If $\vec{a}$ and $\vec{b}$ are unit vectors,then what is the angle between $\vec{a}$ and $\vec{b}$ for $\sqrt{3}\vec{a} - \vec{b}$ to be a unit vector (in $^{\circ}$)?

Let $\vec a = \hat i - \hat j,$ $\vec b = \hat i + \hat j + \hat k$ and $\vec c$ be a vector such that $\vec a \times \vec c + \vec b = 0$ and $\vec a \cdot \vec c = 4$,then ${\left| {\vec c} \right|^2}$ is equal to

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$

In a triangle $ABC$ with usual notations,if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10$,then the projection of $\overline{AB}$ on $\overline{AC}$ is

If $a, b, c$ are non-coplanar vectors,then for what value of $m$ are the three points with position vectors $-2b + 3c$,$2a + mb - 4c$,and $-7b + 10c$ collinear?

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$,find the number of possible vectors $\vec{b}$ such that $\vec{a} \cdot \vec{b} = 10$,where $(x, y, z) \in \mathbb{N}$.