Question

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}$. If $\overrightarrow{c}$ is a vector such that $\overrightarrow{a}\cdot \overrightarrow{c}=|\overrightarrow{c}|,|\overrightarrow{c}-\overrightarrow{a}|=2\sqrt{2}$ and the angle between $\vec{a}\times \vec{b}$ and $\vec{c}$ is $30^{\circ }$, then find value of $|(\vec{a}\times \vec{b})\times \vec{c}|$.

Answer 1

Answer: 1.50

$\vec{a}\times \vec{b}=2\hat{i}-2\hat{j}+\hat{k}$
$|\vec{a}\times \vec{b}|=3$
$|\vec{c}-\vec{a}|=2\sqrt{2}$
$|\vec{c}-\vec{a}{|}^2=8$
$|\vec{c}{|}^2+|\vec{a}{|}^2-2\vec{a}\cdot \vec{c}=8$
$|\vec{c}{|}^2+|\vec{a}{|}^2-2\vec{a}\cdot \vec{c}=8$
$|\vec{c}{|}^2+9-2|\vec{c}|=8$
$\therefore (|\vec{c}|-1{)}^2=0$
$|\vec{c}|=1$
$\therefore |(\vec{a}\times \vec{b})\times \vec{c}|$
$=|\vec{a}\times \vec{b}||\vec{c}|\sin 30^{\circ }=\frac{3}{2}$