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 $\frac{\pi}{4}$?

Coordinate planes and the planes $\pi_1, \pi_2, \pi_3$ which are respectively parallel to $YZ, ZX, XY$ planes at distances $a, b, c$,form a rectangular parallelepiped. $d_1$ is a diagonal of the face on the $XY$-plane not passing through the origin,and $d_2$ is a diagonal of plane $\pi_2$ coterminous with $d_1$. If none of the coordinates of the vertices of the parallelepiped are negative and the angle between $d_1$ and $d_2$ is $\theta$,then $\cos \theta=$

$a, b, c$ are three vectors such that $|a|=1, |b|=2, |c|=3$ and $b \cdot c=0$. If the projection of $b$ along $a$ is equal to the projection of $c$ along $a$,then $|2a+3b-3c|=$

Find the torque of the couple formed by forces $(9, -1, 2)$ and $(3, -2, 1)$ acting at the points $5\hat{i} + \hat{k}$ and $-5\hat{i} - \hat{k}$ respectively.

For any two vectors $\vec{a}$ and $\vec{b}$,we always have $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ (Cauchy-Schwarz inequality). Is this statement true or false?

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}$ and $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, then $\alpha + \beta$ is equal to