$x, y, z$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $a=x \times(y \times z), b=y \times(z \times x)$,$c=x \times y$,then $x=$

If $a = (1, 1, 1)$ and $c = (0, 1, -1)$ are two vectors and $b$ is a vector such that $a \times b = c$ and $a \cdot b = 3$,then $b$ is equal to

Vectors $\bar{a}$ and $\bar{b}$ are such that $|\bar{a}|=1$,$|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

The area of the triangle,whose vertices are $A \equiv(1,-1,2)$,$B \equiv(2,1,-1)$ and $C \equiv(3,-1,2)$,is

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.