NCERT Exemplar Solutions Class 12 Maths Vector Algebra — Objective, Fill in the Blanks and True or False Q19–45

NCERT Exemplar Problems, Class 12 Mathematics — Vector Algebra, Objective Type questions 19 to 33, Fill in the Blanks 34 to 40 and True or False 41 to 45. All 27 questions solved.

Twenty-seven short questions. Most are one line of working once you see which fact they test, and a handful reward a trick worth keeping: squaring $\vec a + \vec b + \vec c = \vec 0$ (Questions 29 and 31), Lagrange’s identity $|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2$ (27, 38, 39), and the scalar triple product as a test for coplanarity (28, 35).

Two answers in the printed key are wrong, Question 34 (a fill-in) and Question 41 (a true-or-false). Each is explained where it occurs.

Key insight. When a question gives magnitudes and a dot product but asks about a cross product, or the other way round, reach for $|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2$. It is just $\sin^2\theta + \cos^2\theta = 1$ multiplied by $|\vec a|^2|\vec b|^2$.

Question 19

The vector in the direction of the vector $\hat i – 2\hat j + 2\hat k$ that has magnitude $9$ is

Solution. The given vector has magnitude $\sqrt{1 + 4 + 4} = 3$, so its unit vector is $\frac13(\hat i – 2\hat j + 2\hat k)$. Multiplying by $9$ gives $3(\hat i – 2\hat j + 2\hat k)$. Option (D) is the trap: it multiplies the original vector, not its unit vector, and has magnitude $27$.

Question 20

The position vector of the point which divides the join of points $2\vec a – 3\vec b$ and $\vec a + \vec b$ in the ratio $3 : 1$ is

Solution. By the section formula (internal division, $m : n = 3 : 1$), the point is $\dfrac{3(\vec a + \vec b) + 1(2\vec a – 3\vec b)}{3 + 1} = \dfrac{5\vec a + 0\,\vec b}{4}$. The $\vec b$ terms cancel exactly. Remember that the $3$ multiplies the second point, the one the ratio’s first number is measured towards.

Question 21

The vector having initial and terminal points as $(2, 5, 0)$ and $(-3, 7, 4)$, respectively is

Solution. Terminal minus initial: $(-3 – 2)\hat i + (7 – 5)\hat j + (4 – 0)\hat k = -5\hat i + 2\hat j + 4\hat k$. Option (B) is initial minus terminal, the same vector reversed.

Question 22

The angle between two vectors $\vec a$ and $\vec b$ with magnitudes $\sqrt3$ and $4$, respectively, and $\vec a\cdot\vec b = 2\sqrt3$ is

Solution. $\cos\theta = \dfrac{2\sqrt3}{\sqrt3\cdot4} = \dfrac12$, so $\theta = \dfrac\pi3$. Option (D) cannot be right in any case: the angle between two vectors lies in $[0, \pi]$.

Question 23

Find the value of $\lambda$ such that the vectors $\vec a = 2\hat i + \lambda\hat j + \hat k$ and $\vec b = \hat i + 2\hat j + 3\hat k$ are orthogonal

Solution. Orthogonal means $\vec a\cdot\vec b = 0$: $2 + 2\lambda + 3 = 0$, so $\lambda = -\dfrac52$.

Question 24

The value of $\lambda$ for which the vectors $3\hat i – 6\hat j + \hat k$ and $2\hat i – 4\hat j + \lambda\hat k$ are parallel is

Solution. Parallel vectors have proportional components: $\dfrac23 = \dfrac{-4}{-6} = \dfrac\lambda1$. The first two ratios already agree at $\frac23$, so $\lambda = \frac23$.

Question 25

The vectors from origin to the points A and B are $\vec a = 2\hat i – 3\hat j + 2\hat k$ and $\vec b = 2\hat i + 3\hat j + \hat k$, respectively, then the area of triangle OAB is

Solution.

$$\vec a\times\vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = (-3 – 6)\hat i – (2 – 4)\hat j + (6 + 6)\hat k = -9\hat i + 2\hat j + 12\hat k$$

Its magnitude is $\sqrt{81 + 4 + 144} = \sqrt{229}$, and the triangle is half the parallelogram. Option (C) forgets the half.

