NCERT Exemplar Solutions Class 12 Maths Vector Algebra — Short and Long Answer Q1–18

NCERT Exemplar Problems, Class 12 Mathematics — Vector Algebra, Short Answer questions 1 to 14 and Long Answer questions 15 to 18. All 18 questions solved.

Eighteen questions that between them use every tool in the chapter: unit vectors and direction cosines (1–3, 6, 7), section and collinearity (4, 5), the dot product for angles and projections (9, 12), and the cross product for perpendicular vectors and areas (8, 11, 13, 17). The last six are proofs, and they are worth the time — Questions 14 to 16 show that results you learnt as geometry fall out of two lines of vector algebra.

One answer in the printed key is wrong, Question 4, and for Question 8 the key gives only one of two valid answers. Both are explained where they occur.

Key insight. The dot product measures how much two vectors agree in direction; the cross product measures the area they span. Almost every question here is one of those two facts: an angle or a projection is a dot product, and a perpendicular vector or an area is a cross product.

Question 1

Find the unit vector in the direction of sum of vectors $\vec a = 2\hat i – \hat j + \hat k$ and $\vec b = 2\hat j + \hat k$.

Solution. Add componentwise: $\vec a + \vec b = 2\hat i + \hat j + 2\hat k$. Its magnitude is $\sqrt{4 + 1 + 4} = 3$. Dividing a vector by its magnitude keeps its direction and makes its length $1$:

Question 2

If $\vec a = \hat i + \hat j + 2\hat k$ and $\vec b = 2\hat i + \hat j – 2\hat k$, find the unit vector in the direction of (i) $6\vec b$ (ii) $2\vec a – \vec b$.

Solution. (i) Multiplying by the positive number $6$ stretches $\vec b$ without turning it, so $6\vec b$ has the same unit vector as $\vec b$. With $|\vec b| = \sqrt{4 + 1 + 4} = 3$, that is $\frac13\left(2\hat i + \hat j – 2\hat k\right)$. (Directly: $|6\vec b| = 18$ and $\frac{6\vec b}{18}$ is the same thing.)

(ii) $2\vec a – \vec b = (2 – 2)\hat i + (2 – 1)\hat j + (4 + 2)\hat k = \hat j + 6\hat k$, with magnitude $\sqrt{1 + 36} = \sqrt{37}$.

Question 3

Find a unit vector in the direction of $\overrightarrow{PQ}$, where P and Q have co-ordinates $(5, 0, 8)$ and $(3, 3, 2)$, respectively.

Solution. A vector from P to Q is “end minus start”: $\overrightarrow{PQ} = (3 – 5)\hat i + (3 – 0)\hat j + (2 – 8)\hat k = -2\hat i + 3\hat j – 6\hat k$. Its magnitude is $\sqrt{4 + 9 + 36} = 7$.

Question 4

If $\vec a$ and $\vec b$ are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that $BC = 1.5\,BA$.

Solution. “BA produced” is the line from B through A, extended beyond A. So C lies on the same side of B as A, one and a half times as far away. In vectors that is

$$\overrightarrow{BC} = 1.5\,\overrightarrow{BA} \quad\Longrightarrow\quad \vec c – \vec b = \frac32\left(\vec a – \vec b\right)$$

$$\vec c = \frac32\vec a – \frac12\vec b$$

Check. A divides BC internally in the ratio $1 : 0.5 = 2 : 1$, so by the section formula $\vec a = \dfrac{2\vec c + \vec b}{3}$, which rearranges to the same $\vec c$.

The printed key gives $\vec c = \dfrac{3\vec b – \vec a}{2}$, and it does not fit the question. That point satisfies $\vec c – \vec a = \frac32(\vec b – \vec a)$, i.e. $AC = 1.5\,AB$ — it lies on AB produced, beyond B. The key has swapped the roles of A and B.

Question 5

Using vectors, find the value of $k$ such that the points $(k, -10, 3)$, $(1, -1, 3)$ and $(3, 5, 3)$ are collinear.

Solution. Call the points A, B, C. They are collinear exactly when $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are parallel, i.e. one is a scalar multiple of the other.

$$\overrightarrow{BC} = 2\hat i + 6\hat j + 0\hat k, \qquad \overrightarrow{BA} = (k – 1)\hat i – 9\hat j + 0\hat k$$

Parallel means proportional components: $\dfrac{k – 1}{2} = \dfrac{-9}{6} = -\dfrac32$, so $k – 1 = -3$.

