Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ where $|\bar{a}|=4, |\bar{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$,then $|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$ has the value

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

Let $p=\hat{i}+2\hat{j}-\hat{k}$ and $q=2\hat{i}-\hat{j}+\hat{k}$. If $a$ and $b$ are two vectors such that $p=a-2b$ and $q=2a+b$, then the angle between $a$ and $b$ is

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=3 \hat{i}-\hat{j}+2 \hat{k}$,then find the angle between the vectors $2 a+b$ and $a+2 b$.