Question

Match List-I with List-II:List-I	List-II
(A) 4î − 2ĵ − 4k̂	(I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂
(B) 4î − 4ĵ + 2k̂	(II) Direction ratios are −2, 1, 2
(C) 2î − 4ĵ + 4k̂	(III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6)
(D) 4î − ĵ − 2k̂	(IV) Dot product with −2î + ĵ + 3k̂ is 10
Choose the correct answer from the options given below:

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

Answer 1

Answer: (B)

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

Answer 2

(A): $4\hat{i} - 2\hat{j} - 4\hat{k}$ has direction ratios $4, -2, -4$. Simplifying these gives the ratios $-2, 1, 2$, which matches with (II).

(B): $4\hat{i} - 4\hat{j} + 2\hat{k}$. To find the dot product with $-2\hat{i} + \hat{j} + 3\hat{k}$:

Dot product = $4(-2) + (-4)(1) + 2(3) = -8 - 4 + 6 = -6$.

This matches with (IV).

(C): $2\hat{i} - 4\hat{j} + 4\hat{k}$ makes an angle with $\hat{i} - 2\hat{j} - \hat{k}$. The angle between two vectors is given by:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}.$

After computation, this matches with (III).

(D): $4\hat{i} - \hat{j} - 2\hat{k}$ is perpendicular to both $\hat{i} + 2\hat{j} + \hat{k}$ and $2\hat{i} + 2\hat{j} + 3\hat{k}$. This matches with (I).