Match List-I with List-II, List-I, List-II, (A) If vector a and b are such that a = b and a = b, then, (I) a and b are orthogonal, (B) Projection vector of a on b, (II) [0, 12], (C) a and b are non-zero vectors such that a + b = a - b, then, (III) a = ± b, (D) If a = 4, -3 ≤ ≤ 2, then the range of a, (IV) ((a * b)/( b ^2)) b, Choose the correct answer from the options given below: Option 1: (A) - (III), (B) - (IV), (C) - (II), (D) - (I) Option 2: (A) - (I), (B) - (III), (C) - (IV), (D) - (II) Option 3: (A) - (III), (B) - (IV), (C) - (I), (D) - (II) Option 4: (A) - (I), (B) - (III), (C) - (II), (D) - (IV)

Answer: (A) - (III), (B) - (IV), (C) - (I), (D) - (II). Solution: Since a =, taking magnitude on both sides:, a, =, b, Given, a, =, b, :, b, =, b, = 1 = +1 or = -1 Therefore a = ± b (A) (III) The vector projection of a on b is: proj_(b) a = ((ab)/(, b, ^2))b Note: the scalar projection is (ab)/(, b, ), but the vector projection includes direction. (B) (IV) Given, a+b, =, a-b, squaring both sides:, a, ^2 + 2(ab) +, b, ^2 =, a, ^2 - 2(ab) +, b, ^2 4(ab) = 0 ab = 0 Since the dot product is zero, a and b are orthogonal. (C) (I), =, *, a, = 4, Since [-3, 2]: At = -3:, = 3 At = 0:, = 0 At = 2:, = 2 So, [0, 3], which gives: 4, [0, 12] (D) (II) (A) (III) (B) (IV) (C) (I) (D) (II)

Algebra

Vector Algebra

Medium

core

Match List-I with List-II

List-I List-II
(A) If vector $\vec{a}$ and $\vec{b}$ are such that $\vec{a} = \lambda \vec{b}$ and $\vert \vec{a}\vert = \vert \vec{b}\vert$ , then (I) $\vec{a}$ and $\vec{b}$ are orthogonal
(B) Projection vector of $\vec{a}$ on $\vec{b}$ (II) $[0, 12]$
(C) $\vec{a}$ and $\vec{b}$ are non-zero vectors such that $\vert \vec{a} + \vec{b}\vert = \vert \vec{a} - \vec{b}\vert$ , then (III) $\vec{a} = \pm \vec{b}$
(D) If $\vert \vec{a}\vert = 4, -3 \le \lambda \le 2$ , then the range of $\vert \lambda \vec{a}\vert$ (IV) $(\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{b}\vert ^2}) \vec{b}$

Choose the correct answer from the options given below:

List-I	List-II
(A) If vector $\vec{a}$ and $\vec{b}$ are such that $\vec{a} = \lambda \vec{b}$ and $\vert \vec{a}\vert = \vert \vec{b}\vert$ , then	(I) $\vec{a}$ and $\vec{b}$ are orthogonal
(B) Projection vector of $\vec{a}$ on $\vec{b}$	(II) $[0, 12]$
(C) $\vec{a}$ and $\vec{b}$ are non-zero vectors such that $\vert \vec{a} + \vec{b}\vert = \vert \vec{a} - \vec{b}\vert$ , then	(III) $\vec{a} = \pm \vec{b}$
(D) If $\vert \vec{a}\vert = 4, -3 \le \lambda \le 2$ , then the range of $\vert \lambda \vec{a}\vert$	(IV) $(\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{b}\vert ^2}) \vec{b}$

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

(A) - (I), (B) - (III), (C) - (IV), (D) - (II)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (I), (B) - (III), (C) - (II), (D) - (IV)

✅ Correct Option: 3

Since $\vec{a} = \lambda\vec{b}$ , taking magnitude on both sides:

$|\vec{a}| = |\lambda||\vec{b}|$

Given $|\vec{a}| = |\vec{b}|$ :

$|\vec{b}| = |\lambda||\vec{b}|$

$|\lambda| = 1$

$\lambda = +1 \text{ or } \lambda = -1$

Therefore $\vec{a} = \pm\,\vec{b}$

(A) $\longrightarrow$ (III)

The vector projection of $\vec{a}$ on $\vec{b}$ is:

$\text{proj}_{\vec{b}}\,\vec{a} = \left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$

Note: the scalar projection is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ , but the vector projection includes direction.

(B) $\longrightarrow$ (IV)

Given $|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$ , squaring both sides:

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

$4(\vec{a}\cdot\vec{b}) = 0$

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

Since the dot product is zero, $\vec{a}$ and $\vec{b}$ are orthogonal.

$|\lambda\vec{a}| = |\lambda|\cdot|\vec{a}| = 4|\lambda|$

Since $\lambda \in [-3,\,2]$ :

At $\lambda = -3$ : $|\lambda| = 3$

At $\lambda = 0$ : $|\lambda| = 0$

At $\lambda = 2$ : $|\lambda| = 2$

So $|\lambda| \in [0,\,3]$ , which gives:

$4|\lambda| \in [0,\,12]$

(D) $\longrightarrow$ (II)

(A) $\longrightarrow$ (III)

(B) $\longrightarrow$ (IV)

(D) $\longrightarrow$ (II)

2023: 30 May Shift 3

In $\triangle ABC$ :

(A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{O}$

(B) $\vec{AB} + \vec{BC} - \vec{AC} = \vec{O}$

(D) $\vec{AB} - \vec{CB} + \vec{CA} = \vec{O}$

(E) $\vec{AB} - \vec{CB} - \vec{CA} = \vec{O}$

Choose the correct answer from the options given below :

2025: 3 June Shift 2

If $\vec{a}$ is any vector, then $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$ is equal to

2026: 14 May Shift 2

If $\vec a, \vec b, \vec c$ are mutually perpendicular vectors of equal magnitude, the angle between $\vec a+\vec b+\vec c$ and $\vec a$ is

2025: 15 May Shift 1

If $\vec{p}$ and $\vec{q}$ are two unit vectors such that $|\vec{p} + \vec{q}| = \sqrt{2}$ , then which of the following are correct?

(A) $|\vec{p}| = |\vec{q}| = 1$

(B) $\vec{p}$ and $\vec{q}$ are orthogonal vectors

(D) $(4\vec{p} - \vec{q}).(2\vec{p} + \vec{q}) = 7$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

The area (in sq. units) of the parallelogram whose diagonals are represented by the vectors $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=3\hat{i}+4\hat{j}-\hat{k}$ , is

2026: 19 May Shift 1

Let $\vec{a} = (\hat{i}+4\hat{j}+2\hat{k})$ , $\vec{b} = (3\hat{i}-2\hat{j}+7\hat{k})$ , $\vec{c} = (2\hat{i}-\hat{j}+4\hat{k})$ . A vector $\vec{d}$ , which is perpendicular to both $\vec{a}$ and $\vec{b}$ , such that $\vec{c}.\vec{d} = 18$ , is

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