If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$,then $|\vec{x}| = $ . . . . . . .

If $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$ and $|\vec{a} \times \vec{b}| = 35$,then $\vec{a} \cdot \vec{b}$ is equal to

The area of the rectangle having vertices $P, Q, R, S$ with position vectors $-\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}, -\hat{i}-\hat{j}+\hat{k}$ respectively is

If $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$,$\bar{b}=\hat{i}-\hat{j}+\hat{k}$,and $\bar{c}=\hat{i}+\hat{j}-\hat{k}$,then a vector in the plane of $\bar{a}$ and $\bar{b}$,whose projection on $\bar{c}$ is $\frac{1}{\sqrt{3}}$,is

Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{|\vec{a} \times \vec{b}|}{2}$. Also,find the area of the parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}-\hat{k}$.