The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (\hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to

Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves such that at any time $t$,the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \sin t + \hat{b} \cos t$. When $P$ is farthest from the origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$. Then:

In a parallelogram $ABCD$,the position vectors of vertices $A$ and $C$ are $3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\hat{i} - 5\hat{j} - 5\hat{k}$ respectively. If $M$ is the midpoint of the diagonal $DB$,find the projection of $\overline{OM}$ on $\overline{OC}$,where $O$ is the origin.

Let $\vec{r}$ be a vector in the plane of $\hat{i} - 2\hat{j} + \hat{k}$ and $\hat{i} - \hat{j} - \hat{k}$ such that $\vec{r} \cdot (\hat{i} + \hat{j}) + 2 = 0$ and the length of the projection of $\vec{r}$ on $\hat{i} - \hat{j}$ is $4\sqrt{2}$. Then,the magnitude of vector $\vec{r}$ is:

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.