What will be the length of the longer diagonal of the parallelogram constructed on vectors $5a + 2b$ and $a - 3b$,given that $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle between $a$ and $b$ is $\frac{\pi}{4}$?

If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$. Then $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$, $\vec{b}$ and $\vec{c}$ is $\beta$, and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

If $a+b+c=0$ and $|a|=5, |b|=3$ and $|c|=7$,then the angle between $a$ and $b$ 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}| = . . . . . . $.