Find the angle between the following pairs of lines:
$\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and
$\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ is:

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ are given vectors. If $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,then $\lambda = . . . . . .$.

If with reference to the right-handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}$,$\vec{\alpha} = 3\hat{i} - \hat{j}$ and $\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}$,then express $\vec{\beta}$ in the form $\vec{\beta} = \vec{\beta}_{1} + \vec{\beta}_{2}$,where $\vec{\beta}_{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$.