If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}$ is equal to:

Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

For non-zero vectors $a$ and $b$,if $|a+b| < |a-b|$,then $a$ and $b$ are

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

In a trapezium $ABCD$,$\vec{BC} = \lambda \vec{AD}$ and $\vec{x} = \vec{AC} + \vec{BD}$. If $\vec{x} = p \vec{AD}$,then $p =$