If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

The area of the parallelogram whose adjacent sides are $\hat{i}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ is

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to

If $|\vec{a}| = 4$,$|\vec{b}| = 2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $|\vec{a} \times \vec{b}|^2 = \dots$

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

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \neq \overline{0}$ and $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,$|\overline{a}| = |\overline{c}| = 1$,$|\overline{b}| = 4$ and $|\overline{b} \times \overline{c}| = \sqrt{15}$. If $\overline{b} - 2 \overline{c} = \lambda \overline{a}$,then $\lambda$ is