$\vec{c}=\hat{j}-\hat{k} \Rightarrow(\text { Given })|\bar{c}| \sqrt{2}$
Now, $\vec{a} \times \vec{b}=\vec{c}$
$\Rightarrow|\vec{a}||\vec{b}| \sin \theta=|\vec{c}|$
$\Rightarrow|\vec{a}||\vec{b}| \sin \theta=\sqrt{2}$ ........$[i]$
Also $\vec{a} \cdot \vec{b}=3$
$\Rightarrow|\vec{a}||\vec{b}| \cos \theta=3 $ ........$[ii]$
Dividing $[i]$ by $[ii],$ we get
$\tan \theta=\frac{\sqrt{2}}{3}$
$\therefore \sin \theta=\frac{\sqrt{2}}{\sqrt{11}}$
Substituting value of $\sin \theta$ in $[i]$ we get
$\sqrt{3}|\vec{b}| \frac{\sqrt{2}}{\sqrt{11}}=\sqrt{2}$
$|\vec{b}|=\frac{\sqrt{11}}{\sqrt{3}}$
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