} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$$

Check: $\tfrac34 + \tfrac14 = 1$.

Question 2

Find the scalar components and magnitude of the vector joining the points $\mathrm{P}(x_1, y_1, z_1)$ and $\mathrm{Q}(x_2, y_2, z_2)$.

Solution. Terminal minus initial, as always.

$$\overrightarrow{\mathrm{PQ}} = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} + (z_2 – z_1)\hat{k}$$

The scalar components are the three coefficients, and the magnitude is the three-dimensional distance formula — which is exactly what this is.

Question 3

A girl walks 4 km towards west, then she walks 3 km in a direction $30°$ east of north and stops. Determine the girl’s displacement from her initial point of departure.

Solution. Displacement is the single vector from start to finish, so add the two legs. Take $\hat{i}$ east and $\hat{j}$ north.

The first leg is 4 km west, which is the negative $\hat{i}$ direction:

$$\vec{u} = -4\hat{i}$$

The second is 3 km at $30°$ east of north — measured from north, turning east. So its north component uses cosine and its east component uses sine:

$$\vec{v} = 3\left(\sin 30°\,\hat{i} + \cos 30°\,\hat{j}\right) = \frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$$

$$\vec{u} + \vec{v} = \left(-4 + \frac{3}{2}\right)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j} = -\frac{5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$$

Question 4

If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.

Solution. No. The triangle inequality says $|\vec{b} + \vec{c}| \leq |\vec{b}| + |\vec{c}|$, with equality **only** when $\vec{b}$ and $\vec{c}$ point in the same direction.

The cleanest justification is a picture in words: let $\vec{b}$ and $\vec{c}$ be two sides of a triangle taken in order, so that $\vec{a} = \vec{b} + \vec{c}$ is the third side. Then $|\vec{a}| = |\vec{b}| + |\vec{c}|$ would say that one side of a triangle equals the sum of the other two, which is impossible unless the triangle is degenerate.

Concretely, take $\vec{b} = 3\hat{i}$ and $\vec{c} = 4\hat{j}$. Then $\vec{a} = 3\hat{i} + 4\hat{j}$ and $|\vec{a}| = 5$, while $|\vec{b}| + |\vec{c}| = 7$.

Question 5

Find the value of $x$ for which $x(\hat{i} + \hat{j} + \hat{k})$ is a unit vector.

Solution. The magnitude must equal 1, and $|x|$ comes out of the magnitude, not $x$ — that is where the second root comes from.

$$|x(\hat{i} + \hat{j} + \hat{k})| = |x|\sqrt{1 + 1 + 1} = |x|\sqrt{3} = 1$$

$$|x| = \frac{1}{\sqrt{3}} \quad\Rightarrow\quad x = \pm\frac{1}{\sqrt{3}}$$

Question 6

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a} = 2\hat{i} + 3\hat{j} – \hat{k}$ and $\vec{b} = \hat{i} – 2\hat{j} + \hat{k}$.

Solution. The two-step pattern: find the direction, then fix the magnitude.

$$\vec{a} + \vec{b} = 3\hat{i} + \hat{j} + 0\hat{k}, \qquad |\vec{a} + \vec{b}| = \sqrt{9 + 1} = \sqrt{10}$$

The unit vector in that direction is $\dfrac{3\hat{i} + \hat{j}}{\sqrt{10}}$, and multiplying by 5 gives the required vector:

$$\frac{5}{\sqrt{10}}\left(3\hat{i} + \hat{j}\right) = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j}$$

Rationalising, $\dfrac{15}{\sqrt{10}} = \dfrac{15\sqrt{10}}{10} = \dfrac{3\sqrt{10}}{2}$ and $\dfrac{5}{\sqrt{10}} = \dfrac{\sqrt{10}}{2}$:

Question 7

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} – \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} – 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} – \vec{b} + 3\vec{c}$.

Solution. Build the combination component by component, watching the signs on $-\vec{b}$.

$$2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}, \quad -\vec{b} = -2\hat{i} + \hat{j} – 3\hat{k}, \quad 3\vec{c} = 3\hat{i} – 6\hat{j} + 3\hat{k}$$

$$2\vec{a} – \vec{b} + 3\vec{c} = (2 – 2 + 3)\hat{i} + (2 + 1 – 6)\hat{j} + (2 – 3 + 3)\hat{k} = 3\hat{i} – 3\hat{j} + 2\hat{k}$$

$$\left|3\hat{i} – 3\hat{j} + 2\hat{k}\right| = \sqrt{9 + 9 + 4} = \sqrt{22}$$

Question 8

Show that the points $\mathrm{A}(1, -2, -8)$, $\mathrm{B}(5, 0, -2)$ and $\mathrm{C}(11, 3, 7)$ are collinear, and find the ratio in which B divides AC.

