Let $\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$, for some real $x$. Then $|\vec{a} \times \vec{b}|=r$ is possible if:

(A) $r \geq 5 \sqrt{\frac{3}{2}}$

(B) $3 \sqrt{\frac{3}{2}}<\mathrm{r}<5 \sqrt{\frac{3}{2}}$

(C) $\sqrt{\frac{3}{2}}<r \leq 3 \sqrt{\frac{3}{2}}$

(D) $0<r \leq \sqrt{\frac{3}{2}}$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Vector Algebra

Let

a = 3 \hat{i} + 2 \hat{j} + x \hat{k}

and

b = \hat{i} - \hat{j} + \hat{k}

, for some real

x

. Then

∣ a \times b ∣ = r

is possible if:

Answer: (A)

Compute the cross product,

a \times b = (2 + x) \hat{i} - (3 - x) \hat{j} - 5 \hat{k}

, so its magnitude squared is

(2 + x)^{2} + (3 - x)^{2} + 25 = 2 x^{2} - 2 x + 38 = 2 (x^{2} - x + 19) = 2 [(x - \frac{1}{2})^{2} + \frac{75}{4}]

. This is smallest at

x = \frac{1}{2}

, where

∣ a \times b ∣ = 2 \cdot \frac{75}{4} = \frac{75}{2} = 5 \frac{3}{2}

. Since that is the least value the magnitude can take, a length

r

is attainable only if

r \geq 5 \frac{3}{2}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option