If $\overrightarrow{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{B} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{C} = 3\hat{i} + \hat{j}$,then the value of $t$ such that $\overrightarrow{A} + t\overrightarrow{B}$ is at a right angle to vector $3\hat{i} + 4\hat{j}$ is

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a},$ for any two nonzero vectors $\vec{a}$ and $\vec{b}.$

If $\overline{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}+3 \hat{j}$ are such that $(\overline{a}+\lambda \overline{b})$ is perpendicular to $\overline{c}$,then the value of $\lambda$ is

$M$ and $N$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$ respectively,then $\overline{AM} + \overline{AN} =$

For any two vectors $\vec{a}$ and $\vec{b}$,we always have $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ (Cauchy-Schwarz inequality). Is this statement true or false?

$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, -\bar{i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+\bar{k}$ are the position vectors of the points $A, B, C, D$ respectively. If $p \bar{i}+q \bar{j}+r \bar{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $ABCD$, then $p+q+r=$