Solution. Form the three vectors and their magnitudes.

$$\overrightarrow{\mathrm{AB}} = 4\hat{i} + 2\hat{j} + 6\hat{k}, \qquad |\overrightarrow{\mathrm{AB}}| = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}$$

$$\overrightarrow{\mathrm{BC}} = 6\hat{i} + 3\hat{j} + 9\hat{k}, \qquad |\overrightarrow{\mathrm{BC}}| = \sqrt{36 + 9 + 81} = \sqrt{126} = 3\sqrt{14}$$

$$\overrightarrow{\mathrm{AC}} = 10\hat{i} + 5\hat{j} + 15\hat{k}, \qquad |\overrightarrow{\mathrm{AC}}| = \sqrt{100 + 25 + 225} = \sqrt{350} = 5\sqrt{14}$$

Since $2\sqrt{14} + 3\sqrt{14} = 5\sqrt{14}$, we have $|\overrightarrow{\mathrm{AB}}| + |\overrightarrow{\mathrm{BC}}| = |\overrightarrow{\mathrm{AC}}|$, so the points are collinear.

B lies between A and C, and divides AC in the ratio of the two parts:

$$\mathrm{AB} : \mathrm{BC} = 2\sqrt{14} : 3\sqrt{14} = 2 : 3$$

Question 9

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\vec{a} + \vec{b})$ and $(\vec{a} – 3\vec{b})$ externally in the ratio $1 : 2$. Also, show that P is the mid point of the line segment RQ.

Solution. External division in the ratio $m : n$ gives $\dfrac{m\vec{q} – n\vec{p}}{m – n}$. With $m = 1$, $n = 2$ the denominator is $-1$, so watch the sign carefully.

$$\vec{r} = \frac{1(\vec{a} – 3\vec{b}) – 2(2\vec{a} + \vec{b})}{1 – 2} = \frac{\vec{a} – 3\vec{b} – 4\vec{a} – 2\vec{b}}{-1} = \frac{-3\vec{a} – 5\vec{b}}{-1}$$

$$\vec{r} = 3\vec{a} + 5\vec{b}$$

For the second part, the midpoint of RQ is the average of their position vectors:

$$\frac{\vec{r} + \vec{q}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} – 3\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \vec{p}$$

which is precisely the position vector of P.

Question 10

The two adjacent sides of a parallelogram are $2\hat{i} – 4\hat{j} + 5\hat{k}$ and $\hat{i} – 2\hat{j} – 3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.

Solution. Two questions, two different tools. The diagonal of a parallelogram is the sum of the adjacent sides; the area is the magnitude of their cross product.

$$\vec{d} = (2\hat{i} – 4\hat{j} + 5\hat{k}) + (\hat{i} – 2\hat{j} – 3\hat{k}) = 3\hat{i} – 6\hat{j} + 2\hat{k}$$

$$|\vec{d}| = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$$

$$\hat{d} = \frac{1}{7}\left(3\hat{i} – 6\hat{j} + 2\hat{k}\right)$$

For the area:

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = \hat{i}(12 + 10) – \hat{j}(-6 – 5) + \hat{k}(-4 + 4)$$

$$= 22\hat{i} + 11\hat{j} + 0\hat{k}$$

$$\text{Area} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5}$$

Question 11

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm\left(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}\right)$.

Solution. “Equally inclined” means the three angles are equal, so their cosines are equal:

$$\alpha = \beta = \gamma \quad\Rightarrow\quad \cos\alpha = \cos\beta = \cos\gamma = l \text{ (say)}$$

Direction cosines always satisfy $l^2 + m^2 + n^2 = 1$, so

$$3l^2 = 1 \quad\Rightarrow\quad l^2 = \frac{1}{3} \quad\Rightarrow\quad l = \pm\frac{1}{\sqrt{3}}$$

Both signs are genuine: a vector along $\hat{i} + \hat{j} + \hat{k}$ and one along $-(\hat{i} + \hat{j} + \hat{k})$ are both equally inclined to the three axes.

Question 12

Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} – 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} – \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c}\cdot\vec{d} = 15$.

