The orthogonal projection of vector $\vec{a}$ on vector $\vec{b}$ is:

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively,having $\vec{a} \cdot \vec{b} = \sqrt{6}$.

If vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ and vector $\vec{b} = -2\hat{i} + 2\hat{j} - \hat{k},$ then $\frac{\text{Projection of vector } \vec{a} \text{ on vector } \vec{b}}{\text{Projection of vector } \vec{b} \text{ on vector } \vec{a}} = $

If $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$,then $|b|$ is equal to

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}|$ and the angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$. If $\frac{1}{8} \vec{a}$ is a unit vector,then $|\vec{b}|$ is equal to :