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 $\overline{a}$ and $\overline{b}$ are two unit vectors such that $5 \overline{a} + 4 \overline{b}$ and $\overline{a} - 2 \overline{b}$ are perpendicular to each other,then the angle between $\overline{a}$ and $\overline{b}$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,$\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$,where $\alpha \in R$,then the value of $\alpha + |\overrightarrow{r}|^{2}$ is equal to:

If $U = [2, -3, 4]$,$X = [0, 2, 3]$,$V = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$,then $UV + XY = $?

If $a$ is a unit vector,then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

The figure formed by the four points $i + j - k$,$2i + 3j$,$3i + 5j - 2k$,and $k - j$ is: