[a + b,\ b + c, a + b + c] is equal to Option 1: -[a, b, c] Option 2: 0 Option 3: 2[a, b, c] Option 4: [a, b, c]

Answer: [a, b, c]. Solution: The scalar triple product equals the determinant of the coefficient matrix times [a,b,c]: vmatrix 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 vmatrix = 1(1-1) - 1(0-1) + 0 = 1. So the expression equals 1 * [a,b,c] = [a,b,c].

Algebra

Vector Algebra

Hard

$[\vec{a} + \vec{b},\ \vec{b} + \vec{c}, \vec{a} + \vec{b} + \vec{c}]$ is equal to

$-[\vec{a}, \vec{b}, \vec{c}]$

$2[\vec{a}, \vec{b}, \vec{c}]$

$[\vec{a}, \vec{b}, \vec{c}]$

✅ Correct Option: 4

The scalar triple product equals the determinant of the coefficient matrix times $[\vec{a},\vec{b},\vec{c}]$ :

$\begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 1(1-1) - 1(0-1) + 0 = 1$ .

So the expression equals $1 \cdot [\vec{a},\vec{b},\vec{c}] = [\vec{a},\vec{b},\vec{c}]$ .

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
(A) The vectors $\lambda\hat{i}$ $+ \hat{j} + 2\hat{k}$ and $\vec{i} + \lambda\hat{j} + \hat{k}$ are perpendicular if λ is equal to	(I) 1
(B) The vectors $3\hat{i} + 6\hat{j} - \hat{k}$ and $2\hat{i} + 4\hat{j} - \lambda\hat{k}$ are collinear if λ is equal to	(II) -1
(C) The number of vectors of unit-length which are perpendicular to both the vectors $\vec{a}$ = $\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b}$ = $3\hat{i} - \hat{j} + 5\hat{k}$ is	(III) 2/3
(D) If $\lvert \vec{a} \rvert = 1$ and $\vec{a} + \vec{b} = \vec{0}$ , then $\lvert \vec{b} \rvert$ is equal to	(IV) 2

Choose the correct answer from the options given below:

2026: 25 May Shift 1

If $\lvert\vec{a}\rvert = 4, \lvert\vec{b}\rvert = 3$ and $\lvert\vec{a}+\vec{b}\rvert = 6$ , then $\lvert\vec{a}-\vec{b}\rvert =$

2025: 26 May Shift 2

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

2026: 22 May Shift 1

If $|\vec a|=2, |\vec b|=7$ and $\vec a\times\vec b=3\hat i+2\hat j+6\hat k$ , then angle between $\vec a$ and $\vec b$ is:

2026: 18 May Shift 2

There are two unit vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}+\vec{b}|=\sqrt{3}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ , then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\vert \vec{a} - \vec{b} \vert$	I.	$0$
B.	$\cos\theta$	II.	$1$
C.	$\theta$	III.	$\dfrac{1}{2}$
D.	$\sin 3\theta$	IV.	$\dfrac{\pi}{3}$

Choose the correct answer from the options given below:

2025: 14 May Shift 1

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$ , then the angle between $\vec{a}$ and $\vec{b}$ is

[a⃗+b⃗, b⃗+c⃗,a⃗+b⃗+c⃗][\vec{a} + \vec{b},\ \vec{b} + \vec{c}, \vec{a} + \vec{b} + \vec{c}][a+b, b+c,a+b+c] is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$[\vec{a} + \vec{b},\ \vec{b} + \vec{c}, \vec{a} + \vec{b} + \vec{c}]$ is equal to