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
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$

Let the unit vectors $a$ and $b$ be perpendicular and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c = \alpha a + \beta b + \gamma (a \times b)$,then

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} < 0$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

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

$\bar{a}, \bar{b}, \bar{c}$ 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 $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

The vector $c$ directed along the internal bisector of the angle between the vectors $a = 7i - 4j - 4k$ and $b = -2i - j + 2k$ with $|c| = 5\sqrt{6}$ is