Solution. Two conditions, so use the two-step pattern. A vector perpendicular to both $\vec{a}$ and $\vec{b}$ must be a multiple of $\vec{a}\times\vec{b}$ — that is the whole point of the cross product.

$$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = \hat{i}(28 + 4) – \hat{j}(7 – 6) + \hat{k}(-2 – 12)$$

$$= 32\hat{i} – \hat{j} – 14\hat{k}$$

So $\vec{d} = \lambda\left(32\hat{i} – \hat{j} – 14\hat{k}\right)$ for some scalar $\lambda$, and the second condition fixes $\lambda$:

$$\vec{c}\cdot\vec{d} = \lambda\big[(2)(32) + (-1)(-1) + (4)(-14)\big] = \lambda(64 + 1 – 56) = 9\lambda$$

$$9\lambda = 15 \quad\Rightarrow\quad \lambda = \frac{5}{3}$$

$$\vec{d} = \frac{5}{3}\left(32\hat{i} – \hat{j} – 14\hat{k}\right) = \frac{1}{3}\left(160\hat{i} – 5\hat{j} – 70\hat{k}\right)$$

Verifying all three conditions, since this answer differs from the textbook’s:

$$\vec{d}\cdot\vec{a} = \tfrac13(160 – 20 – 140) = 0, \qquad \vec{d}\cdot\vec{b} = \tfrac13(480 + 10 – 490) = 0$$

$$\vec{c}\cdot\vec{d} = \tfrac13(320 + 5 – 280) = \tfrac13(45) = 15$$

A note on the printed answer. The answer at the back of the textbook gives $\tfrac13(160\hat{i} – 5\hat{j} + 70\hat{k})$, with a plus sign on the $\hat{k}$ term. That vector does not satisfy the question: its dot product with $\vec{a}$ is $93$ and with $\vec{b}$ is $327$, both of which must be zero for perpendicularity, and $\vec{c}\cdot\vec{d}$ comes to $202$ rather than $15$. The sign is a misprint; the answer above satisfies all three conditions exactly.

Question 13

The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} – 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.

Solution. Form the sum, keeping $\lambda$ symbolic.

$$\vec{s} = (2 + \lambda)\hat{i} + 6\hat{j} – 2\hat{k}, \qquad |\vec{s}| = \sqrt{(2+\lambda)^2 + 36 + 4}$$

The unit vector along $\vec{s}$ is $\vec{s}/|\vec{s}|$, so the condition reads

$$\frac{(\hat{i} + \hat{j} + \hat{k})\cdot\vec{s}}{|\vec{s}|} = 1 \quad\Rightarrow\quad \frac{(2 + \lambda) + 6 – 2}{\sqrt{(2+\lambda)^2 + 40}} = 1$$

$$\lambda + 6 = \sqrt{(2+\lambda)^2 + 40}$$

Squaring both sides — legitimate here because the left side must be positive for the equation to hold:

$$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 4 + 40$$

$$8\lambda = 8 \quad\Rightarrow\quad \lambda = 1$$

Question 14

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}$, $\vec{b}$ and $\vec{c}$.

Solution. (In the current reprint this question ends “show that the vector $\vec{c}\cdot\vec{d} = 15$ is equally inclined to…”, which is a fragment duplicated by mistake from question 12. The vector intended is $\vec{a} + \vec{b} + \vec{c}$, and that is what is proved here.)

Write $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Mutually perpendicular means

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

First find the magnitude of the sum, by squaring it:

$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 3k^2$$

$$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}\,k$$

Now the angle $\theta_1$ between the sum and $\vec{a}$. Every cross term vanishes, leaving only $\vec{a}\cdot\vec{a}$:

$$\cos\theta_1 = \frac{(\vec{a} + \vec{b} + \vec{c})\cdot\vec{a}}{|\vec{a} + \vec{b} + \vec{c}|\,|\vec{a}|} = \frac{k^2 + 0 + 0}{\sqrt{3}\,k \cdot k} = \frac{1}{\sqrt{3}}$$

The calculation for $\vec{b}$ and for $\vec{c}$ is identical — nothing in it distinguished $\vec{a}$ — so $\cos\theta_2 = \cos\theta_3 = \tfrac{1}{\sqrt{3}}$ as well.

Question 15

Prove that $(\vec{a} + \vec{b})\cdot(\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}$, $\vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}$, $\vec{b} \neq \vec{0}$.

Solution. “If and only if” needs both directions, but expanding once gives both at the same time.

$$(\vec{a} + \vec{b})\cdot(\vec{a} + \vec{b}) = \vec{a}\cdot\vec{a} + 2\,\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b} = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2$$

So the stated equation holds exactly when the leftover term is zero:

$$|\vec{a}|^2 + |\vec{b}|^2 + 2\,\vec{a}\cdot\vec{b} = |\vec{a}|^2 + |\vec{b}|^2 \iff 2\,\vec{a}\cdot\vec{b} = 0 \iff \vec{a}\cdot\vec{b} = 0$$

Since neither vector is zero, $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0$ forces $\cos\theta = 0$, that is $\theta = 90°$. Each step is reversible, which is what makes it an “if and only if” — and this is exactly where the condition $\vec{a}, \vec{b} \neq \vec{0}$ is needed, since otherwise the dot product could vanish for a different reason.

Question 16

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a}\cdot\vec{b} \geq 0$ only when

Solution. Since $|\vec{a}|$ and $|\vec{b}|$ are never negative, $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \geq 0$ exactly when $\cos\theta \geq 0$, which on $[0, \pi]$ means $0 \leq \theta \leq \tfrac{\pi}{2}$.

The choice between (A) and (B) is entirely about the endpoints, and the inequality in the question is $\geq$, not $>$. At $\theta = 0$ the dot product is $|\vec{a}||\vec{b}| > 0$ and at $\theta = \tfrac{\pi}{2}$ it is exactly $0$ — both satisfy $\geq 0$, so both endpoints belong.

Question 17

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if

Solution. Square the magnitude of the sum, which turns the condition into one about $\cos\theta$.

$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\,\vec{a}\cdot\vec{b} = 1 + 1 + 2\cos\theta$$

using $|\vec{a}| = |\vec{b}| = 1$. For $\vec{a} + \vec{b}$ to be a unit vector, this must equal $1$:

$$2 + 2\cos\theta = 1 \quad\Rightarrow\quad \cos\theta = -\frac{1}{2} \quad\Rightarrow\quad \theta = \frac{2\pi}{3}$$

The negative cosine is worth noticing: two unit vectors must point away from each other for their sum to shrink back to length one.

Question 18

The value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$ is

Solution. Evaluate each cross product first, using the cyclic rule $\hat{i}\times\hat{j} = \hat{k}$, $\hat{j}\times\hat{k} = \hat{i}$, $\hat{k}\times\hat{i} = \hat{j}$ — and remembering that reversing the order flips the sign.

$$\hat{j}\times\hat{k} = \hat{i} \quad\Rightarrow\quad \hat{i}\cdot\hat{i} = 1$$

$$\hat{i}\times\hat{k} = -(\hat{k}\times\hat{i}) = -\hat{j} \quad\Rightarrow\quad \hat{j}\cdot(-\hat{j}) = -1$$

$$\hat{i}\times\hat{j} = \hat{k} \quad\Rightarrow\quad \hat{k}\cdot\hat{k} = 1$$

$$1 + (-1) + 1 = 1$$

The middle term is the whole question: $\hat{i}\times\hat{k}$ runs against the cyclic order, so it carries a minus sign. Taking it as $+\hat{j}$ gives 3, which is why (D) is offered.

Question 19

If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a}\cdot\vec{b}| = |\vec{a}\times\vec{b}|$ when $\theta$ is equal to

Solution. Write both sides out. The modulus on the left matters, because $\cos\theta$ can be negative while a magnitude cannot.

$$|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}|\,|\cos\theta|, \qquad |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$$

Cancelling the magnitudes (neither vector being zero):

$$|\cos\theta| = \sin\theta$$

On $0 \leq \theta \leq \pi$ the sine is never negative, so this gives $\tan\theta = 1$ and $\theta = \tfrac{\pi}{4}$.

Checking the alternatives quickly: at $\theta = 0$ or $\pi$ the cross product is zero but the dot product is not; at $\theta = \tfrac{\pi}{2}$ it is the other way round.

Common mistakes

• Measuring a compass bearing from the wrong axis. “$30°$ east of north” turns from north towards east, so the north component takes the cosine. Question 3 is unmarkable if this is reversed.
• Assuming magnitudes add. $|\vec{b} + \vec{c}| = |\vec{b}| + |\vec{c}|$ only for parallel vectors in the same direction — question 4 exists to break that habit, and question 17 shows the sum of two unit vectors can be shorter than either.
• Dropping the modulus when solving for a scalar. In question 5 the magnitude gives $|x|\sqrt{3} = 1$, so there are two answers, not one. Same in question 11.
• Mixing up diagonal and area in question 10. The diagonal is the sum of the adjacent sides; the area is the magnitude of their cross product. One question asks for both.
• Trying to satisfy two conditions at once in question 12. Find the direction first from $\vec{a}\times\vec{b}$, then scale by $\lambda$ to meet $\vec{c}\cdot\vec{d} = 15$. Attempting both simultaneously means three unknowns and three equations for no reason.
• Getting the sign of $\hat{i}\times\hat{k}$ wrong. It is $-\hat{j}$, not $+\hat{j}$, because it runs against the cyclic order. That single sign is the difference between (C) and (D) in question 18.
• Forgetting the modulus on the dot product in question 19. Without it the equation becomes $\cos\theta = \sin\theta$, which happens to give the same answer here but is not what the question states.

Practise next

• Chapter 11, Three Dimensional Geometry — where these vectors become the standard language for lines and planes, and the dot and cross products do the work of finding distances and angles.
• Exercise 10.3 and 10.4 — worth a second pass if the choice between dot and cross product still feels arbitrary.