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# The value of

Question:

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

1. $3 \sqrt{\frac{3}{2}} 2.$0<\mathrm{r} \leq \sqrt{\frac{3}{2}}$3.$\sqrt{\frac{3}{2}}

4. $r \geq 5 \sqrt{\frac{3}{2}}$

Correct Option: 4,

Solution:

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1\end{array}\right|$

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

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

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

$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \geq \sqrt{\frac{75}{2}}$

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