Let $F=2 \hat{i}+2 \hat{j}+5 \hat{k}$,$A=(1,2,5)$,$B=(-1,-2,-3)$ and $BA \times F=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$,then $\lambda=$

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

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors such that $\bar{a} \cdot \bar{b} = \frac{1}{2}$,$\bar{c} \cdot \bar{d} = \frac{1}{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$,then the value of $|[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}| = $

The unit vector perpendicular to both the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ is

$r \times a = b \times a;\,\,r \times b = a \times b;\,\,a \ne 0;\,\,b \ne 0;\,\,a \ne \lambda b;\,\,a$ is not perpendicular to $b,$ then $r = $

If $a, b$ and $c$ are position vectors of the vertices of $\triangle ABC$,then $\frac{(a-c) \times (b-a)}{(b-a) \cdot (c-a)} = $