Let a→=3i^+2j^+xk^ and b→=i^-j^+k^, for some real x. Then the condition for a→×b→ =r to follow

Correct answer: (B) r≥ 5√32. As, We know that a→×b→= i^ j^ k^ 3 2 x 1 -1 1 =2+xi^+x-3j^-5k^ a→×b→=√(4+x^2+4x+x^2+9-6x+25) =√(2x^2-2x+38) Therefore, we have ⇒a→×b→≥√752 ⇒a→×b→≥5√32

Q47 / 90 · JEE Main 2019 (08 Apr Shift 2) · Mathematics

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 the condition for $|\vec{a} \times \vec{b}| = r$ to follow

A
$0 < r \leq \sqrt{\frac{3}{2}}$
Correct answer
B
$r \geq 5 \sqrt{\frac{3}{2}}$
C
$\sqrt{\frac{3}{2}} < r \leq 3 \sqrt{\frac{3}{2}}$
D
$3 \sqrt{\frac{3}{2}} < r < 5 \sqrt{\frac{3}{2}}$

Solution

As, We know that $\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & - 1 & 1 \end{matrix}|$
$= (2 + x) \hat{i} + (x - 3) \hat{j} - 5 \hat{k}$

$|\vec{a} \times \vec{b}| = \sqrt{4 + x^{2} + 4 x + x^{2} + 9 - 6 x + 25}$

$= \sqrt{2 x^{2} - 2 x + 38}$

Therefore, we have
$\Rightarrow |\vec{a} \times \vec{b}| \geq \sqrt{\frac{75}{2}}$

$\Rightarrow |\vec{a} \times \vec{b}| \geq 5 \sqrt{\frac{3}{2}}$

Hence, the correct answer is option: 'B'.

Let a→=3i^+2j^+xk^ and b→=i^-j^+k^, for some real x. Then the condition for a→×b→ =r to follow

Solution

Relevant Questions

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 the condition for $|\vec{a} \times \vec{b}| = r$ to follow