Let $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

If the vector $\vec{b} = 3\hat{j} + 4\hat{k}$ is written as the sum of a vector $\vec{b_1}$,parallel to $\vec{a} = \hat{i} + \hat{j}$ and a vector $\vec{b_2}$,perpendicular to $\vec{a}$,then $\vec{b_1} \times \vec{b_2}$ is equal to

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ is

Let $\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes passing through the origin and containing the vectors $(\hat{i}+\hat{j}, \hat{i}+\hat{k})$ and $(\hat{i}-\hat{j}, \hat{j}-\hat{k})$ respectively. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$,then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to

If $a=2\hat{i}+\hat{j}-3\hat{k}$,$b=\hat{i}-2\hat{j}+\hat{k}$,$c=-\hat{i}+\hat{j}-4\hat{k}$ and $d=\hat{i}+\hat{j}+\hat{k}$,then $|(a \times b) \times(c \times d)|=$

Let $\bar{a} = \hat{i} + \hat{j} - \hat{k}$ and $\bar{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$. If $\bar{b} \times \bar{c} = \bar{a}$,then find $|\bar{b}|$.