Let $\vec{a}=-\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=\hat{i}-\hat{j}-3\hat{k}$, $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}=\vec{c}\times\vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

Let $\vec{a} = \hat{i} + \hat{j}$, $\vec{c} = \hat{i} - \hat{j}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}| = $

If $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$,$\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,then find $\vec{a} \times (\vec{b} \times \vec{c})$.

If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$,then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

If $a, b, c$ are three unit vectors such that $a \times (b \times c) = \frac{\sqrt{3}}{2} b + \frac{1}{2} c$,then the angles between $a, b$ and $a, c$ respectively are

If $\vec{A} = \hat{i} - 2\hat{j} - 3\hat{k}$,$\vec{B} = 2\hat{i} + \hat{j} - \hat{k}$,and $\vec{C} = \hat{i} + 3\hat{j} - 2\hat{k}$,then $(\vec{A} \times \vec{B}) \times \vec{C} = \dots$