If = = +b = 1, then Match List-I with List-II, LIST-I, LIST-II, A. -b, I. (2)/3, B. Angle between a and b, II. -1/2, C. Angle between a and a+b, III. 3, D. a.b, IV. ()/3, Choose the correct answer from the options given below: Option 1: A-III, B-I, C-IV, D-II Option 2: A-II, B-III, C-IV, D-I Option 3: A-III, B-IV, C-I, D-II Option 4: A-II, B-IV, C-III, D-I

Answer: A-III, B-I, C-IV, D-II. Solution: From 1=1+1+2 a* b, a* b=-12 (D-II), so the angle is (2)/3 (B-I)., a- b, ^2=1+1+1=3, giving 3 (A-III). a*( a+ b)=12 with both magnitudes 1, so that angle is ()/3 (C-IV).

Algebra

Vector Algebra

Hard

core

If $\lvert\vec{a}\rvert = \lvert\vec{b}\rvert = \lvert\vec{a}+\vec{b}\rvert = 1$ , then

Match List-I with List-II

LIST-I LIST-II
A. $\lvert\vec{a}-\vec{b}\rvert$ I. $\dfrac{2\pi}{3}$
B. Angle between $\vec{a}$ and $\vec{b}$ II. $-\dfrac{1}{2}$
C. Angle between $\vec{a}$ and $\vec{a}+\vec{b}$ III. $\sqrt3$
D. $\vec{a}.\vec{b}$ IV. $\dfrac{\pi}{3}$

Choose the correct answer from the options given below:

LIST-I	LIST-II
A. $\lvert\vec{a}-\vec{b}\rvert$	I. $\dfrac{2\pi}{3}$
B. Angle between $\vec{a}$ and $\vec{b}$	II. $-\dfrac{1}{2}$
C. Angle between $\vec{a}$ and $\vec{a}+\vec{b}$	III. $\sqrt3$
D. $\vec{a}.\vec{b}$	IV. $\dfrac{\pi}{3}$

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

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

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

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

✅ Correct Option: 1

From $1=1+1+2\vec a\cdot\vec b$ , $\vec a\cdot\vec b=-\dfrac12$ (D-II), so the angle is $\dfrac{2\pi}{3}$ (B-I).

$|\vec a-\vec b|^2=1+1+1=3$ , giving $\sqrt3$ (A-III).

$\vec a\cdot(\vec a+\vec b)=\dfrac12$ with both magnitudes 1, so that angle is $\dfrac{\pi}{3}$ (C-IV).

2025: 30 May Shift 1

If the points P, Q, R with position vectors $5\hat{i} + \lambda\hat{j}$ , $20\hat{i} - \hat{j}$ and $15\hat{i} - 6\hat{j}$ respectively are collinear, then the value of $\lambda$ is

2026: 19 May Shift 1

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:

2026: 29 May Shift 2

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 perpendicular is

2025: 14 May Shift 2

Match List-I with List-II

Consider two vectors $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -3\hat{i} - 6\hat{j} + 3\hat{k}$ , then

List-I	List-II
(A) Angle between $\vec{a}$ and $\vec{b}$ is	(I) $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$
(B) Angle between $\vec{a}$ and $x$ -axis is	(II) $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$
(C) Angle between $\vec{b}$ and $x$ -axis is	(III) $\pi$
(D) Angle between $\vec{a}$ and $y$ -axis is	(IV) $\cos^{-1}\left(-\frac{1}{\sqrt{6}}\right)$

Choose the correct answer from the options given below:

2026: 21 May Shift 2

Let $\vec{a}$ and $\vec{b}$ be any two vectors such that $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$ then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\vert \vec{a} \vert$	I.	$5$
B.	$\vert \vec{b} \vert$	II.	$\sqrt{507}$
C.	$\vert \vec{a}.\vec{b} \vert$	III.	$\sqrt{38}$
D.	$\vert \vec{a} \times \vec{b} \vert$	IV.	$\sqrt{14}$

Choose the correct answer from the options given below:

2026: 29 May Shift 1

If $\overrightarrow{AB}=2\hat i+3\hat j-\hat k$ and $\overrightarrow{AC}=3\hat i-2\hat j+4\hat k$ are the two sides of a triangle ABC, then the length of the median through vertex A is equal to

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