(Equivalently, $\overrightarrow{BA} \times \overrightarrow{BC} = \vec 0$; its only non-zero component is $6(k – 1) + 18$, which vanishes at the same $k$.)

Question 6

A vector $\vec r$ is inclined at equal angles to the three axes. If the magnitude of $\vec r$ is $2\sqrt3$ units, find $\vec r$.

Solution. Equal angles with the axes means equal direction cosines, $l = m = n$. Since $l^2 + m^2 + n^2 = 1$, each is $\pm\frac{1}{\sqrt3}$, all with the same sign. So

$$\vec r = |\vec r|\left(l\hat i + m\hat j + n\hat k\right) = 2\sqrt3\cdot\left(\pm\frac{1}{\sqrt3}\right)\left(\hat i + \hat j + \hat k\right)$$

Both signs are genuine answers: $-2(\hat i + \hat j + \hat k)$ makes equal (obtuse) angles with all three axes too.

Question 7

A vector $\vec r$ has magnitude $14$ and direction ratios $2, 3, -6$. Find the direction cosines and components of $\vec r$, given that $\vec r$ makes an acute angle with $x$-axis.

Solution. Direction cosines are the direction ratios divided by $\sqrt{2^2 + 3^2 + 6^2} = 7$, with an overall sign still to choose: $\pm\left(\frac27, \frac37, -\frac67\right)$. An acute angle with the $x$-axis needs $l = \cos\alpha > 0$, so we take the positive sign. Then the components are $14$ times the direction cosines.

Question 8

Find a vector of magnitude $6$, which is perpendicular to both the vectors $2\hat i – \hat j + 2\hat k$ and $4\hat i – \hat j + 3\hat k$.

Solution. The cross product of two vectors is perpendicular to both, so start there:

$$\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix} = (-3 + 2)\hat i – (6 – 8)\hat j + (-2 + 4)\hat k = -\hat i + 2\hat j + 2\hat k$$

Its magnitude is $3$, so the unit vectors perpendicular to both are $\pm\frac13(-\hat i + 2\hat j + 2\hat k)$; multiply by $6$.

The printed key gives only $-2\hat i + 4\hat j + 4\hat k$. The opposite vector is just as perpendicular and just as long, so a full answer names both.

Question 9

Find the angle between the vectors $2\hat i – \hat j + \hat k$ and $3\hat i + 4\hat j – \hat k$.

Solution. $\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$. Here $\vec a\cdot\vec b = 6 – 4 – 1 = 1$, $|\vec a| = \sqrt6$ and $|\vec b| = \sqrt{26}$, so $\cos\theta = \dfrac{1}{\sqrt{156}}$.

The dot product is small and positive, so the angle is just under $90^\circ$ (about $85.4^\circ$).

Question 10

If $\vec a + \vec b + \vec c = \vec 0$, show that $\vec a\times\vec b = \vec b\times\vec c = \vec c\times\vec a$. Interpret the result geometrically.

Solution. Cross the given equation with $\vec b$ on the right. Since $\vec b\times\vec b = \vec 0$,

$$\vec a\times\vec b + \vec c\times\vec b = \vec 0 \quad\Longrightarrow\quad \vec a\times\vec b = -\vec c\times\vec b = \vec b\times\vec c$$

Now cross it with $\vec c$ on the right: $\vec a\times\vec c + \vec b\times\vec c = \vec 0$, so $\vec b\times\vec c = -\vec a\times\vec c = \vec c\times\vec a$. Together these give the result.

The equality is of vectors, not only of magnitudes, so the three parallelograms also face the same way: their normals all point to the same side of the triangle’s plane.

Question 11

Find the sine of the angle between the vectors $\vec a = 3\hat i + \hat j + 2\hat k$ and $\vec b = 2\hat i – 2\hat j + 4\hat k$.

Solution. Use $\sin\theta = \dfrac{|\vec a\times\vec b|}{|\vec a||\vec b|}$.

$$\vec a\times\vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix} = (4 + 4)\hat i – (12 – 4)\hat j + (-6 – 2)\hat k = 8\hat i – 8\hat j – 8\hat k$$

So $|\vec a\times\vec b| = 8\sqrt3$, while $|\vec a| = \sqrt{14}$ and $|\vec b| = \sqrt{24} = 2\sqrt6$. Then $\sin\theta = \dfrac{8\sqrt3}{2\sqrt{84}} = \dfrac{8\sqrt3}{4\sqrt{21}} = \dfrac{2}{\sqrt7}$.

Check through the dot product. $\vec a\cdot\vec b = 6 – 2 + 8 = 12$, so $\cos\theta = \dfrac{12}{\sqrt{14}\cdot2\sqrt6} = \sqrt{\dfrac37}$ and $\sin^2\theta = 1 – \dfrac37 = \dfrac47$. The two routes agree.

Question 12

If A, B, C, D are the points with position vectors $\hat i + \hat j – \hat k$, $2\hat i – \hat j + 3\hat k$, $2\hat i – 3\hat k$, $3\hat i – 2\hat j + \hat k$, respectively, find the projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}$.

Solution. $\overrightarrow{AB} = \hat i – 2\hat j + 4\hat k$ and $\overrightarrow{CD} = \hat i – 2\hat j + 4\hat k$. The projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}$ is

$$\frac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{CD}|} = \frac{1 + 4 + 16}{\sqrt{21}} = \frac{21}{\sqrt{21}}$$

The two vectors turn out to be equal, so the projection is simply the full length of $\overrightarrow{AB}$ — and ABDC is a parallelogram.

Question 13

Using vectors, find the area of the triangle ABC with vertices $A(1, 2, 3)$, $B(2, -1, 4)$ and $C(4, 5, -1)$.

Solution. The area is half the magnitude of the cross product of two sides from the same vertex. $\overrightarrow{AB} = \hat i – 3\hat j + \hat k$ and $\overrightarrow{AC} = 3\hat i + 3\hat j – 4\hat k$, so

$$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix} = (12 – 3)\hat i – (-4 – 3)\hat j + (3 + 9)\hat k = 9\hat i + 7\hat j + 12\hat k$$

with magnitude $\sqrt{81 + 49 + 144} = \sqrt{274}$.

Question 14

Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Solution. Let ABCD and ABEF be parallelograms on the same base AB, between the same parallel lines AB and DE (so D, C, F, E all lie on one line parallel to AB). Put $\vec a = \overrightarrow{AB}$ and $\vec b = \overrightarrow{AD}$.

Because F lies on the line through D parallel to AB, $\overrightarrow{DF}$ is a multiple of $\vec a$, say $\overrightarrow{DF} = t\vec a$. Then $\overrightarrow{AF} = \vec b + t\vec a$.

$$\text{Area}(ABEF) = \left|\overrightarrow{AB}\times\overrightarrow{AF}\right| = \left|\vec a\times(\vec b + t\vec a)\right| = \left|\vec a\times\vec b + t\,\vec a\times\vec a\right| = |\vec a\times\vec b|$$

since $\vec a\times\vec a = \vec 0$. And $|\vec a\times\vec b| = \left|\overrightarrow{AB}\times\overrightarrow{AD}\right|$ is the area of ABCD.

The proof shows why it is true: sliding the top side along its line adds a multiple of $\vec a$ to the second side, and the cross product ignores any part parallel to $\vec a$.

Question 15

Prove that in any triangle ABC, $\cos A = \dfrac{b^2 + c^2 – a^2}{2bc}$, where $a, b, c$ are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Solution. Let $\vec u = \overrightarrow{AB}$ and $\vec v = \overrightarrow{AC}$, so $|\vec u| = c$, $|\vec v| = b$, and the angle between them is A, giving $\vec u\cdot\vec v = bc\cos A$. The third side is $\overrightarrow{BC} = \vec v – \vec u$, with $|\overrightarrow{BC}| = a$. Square it with the dot product:

$$a^2 = (\vec v – \vec u)\cdot(\vec v – \vec u) = |\vec v|^2 + |\vec u|^2 – 2\,\vec u\cdot\vec v = b^2 + c^2 – 2bc\cos A$$

Rearranging gives the result.

Choosing both vectors to start at A is the step that matters: it makes the angle between them exactly A, not its supplement.

Question 16

If $\vec a, \vec b, \vec c$ determine the vertices of a triangle, show that $\frac12\left[\vec b\times\vec c + \vec c\times\vec a + \vec a\times\vec b\right]$ gives the vector area of the triangle. Hence deduce the condition that the three points $\vec a, \vec b, \vec c$ are collinear. Also find the unit vector normal to the plane of the triangle.

