In quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$. If $M$ is the midpoint of $BC$ and $N$ is a point on $DM$ such that $DN=\left(\frac{4}{5}\right) DM$,then $5 AN=$

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

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

If the vector $a + b$ makes equal angles with vectors $a$ and $b$,then:

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}|=5$, $|\bar{b}|=12$ and $|\bar{a}-\bar{b}|=13$, then $|2\bar{a}+\bar{b}|=$

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$,then $|\vec{a} \times \vec{c}|$ is equal to