Question

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

• A. $0<r\le \sqrt{\frac{3}{2}}$
• B. $\sqrt{\frac{3}{2}}<r\le 3\sqrt{\frac{3}{2}}$
• C. $3\sqrt{\frac{3}{2}}<r<5\sqrt{\frac{3}{2}}$
• D. $r\ge 5\sqrt{\frac{3}{2}}$

Answer 1

The correct option is D $r\ge 5\sqrt{\frac{3}{2}}$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& x\\ 1& -1& 1\end{array}\right|=(x+2)\hat{i}+(x-3)\hat{j}-5\hat{k}$

$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{(x+2{)}^2+(x-3{)}^2+(-5{)}^2}$
$=\sqrt{2x^2-2x+38}$
$=\sqrt{2[{(x-\frac{1}{2})}^2+\frac{75}{4}]}$

$\therefore$ Minimum value of $|\overrightarrow{a}\times \overrightarrow{b}|$ is $\sqrt{\frac{75}{2}}=5\sqrt{\frac{3}{2}}$
$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|\ge 5\sqrt{\frac{3}{2}}$
or, $r\ge 5\sqrt{\frac{3}{2}}$