$A, B, C, D$ are any four points,then $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

If $|\vec{a}| = 2\sqrt{2}$,$|\vec{b}| = 3$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$,find the length of the longer diagonal of the parallelogram whose sides are $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$.

The values of $x$ for which the angle between the vectors $\bar{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k}$ and $\bar{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ is obtuse,are