Let $\vec a =2\hat i+\hat j-2\hat k$ and $\vec b =\hat i+\hat j$. Let $\vec c$ be a vector such that $|\vec c-\vec a=3,|(\vec a×\vec b)×\vec c|=3$ and the angle between $\vec c$ and $\vec a ×\vec b$ be 30°. Then, $\vec a.\vec c$ is equal to

2

We have, $\vec a=2\hat i+\hat j-2\hat k$ and $\vec b=\hat i+\hat j$.

$∴|\vec a|=3,|\vec b|=\sqrt{2},\vec a×\vec b=2\hat i-2\hat j+\hat k$ and $|\vec a×\vec b|=3$

Now, $|(\vec a×\vec b)×\vec c|$

$⇒|(\vec a×\vec b)||\vec c|\sin 30°=3$

$⇒\frac{3}{2}|\vec c|=3⇒|\vec c|=2$

Now,

$|\vec c-\vec a|=3$

$⇒|\vec c|^2+\vec a|^2-2(\vec a.\vec c)=9$

$⇒4+9-2(\vec a.\vec c)=9⇒\vec a.\vec c=2$