In a triangle $ABC,$ if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10,$ then the projection of the vector $\overline{AB}$ on $\overline{AC}$ is equal to ....... .

Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is:

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

If $\bar{a}, \bar{b}, \bar{c}$ are perpendicular to $\bar{b}+\bar{c}, \bar{c}+\bar{a}$ and $\bar{a}+\bar{b}$ respectively and $|\bar{a}+\bar{b}|=2, |\bar{b}+\bar{c}|=6, |\bar{c}+\bar{a}|=4$,then $|\bar{a}+\bar{b}+\bar{c}|=$

If $a, b, c$ are mutually perpendicular unit vectors,then $|a + b + c| = $

The projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is: