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

If $ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $,then the angle between $ \vec{a} $ and $ \vec{b} $ is: (in $^{\circ}$)

Let $\bar{a}$ and $\bar{b}$ be two non-collinear unit vectors. If $\bar{u}=\bar{a}-(\bar{a} \cdot \bar{b}) \bar{b}$ and $\bar{v}=\bar{a} \times \bar{b}$,then $|\bar{v}|=$

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices $A, B, C$ of a triangle respectively. Find the area of the triangle.