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

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$,then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 2$, $|\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 4$, then $|\vec{a} - \vec{b}| = . . . . . . $.

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.

If $x$ and $y$ are two unit vectors and the angle between them is $\phi$,then $\frac{1}{2} |x - y| = $

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = \dots$