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 :
Correct Option: , 4
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1\end{array}\right|$
$=(2+x) \hat{i}+(x-3) \hat{j}-5 k$
$|\vec{a} \times \vec{b}|=\sqrt{4+x^{2}+4 x+x^{2}+9-6 x+25}$
$|\vec{a} \times \vec{b}|=\sqrt{4+x^{2}+4 x+x^{2}+9-6 x+25}$
$=\sqrt{2 x^{2}-2 x+38}$
$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \geq \sqrt{\frac{75}{2}}$
$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \geq 5 \sqrt{\frac{3}{2}}$
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