If $a, b, c$ are position vectors of vertices of a triangle $ABC$,then the unit vector perpendicular to its plane is:

$x, y, z$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $a=x \times(y \times z), b=y \times(z \times x)$, $c=x \times y$, then $x=$

Let the vectors $\overline{a}, \overline{b}, \overline{c}$ and $\overline{d}$ be such that $(\overline{a} \times \overline{b}) \times(\overline{c} \times \overline{d})=\overline{0}$. Let $P_1$ and $P_2$ be the planes determined by the pair of vectors $\overline{a}, \overline{b}$ and $\overline{c}, \overline{d}$ respectively,then the angle between $P_1$ and $P_2$ is

Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$. Let $P_1$ and $P_2$ be planes determined by the pairs of vectors $(a, b)$ and $(c, d)$ respectively. Then the angle between $P_1$ and $P_2$ is:

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

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