$ABCD$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3:1$. If the line $BP$ meets the diagonal $AC$ in $Q$,then $AQ:QC$ equals

If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}, 2\bar{i}-\bar{j}, \bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively,then the quadrilateral $ABCD$ is a

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 $|\vec{a}|^2=$

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

$a, b, c$ are three vectors such that $a + b + c = 0$,$|a| = 1, |b| = 2, |c| = 3$. Then $a \cdot b + b \cdot c + c \cdot a$ is equal to:

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.