If $\overline{a}, \overline{b}, \overline{c}$ are unit vectors and $\theta$ is the angle between $\overline{a}$ and $\overline{c}$ and $\overline{a}+2 \overline{b}+2 \overline{c}=\overline{0}$,then $|\overline{a} \times \overline{c}|=$

Let $ABC$ be an acute scalene triangle,and $O$ and $H$ be its circumcentre and orthocentre respectively. Further,let $N$ be the mid-point of $OH$. The value of the vector sum $\overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}$ is

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then: