If $p$-th, $q$-th, and $r$-th terms of a geometric progression are the positive numbers $a, b,$ and $c$ respectively, then the angle between the vectors $(\log a^2) i + (\log b^2) j + (\log c^2) k$ and $(q-r) i + (r-p) j + (p-q) k$ is

The component of $\vec{i} + \vec{j}$ along $\vec{j} + \vec{k}$ is:

Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2},$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular,given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}.$

Let $\vec{a} = 2\hat{i} + \lambda_{1}\hat{j} + 3\hat{k}$,$\vec{b} = 4\hat{i} + (3 - \lambda_{2})\hat{j} + 6\hat{k}$,and $\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_{3} - 1)\hat{k}$ be three vectors such that $\vec{b} = 2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. Then a possible value of $(\lambda_{1}, \lambda_{2}, \lambda_{3})$ is

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 $|\overline{a} \times \overline{c}|=$

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?