If $\bar{a} = \hat{i} + \hat{j}$ and $\bar{b} = 2\hat{i} - \hat{k}$,then the point of intersection of the lines $\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b} = \bar{a} \times \bar{b}$ is

If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k},$ then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

The angle between any two diagonals of a cube is

If $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ are the position vectors of points $A, B, C$ and $D$ respectively,then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)