Solution. Let the vertices be A, B, C with position vectors $\vec a, \vec b, \vec c$. The vector area is $\frac12\,\overrightarrow{AB}\times\overrightarrow{AC}$, whose magnitude is the area and whose direction is normal to the triangle. Expand:

$$\overrightarrow{AB}\times\overrightarrow{AC} = (\vec b – \vec a)\times(\vec c – \vec a) = \vec b\times\vec c – \vec b\times\vec a – \vec a\times\vec c + \vec a\times\vec a$$

The last term is $\vec 0$; $-\vec b\times\vec a = \vec a\times\vec b$ and $-\vec a\times\vec c = \vec c\times\vec a$. So the vector area is $\frac12\left[\vec b\times\vec c + \vec c\times\vec a + \vec a\times\vec b\right]$.

Collinearity. Three points are collinear exactly when the “triangle” they form has zero area, so the condition is $\vec a\times\vec b + \vec b\times\vec c + \vec c\times\vec a = \vec 0$.

Unit normal. The vector area is perpendicular to the plane of the triangle; divide it by its magnitude.

Question 17

Show that area of the parallelogram whose diagonals are given by $\vec a$ and $\vec b$ is $\dfrac{|\vec a\times\vec b|}{2}$. Also find the area of the parallelogram whose diagonals are $2\hat i – \hat j + \hat k$ and $\hat i + 3\hat j – \hat k$.

Solution. Let the sides be $\vec p$ and $\vec q$, so the area is $|\vec p\times\vec q|$ and the diagonals are $\vec a = \vec p + \vec q$ and $\vec b = \vec q – \vec p$. Then

$$\vec a\times\vec b = (\vec p + \vec q)\times(\vec q – \vec p) = \vec p\times\vec q – \vec p\times\vec p + \vec q\times\vec q – \vec q\times\vec p = 2\,\vec p\times\vec q$$

so $|\vec p\times\vec q| = \frac12|\vec a\times\vec b|$, as required. (If the second diagonal is taken as $\vec p – \vec q$ instead, only the sign changes.)

For the given diagonals:

$$\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = (1 – 3)\hat i – (-2 – 1)\hat j + (6 + 1)\hat k = -2\hat i + 3\hat j + 7\hat k$$

with magnitude $\sqrt{4 + 9 + 49} = \sqrt{62}$.

Question 18

If $\vec a = \hat i + \hat j + \hat k$ and $\vec b = \hat j – \hat k$, find a vector $\vec c$ such that $\vec a\times\vec c = \vec b$ and $\vec a\cdot\vec c = 3$.

Solution. Let $\vec c = x\hat i + y\hat j + z\hat k$. Then

$$\vec a\times\vec c = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = (z – y)\hat i + (x – z)\hat j + (y – x)\hat k$$

Matching with $\vec b = 0\hat i + \hat j – \hat k$: $z = y$, $x – z = 1$, $y – x = -1$ (the third repeats the first two). So $x = y + 1$ and $z = y$. The dot product condition gives $x + y + z = 3$, i.e. $3y + 1 = 3$, so $y = \frac23$, $z = \frac23$, $x = \frac53$.

The cross product alone cannot fix $\vec c$: adding any multiple of $\vec a$ leaves $\vec a\times\vec c$ unchanged. That is why the question also gives $\vec a\cdot\vec c$ — it picks one vector from that whole line of solutions.

Common mistakes

• Reading “BA produced” as “AB produced” in Question 4. The first letter is where you start; the line is extended beyond the second. The printed key makes exactly this slip.
• Giving one sign when the question allows two. Questions 6 and 8 each have a pair of opposite answers. Question 7 is different: “acute angle with the $x$-axis” is there precisely to pick one sign, so use it.
• Expanding a $3\times3$ determinant with the wrong sign on the $\hat j$ term. The middle term is subtracted; in Questions 8, 11, 13 and 17 a slip there changes both the direction and, usually, the magnitude.
• Using $\vec a\times\vec b = \vec b\times\vec a$. The cross product changes sign when the order is swapped. Questions 10 and 16 depend on handling those signs exactly.
• Measuring the angle at A with $\overrightarrow{BA}$ and $\overrightarrow{AC}$ in Question 15. Both vectors must start at A, or you get $\cos(180^\circ – A)$ and the sign of the middle term flips.

Practise next

• Exemplar Objective, Fill in the Blanks and True or False Q19–45, the rest of the unit — fifteen multiple-choice questions, seven blanks and five true-or-false statements, two of which the key gets wrong.
• NCERT textbook Exercise 10.4 for more cross products and areas of triangles and parallelograms.
• NCERT textbook Exercise 10.3 for the dot product, angles and projections.
• NCERT textbook Miscellaneous Exercise on Vector Algebra, which mixes the section formula with dot and cross products in the same way as Questions 4 and 18.