If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{2}$ and $\vec{a} \cdot \vec{b}=-1$,then the angle between $\vec{a}$ and $\vec{b}$ is

Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$ is equal to

Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.

If $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

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: