Match List-I with List-II, List-I, List-II, (A) (a-b) x (a+b) equals, (I) 0, (B) a = (3i+j-4k), b = (6i+5j-2k), then axb equals, (II) 5, (C) ax(b+c) + bx(c+a) + cx(a+b) =, (III) 2(axb), (D) Projection of vector (i+3j+7k) on vector (2i-3j+6k) is, (IV) 27, Choose the correct answer from the options given below: Option 1: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 2: (A) - (II), (B) - (I), (C) - (III), (D) - (IV) Option 3: (A) - (IV), (B) - (II), (C) - (III), (D) - (I) Option 4: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Answer: (A) - (III), (B) - (IV), (C) - (I), (D) - (II). Solution: (A) expands to 2(axb). (B) axb = 18i-18j+9k with magnitude 27. (C) every cross product cancels with its opposite, giving 0. (D) projection = (2-9+42)/7 = 5.

Algebra

Vector Algebra

Medium

core

Match List-I with List-II

List-I List-II
(A) $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$ equals (I) $\vec{0}$
(B) $\vec{a} = (3\hat{i}+\hat{j}-4\hat{k}), \vec{b} = (6\hat{i}+5\hat{j}-2\hat{k})$ , then $\left\vert \vec{a}\times\vec{b} \right\vert$ equals (II) $5$
(C) $\vec{a}\times(\vec{b}+\vec{c}) + \vec{b}\times(\vec{c}+\vec{a}) + \vec{c}\times(\vec{a}+\vec{b}) =$ (III) $2(\vec{a}\times\vec{b})$
(D) Projection of vector $(\hat{i}+3\hat{j}+7\hat{k})$ on vector $(2\hat{i}-3\hat{j}+6\hat{k})$ is (IV) $27$

Choose the correct answer from the options given below:

List-I	List-II
(A) $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$ equals	(I) $\vec{0}$
(B) $\vec{a} = (3\hat{i}+\hat{j}-4\hat{k}), \vec{b} = (6\hat{i}+5\hat{j}-2\hat{k})$ , then $\left\vert \vec{a}\times\vec{b} \right\vert$ equals	(II) $5$
(C) $\vec{a}\times(\vec{b}+\vec{c}) + \vec{b}\times(\vec{c}+\vec{a}) + \vec{c}\times(\vec{a}+\vec{b}) =$	(III) $2(\vec{a}\times\vec{b})$
(D) Projection of vector $(\hat{i}+3\hat{j}+7\hat{k})$ on vector $(2\hat{i}-3\hat{j}+6\hat{k})$ is	(IV) $27$

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

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

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

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

✅ Correct Option: 4

(A) expands to $2(\vec{a}\times\vec{b})$ . (B) $\vec{a}\times\vec{b} = 18\hat{i}-18\hat{j}+9\hat{k}$ with magnitude 27. (C) every cross product cancels with its opposite, giving $\vec{0}$ . (D) projection $= \frac{2-9+42}{7} = 5$ .

2024: 16 May Shift 1

If $(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})=27$ and $|\vec{a}|=2|\vec{b}|$ , then $|\vec{b}|$ is :

2025: 13 May Shift 2

The length of line segment joining the points with position vectors $2\hat{i} - 2\hat{j} + 3\hat{k}$ and $5\hat{i} + 2\hat{j} + 3\hat{k}$ is

2025: 27 May Shift 1

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:

2025: 3 June Shift 1

Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$ , $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 4\hat{k}$ . A vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ , and $\vec{c} \cdot \vec{d} = 14$ , is:

2026: 30 May Shift 1

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 (in Sq. unit) of triangle OAB is:

2025: 21 May Shift 2

If $\vec{a}$ and $\vec{b}$ are two non-zero vectors such that $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ , then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is

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