Let $u$ and $v$ be two non-zero vectors. If $|u+v|=|u-v|$,then:

If $a$ and $b$ are two vectors,then $(a \times b)^2$ equals

Let $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}$,$\vec{c}=\hat{i}-3 \hat{j}-7 \hat{k}$. If $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$,$\vec{r} \cdot \vec{a}=9$,$\vec{r} \cdot \vec{b}=7$,$\vec{r} \cdot \vec{c}=6$,then $(x, y, z) = $

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$. Then $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively, then $\frac{a \times b}{a \cdot b}=$

$A, B, C, D$ are any four points,then $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $