If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :

$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

If the projection of $\bar{a}$ on $\bar{b}+\bar{c}$ is twice the projection of $\bar{b}+\bar{c}$ on $\bar{a}$,and if $|\bar{b}|=2 \sqrt{2}$,$|\bar{c}|=4$,and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$,then $|\bar{a}|=$

If $a = (1, -1, 2)$,$b = (-2, 3, 5)$,$c = (2, -2, 4)$ and $i$ is the unit vector in the $x$-direction,then $(a - 2b + 3c) \cdot i = $

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if