The angle between the vectors $\bar{a} = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and $\bar{b} = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ is . . . . . . .

If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

Let $A=(3,4,0), B=(4,4,4), C=(-6,2,3)$ and $D=(1,1,2)$. If $\theta$ is the acute angle between the lines $AB$ and $CD$, then $\cos \theta=$

If $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$,express $\beta$ in the form $\beta=\beta_1+\beta_2$,where $\beta_1$ is parallel to $\alpha$ and $\beta_2$ is perpendicular to $\alpha$. Then $\beta_1$ is given by:

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$:

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a},$ for any two nonzero vectors $\vec{a}$ and $\vec{b}.$