If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\pi/6$,then find $|a \times b|^{2}$.

Let $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$b=7 \hat{i}+2 \hat{j}-3 \hat{k}$,and $c=\hat{i}+\hat{j}+\hat{k}$. The vector $x$ such that $x \cdot c=60$ and $x$ is perpendicular to both $a$ and $b$ is:

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$ and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $60^{\circ}$,then the value of $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be:

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is

$A$ unit vector orthogonal to the vector $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and coplanar with the vectors $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is: