Find $|a \times b|^2$,if $|a|=2, |b|=3$ and the angle between $a$ and $b$ is $\theta = \frac{\pi}{6}$.

For vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle,passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

The number of unit vectors perpendicular to $\overline{a}=(1,1,0)$ and $\overline{b}=(0,1,1)$ is

$A$ unit vector coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is

$A$ vector perpendicular to both of the vectors $i + j + k$ and $i + j$ is