Let a = i + 4j + 2k, b = 3i - 2j + 7k and c = 2i + j + 4k. A vector d which is perpendicular to both a and b, and c * d = 14, is: Option 1: 64i + 2j - 28k Option 2: 64i - 2j - 28k Option 3: 64i + 2j + 28k Option 4: 32i + j + 14k

Answer: 64i - 2j - 28k. Solution: If d is perpendicular to both a and b, then d must be parallel to a x b, so: d = (a x b) a x b = vmatrix i & j & k \\ 1 & 4 & 2 \\ 3 & -2 & 7 vmatrix = i(28 + 4) - j(7 - 6) + k(-2 - 12) a x b = 32i - j - 14k d = (32i - j - 14k) Using c * d = 14 with c = 2i + j + 4k: c * d = [(2)(32) + (1)(-1) + (4)(-14)] = (64 - 1 - 56) = 7 7 = 14 = 2 d = 2(32i - j - 14k) d = 64i - 2j - 28k

Algebra

Vector Algebra

Medium

core

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:

$64\hat{i} + 2\hat{j} - 28\hat{k}$

$64\hat{i} - 2\hat{j} - 28\hat{k}$

$64\hat{i} + 2\hat{j} + 28\hat{k}$

$32\hat{i} + \hat{j} + 14\hat{k}$

✅ Correct Option: 2

If $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ , then $\vec{d}$ must be parallel to $\vec{a} \times \vec{b}$ , so:

$\vec{d} = \lambda(\vec{a} \times \vec{b})$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix}$

$= \hat{i}(28 + 4) - \hat{j}(7 - 6) + \hat{k}(-2 - 12)$

$\vec{a} \times \vec{b} = 32\hat{i} - \hat{j} - 14\hat{k}$

$\vec{d} = \lambda(32\hat{i} - \hat{j} - 14\hat{k})$

Using $\vec{c} \cdot \vec{d} = 14$ with $\vec{c} = 2\hat{i} + \hat{j} + 4\hat{k}$ :

$\vec{c} \cdot \vec{d} = \lambda\big[(2)(32) + (1)(-1) + (4)(-14)\big]$

$= \lambda(64 - 1 - 56)$

$= 7\lambda$

$7\lambda = 14$

$\lambda = 2$

$\vec{d} = 2(32\hat{i} - \hat{j} - 14\hat{k})$

$\vec{d} = 64\hat{i} - 2\hat{j} - 28\hat{k}$

2026: 22 May Shift 2

A unit vector perpendicular to both the vectors $4\hat{i}-\hat{j}+3\hat{k}$ and $-2\hat{i}+\hat{j}-2\hat{k}$ is:

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:

2025: 13 May Shift 1

If $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular unit vectors, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

2022: 6 Aug Shift 2

The value of $\hat{i} \cdot (\hat{k} \times \hat{j}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{j} \times \hat{i})$ is

2022: 6 Aug Shift 2

Let $\vec{OA} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{OB} = \hat{i} + \hat{j} - \hat{k}$ . Then

A. The magnitude of vector $\vec{OA}$ is 6

B. The magnitude of vector $\vec{OB}$ is $\sqrt{3}$

C. The vector $\vec{AB}$ is $(-\hat{i} + 2\hat{j} - 2\hat{k})$

D. $\vec{OA} \cdot \vec{OB} = 0$

E. $\vec{OA} \parallel \vec{OB}$

Choose the correct answer from the options given below:

2026: 19 May Shift 1

If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following statements is/are incorrect?

(A) $\vec{b} = \lambda\vec{a}$ , for some scalar $\lambda$ .

(B) Always $\vec{a} = \pm\vec{b}$ .

(D) both vectors $\vec{a}$ and $\vec{b}$ always have same direction.

Choose the correct answer from the options given below:

Solution

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