If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$, then $\sin 2 \theta=$

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 $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}.$

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

Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$, then $|\vec{a}|^2=$

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

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