Mathematics - Vector Question with Solution | TestHub

$\vec{a}=2\hat{i}+\hat{j}-2\hat{k},\vec{b}=\hat{i}+\hat{j}$ $\Rightarrow |\vec{a}|=3$ and $\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\right|=2\hat{i}-2\hat{j}+\hat{k}$ $|\vec{a}\times \vec{b}|=\sqrt{4+4+1}=3$ Now, $|\vec{c}-\vec{a}|=2\sqrt{2}\Rightarrow |\vec{c}-\vec{a}{|}^2=8$ $\Rightarrow |\vec{c}-\vec{a}|\cdot (\vec{c}-\vec{a})=8$ $\Rightarrow |\vec{c}{|}^2+|\vec{a}{|}^2-2\vec{c}\cdot \vec{a}=8$ $\Rightarrow |\vec{c}{|}^2+9-2|\vec{c}|=8$ $\Rightarrow (|\vec{c}|-1{)}^2=0\Rightarrow |\vec{c}|=1$ $\therefore |(\vec{a}\times \vec{b})\times \vec{c}|=|\vec{a}\times \vec{b}||\vec{c}|\sin 30^{\circ }=3\times 1\times \frac{1}{2}=\frac{3}{2}$