Question 26

For any vector $\vec a$, the value of $(\vec a\times\hat i)^2 + (\vec a\times\hat j)^2 + (\vec a\times\hat k)^2$ is equal to

Solution. By Lagrange’s identity, $|\vec a\times\hat i|^2 = |\vec a|^2|\hat i|^2 – (\vec a\cdot\hat i)^2 = |\vec a|^2 – a_1^2$, and similarly for $\hat j$ and $\hat k$. Adding, $3|\vec a|^2 – (a_1^2 + a_2^2 + a_3^2) = 3|\vec a|^2 – |\vec a|^2 = 2|\vec a|^2$.

Question 27

If $|\vec a| = 10$, $|\vec b| = 2$ and $\vec a\cdot\vec b = 12$, then value of $|\vec a\times\vec b|$ is

Solution. $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 – (\vec a\cdot\vec b)^2 = 400 – 144 = 256$, so $|\vec a\times\vec b| = 16$.

Question 28

The vectors $\lambda\hat i + \hat j + 2\hat k$, $\hat i + \lambda\hat j – \hat k$ and $2\hat i – \hat j + \lambda\hat k$ are coplanar if

Solution. Three vectors are coplanar exactly when their scalar triple product is zero:

$$\begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{vmatrix} = \lambda(\lambda^2 – 1) – 1(\lambda + 2) + 2(-1 – 2\lambda) = \lambda^3 – 6\lambda – 4$$

This factorises as $(\lambda + 2)(\lambda^2 – 2\lambda – 2)$, so $\lambda = -2$ or $\lambda = 1 \pm\sqrt3$. Only $-2$ is offered. (Substituting the options is quicker: only $\lambda = -2$ makes $\lambda^3 – 6\lambda – 4$ vanish.)

Question 29

If $\vec a, \vec b, \vec c$ are unit vectors such that $\vec a + \vec b + \vec c = \vec 0$, then the value of $\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a$ is

Solution. Dot the equation with itself: $|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0$. With each magnitude $1$, $3 + 2S = 0$, so $S = -\frac32$. (Geometrically, the three unit vectors make an equilateral triangle head to tail, so each pair is at $120^\circ$ and each dot product is $-\frac12$.)

Question 30

Projection vector of $\vec a$ on $\vec b$ is

Solution. The scalar projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$; the projection vector is that length times the unit vector $\hat b = \dfrac{\vec b}{|\vec b|}$, giving $\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b$. Option (B) is the scalar projection — a number, not a vector.

Question 31

If $\vec a, \vec b, \vec c$ are three vectors such that $\vec a + \vec b + \vec c = \vec 0$ and $|\vec a| = 2$, $|\vec b| = 3$, $|\vec c| = 5$, then value of $\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a$ is

Solution. The same squaring as Question 29: $4 + 9 + 25 + 2S = 0$, so $S = -19$. (Here $2 + 3 = 5$, so the three vectors are collinear, with $\vec a$ and $\vec b$ pointing the same way and $\vec c$ opposite — a degenerate triangle. The identity does not care.)

Question 32

If $|\vec a| = 4$ and $-3 \le \lambda \le 2$, then the range of $|\lambda\vec a|$ is

Solution. $|\lambda\vec a| = |\lambda|\cdot4$. On $[-3, 2]$, $|\lambda|$ runs from $0$ (at $\lambda = 0$) up to $3$ (at $\lambda = -3$), so $|\lambda\vec a|$ runs over $[0, 12]$. Option (A) uses the right end-point $2$ instead of the larger $|{-3}|$; option (B) forgets that a magnitude is never negative.

