Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $\frac{1}{8} \vec{a}$ is a unit vector,then $|\vec{b}|$ is equal to :

If $a \cdot b = 0$ and $a + b$ makes an angle $60^{\circ}$ with $a$,then

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,and $|\vec{a} + \vec{b} + \vec{c}| = 1$,then the angle between $\vec{b}$ and $\vec{c}$ is:

If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$, then $\sin 2 \theta=$

Let $u = -2 \hat{i} + 2 \hat{j} + \hat{k}$ and $v = \hat{i} - 2 \hat{j} + 2 \hat{k}$. Then the angle between $u$ and $v$ is

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$ be three vectors. If $\vec{v}$ is a vector in the plane of $\vec{a}$ and $\vec{b}$ such that the projection of $\vec{v}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,then $\vec{v} = $