If $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$,then $|b|$ is equal to

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

In a $\triangle ABC$,let $G$ denote its centroid and let $M, N$ be points in the interiors of the segments $AB, AC$,respectively,such that $M, G, N$ are collinear. If $r$ denotes the ratio of the area of $\triangle AMN$ to the area of $\triangle ABC$,then

The angle between the vectors $2 \hat{k} - 3 \hat{j}$ and $\hat{i} - 2 \hat{k}$ is

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

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 $C$: