If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$, then the angle between $a$ and $b$ is

If $\overline{a}+\overline{b}+\overline{c}=\overline{0}$ with $|\overline{a}|=3, |\overline{b}|=5$ and $|\overline{c}|=7$,then the angle between $\overline{a}$ and $\overline{b}$ is:

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$, $\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$, $\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then $x^2+y^2=$

In a quadrilateral $ABCD$,if $P$ and $Q$ are the midpoints of $\overline{BC}$ and $\overline{AD}$ respectively,then $\vec{AB} + \vec{DC} = \dots$

If $\vec{a}$ and $\vec{b}$ are unit vectors,then what is the angle between $\vec{a}$ and $\vec{b}$ for $\sqrt{3}\vec{a} - \vec{b}$ to be a unit vector (in $^{\circ}$)?

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}=2 \hat{i}+2 \hat{j}+p \hat{k}$, $|\vec{b}|=7$, $\vec{a} \cdot \vec{b}=4$ and $|\vec{a} \times \vec{b}|=5 \sqrt{17}$, then $p=$