Question 33

The number of vectors of unit length perpendicular to the vectors $\vec a = 2\hat i + \hat j + 2\hat k$ and $\vec b = \hat j + \hat k$ is

Solution. $\vec a$ and $\vec b$ are not parallel, so the directions perpendicular to both form a single line, along $\vec a\times\vec b = -\hat i – 2\hat j + 2\hat k$ (magnitude $3$). A line has two unit vectors along it: $\pm\frac13(-\hat i – 2\hat j + 2\hat k)$.

Question 34

The vector $\vec a + \vec b$ bisects the angle between the non-collinear vectors $\vec a$ and $\vec b$ if $\rule{2cm}{0.4pt}$

Solution. $\vec a + \vec b$ is the diagonal of the parallelogram on $\vec a$ and $\vec b$. A diagonal bisects the angle at its vertex exactly when the parallelogram is a rhombus, i.e. when the sides are equal in length. In vectors: the angles of $\vec a + \vec b$ with $\vec a$ and with $\vec b$ are equal when $\dfrac{(\vec a + \vec b)\cdot\vec a}{|\vec a|} = \dfrac{(\vec a + \vec b)\cdot\vec b}{|\vec b|}$, which simplifies to $\left(|\vec a| – |\vec b|\right)\left(|\vec a||\vec b| – \vec a\cdot\vec b\right) = 0$; the second factor is non-zero for non-collinear vectors, so $|\vec a| = |\vec b|$.

The printed key says “if $\vec a$ and $\vec b$ are equal vectors”, and that cannot be right. Equal vectors point the same way, so they are collinear — which the question rules out — and there would be no angle to bisect. What has to be equal is their magnitudes, not the vectors.

Question 35

If $\vec r\cdot\vec a = 0$, $\vec r\cdot\vec b = 0$, and $\vec r\cdot\vec c = 0$ for some non-zero vector $\vec r$, then the value of $\vec a\cdot(\vec b\times\vec c)$ is $\rule{2cm}{0.4pt}$

Solution. All three vectors are perpendicular to the same non-zero $\vec r$, so all three lie in the plane perpendicular to $\vec r$. Coplanar vectors have scalar triple product zero.

Question 36

The vectors $\vec a = 3\hat i – 2\hat j + 2\hat k$ and $\vec b = -\hat i – 2\hat k$ are the adjacent sides of a parallelogram. The acute angle between its diagonals is $\rule{2cm}{0.4pt}$

Solution. The diagonals are $\vec a + \vec b = 2\hat i – 2\hat j$ and $\vec a – \vec b = 4\hat i – 2\hat j + 4\hat k$, with magnitudes $2\sqrt2$ and $6$. Their dot product is $8 + 4 + 0 = 12$, so $\cos\theta = \dfrac{12}{2\sqrt2\cdot6} = \dfrac{1}{\sqrt2}$.

Question 37

The values of $k$ for which $|k\vec a| < |\vec a|$ and $k\vec a + \frac12\vec a$ is parallel to $\vec a$ holds true are $\rule{2cm}{0.4pt}$

Solution. The first condition is $|k||\vec a| < |\vec a|$, i.e. $-1 < k < 1$ (with $\vec a \ne \vec 0$). The second: $k\vec a + \frac12\vec a = \left(k + \frac12\right)\vec a$ is a multiple of $\vec a$, so it is parallel to $\vec a$ — unless the multiple is $0$, when it is the zero vector, which has no direction. That excludes $k = -\frac12$.

Question 38

The value of the expression $|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2$ is $\rule{2cm}{0.4pt}$

Solution. $|\vec a|^2|\vec b|^2\sin^2\theta + |\vec a|^2|\vec b|^2\cos^2\theta = |\vec a|^2|\vec b|^2$.

Question 39

If $|\vec a\times\vec b|^2 + |\vec a\cdot\vec b|^2 = 144$ and $|\vec a| = 4$, then $|\vec b|$ is equal to $\rule{2cm}{0.4pt}$

Solution. By Question 38 the left side is $|\vec a|^2|\vec b|^2 = 16|\vec b|^2$. So $|\vec b|^2 = 9$.

