Let $\vec a=2=2\hat i+\hat j-2\hat k$ 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, $|(\vec a ×\vec b)×\vec c|$ is equal to
Answer & explanation
Correct answer: option 2
We have, $\vec a ×\vec b=2\hat i-2\hat j+\hat k$
$∴|(\vec a ×\vec b)×\vec c|=|\vec a ×\vec b||\vec c|\sin 30°=\frac{3}{2}|\vec c|$
Now,
$|\vec c-\vec a|=2\sqrt{2}$
$⇒|\vec c-\vec a|^2=8$
$⇒|\vec c|^2+|\vec a|^2-2(\vec a.\vec c)=8$
$⇒|\vec c|^2+9-2|\vec c|=8$
so $|\vec c|^2-2|\vec c|+1=0$
$|\vec c|=1$
Hence, $|(\vec a ×\vec b)×\vec c|=\frac{3}{2}$