If the volume of a parallelepiped, whose coterminus edges are given by the vectors $\vec{a}=\hat{i}+\hat{j}+n \hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-n \hat{k}$ and $\vec{c}=\hat{i}+n \hat{j}+3 \hat{k}(n \geq 0)$, is 158 cu.units, then:

(C) $\overrightarrow{b} \cdot \overrightarrow{c}=10$

(A) $\overrightarrow{a} \cdot \overrightarrow{c}=14$

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Vector Algebra

If the volume of a parallelepiped, whose coterminus edges are given by the vectors

a = \hat{i} + \hat{j} + n \hat{k}, b = 2 \hat{i} + 4 \hat{j} - n \hat{k}

and

c = \hat{i} + n \hat{j} + 3 \hat{k} (n \geq 0)

, is 158 cu.units, then:

Answer: (C)

The volume is the magnitude of the scalar triple product,

121 14 n n - n 3 = 1 (12 + n^{2}) - 1 (6 + n) + n (2 n - 4) = 3 n^{2} - 5 n + 6

, set equal to

158

. Solving

3 n^{2} - 5 n - 152 = 0

gives the non-negative root

n = 8

. With

b = 2 \hat{i} + 4 \hat{j} - 8 \hat{k}

and

c = \hat{i} + 8 \hat{j} + 3 \hat{k}

, then

b \cdot c = (2) (1) + (4) (8) + (- 8) (3) = 2 + 32 - 24 = 10

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option