The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

Find the angle between the following pairs of lines:
$\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

If $P=(0,1,2)$,$Q=(4,-2,1)$,and $O=(0,0,0)$,then $\angle POQ$ is equal to

If three vectors $a, b, c$ satisfy $a + b + c = 0$ and $|a| = 3, |b| = 5, |c| = 7,$ then the angle between $a$ and $b$ is .............. $^o$

Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$,then $\sin (\alpha + \beta) =$