Question 40

If $\vec a$ is any non-zero vector, then $(\vec a\cdot\hat i)\hat i + (\vec a\cdot\hat j)\hat j + (\vec a\cdot\hat k)\hat k$ equals $\rule{2cm}{0.4pt}$

Solution. With $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$, the three dot products are $a_1$, $a_2$, $a_3$, so the expression rebuilds $\vec a$ from its components.

Question 41

State True or False: If $|\vec a| = |\vec b|$, then necessarily it implies $\vec a = \pm\vec b$.

Solution. Equal magnitudes say nothing about direction. $\hat i$ and $\hat j$ both have magnitude $1$, but $\hat i$ is neither $\hat j$ nor $-\hat j$. One counter-example is enough to make a “necessarily” statement false.

The printed key says True, and it is wrong. The implication runs only the other way: $\vec a = \pm\vec b$ does imply $|\vec a| = |\vec b|$.

Question 42

State True or False: Position vector of a point P is a vector whose initial point is origin.

Solution. That is the definition: the position vector of P is $\overrightarrow{OP}$, starting at the origin O and ending at P.

Question 43

State True or False: If $|\vec a + \vec b| = |\vec a – \vec b|$, then the vectors $\vec a$ and $\vec b$ are orthogonal.

Solution. Square both sides: $|\vec a|^2 + 2\vec a\cdot\vec b + |\vec b|^2 = |\vec a|^2 – 2\vec a\cdot\vec b + |\vec b|^2$, so $4\vec a\cdot\vec b = 0$. Geometrically, a parallelogram with equal diagonals is a rectangle. (For non-zero vectors this is orthogonality; if one is zero the dot product is still zero, which is the usual convention.)

Question 44

State True or False: The formula $(\vec a + \vec b)^2 = \vec a^2 + \vec b^2 + 2\vec a\times\vec b$ is valid for non-zero vectors $\vec a$ and $\vec b$.

Solution. The left side and $\vec a^2 + \vec b^2$ are numbers, while $\vec a\times\vec b$ is a vector, so the formula does not even make sense as written. The correct expansion is $(\vec a + \vec b)^2 = \vec a^2 + \vec b^2 + 2\,\vec a\cdot\vec b$.

Question 45

State True or False: If $\vec a$ and $\vec b$ are adjacent sides of a rhombus, then $\vec a\cdot\vec b = 0$.

Solution. $\vec a\cdot\vec b = 0$ would make the adjacent sides perpendicular, and a rhombus need not have right angles — only a square does. What is always true of a rhombus is that its diagonals are perpendicular: $(\vec a + \vec b)\cdot(\vec a – \vec b) = |\vec a|^2 – |\vec b|^2 = 0$.

Common mistakes

• Confusing equal magnitudes with equal vectors. It produces both key errors on this page: Question 34 needs $|\vec a| = |\vec b|$, not $\vec a = \vec b$, and Question 41’s $|\vec a| = |\vec b|$ does not give $\vec a = \pm\vec b$.
• Scaling the original vector instead of its unit vector in Question 19. Find the unit vector first, then multiply.
• Mixing up the scalar and vector projections in Question 30. One is a length, the other is that length times $\hat b$.
• Taking the right end-point instead of the largest $|\lambda|$ in Question 32. A magnitude depends on $|\lambda|$, and on $[-3, 2]$ the largest $|\lambda|$ is $3$.
• Forgetting the zero vector in Question 37. $\left(k + \frac12\right)\vec a$ is parallel to $\vec a$ for every $k$ except the one that makes it zero.
• Treating a rhombus as a square in Question 45. Its diagonals are perpendicular; its sides need not be.

Practise next

• Exemplar Short and Long Answer Q1–18, the first half of the unit — unit vectors, the section formula, areas and three proofs.
• NCERT textbook Exercise 10.3 for more dot products, including the $|\vec a + \vec b| = |\vec a – \vec b|$ type behind Question 43.
• NCERT textbook Exercise 10.4, where Questions 25, 27 and 33 come from.
• NCERT textbook Exercise 10.2 for components, unit vectors and the section formula used in Questions 19 to 21.