Let $\vec a = 2\hat i +\hat j-2\hat k$ and $\vec b =\hat i+\hat j$. If $\vec c$ is a vector such that $\vec a. \vec c=|\vec c|,|\vec c-\vec a|=2\sqrt{2}$ and the angle between $\vec a×\vec b$ and $\vec c$ is 30°, then $|(\vec a×\vec b)×\vec c|=$
Answer & explanation
Correct answer: option 2
We have,
$\vec a. \vec c=|\vec c|$ and $|\vec c-\vec a|=2\sqrt{2}$
$⇒\vec a. \vec c=|\vec c|$ and $|\vec c|^2+|\vec a|^2-2(\vec a. \vec c)=8$
$⇒|\vec c|^2+9-2|\vec c|=8⇒(|\vec c|-1)^2=0⇒|\vec c|=1$
Now,
$|(\vec a×\vec b)×\vec c|=|\vec a×\vec b||\vec c|\sin 30°$
$⇒|(\vec a×\vec b)×\vec c|=\frac{1}{2}|\vec a×\vec b|=\frac{3}{2}$ $[∵\vec a×\vec b=2\hat i-2\hat j+\hat k]$