$OA$ and $OB$ are two vectors of magnitudes $5$ and $6$ respectively. If $\angle BOA = 60^{\circ}$,then $OA \cdot OB$ is equal to

What will be the length of the longer diagonal of the parallelogram constructed on vectors $5a + 2b$ and $a - 3b$,given that $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle between $a$ and $b$ is $\frac{\pi}{4}$?

Given three vectors $a=2 \hat{i}-\hat{j}+\hat{k}$,$b=\hat{i}+2 \hat{j}-\hat{k}$,and $c=\hat{i}+\hat{j}-2 \hat{k}$,a vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{\frac{2}{3}}$ is

In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$

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

For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$,if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$,then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = $ . . . . . . .