Given vertices $A(3, -1)$,$B(2, 3)$,and $C(5, 1)$,find $m \angle A$.

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.

Let $\vec{u} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{v} = -3\hat{i} + 2\hat{j}$ and $\vec{w} = \hat{i} - \hat{j} + 4\hat{k}$. Then which of the following statements is true?

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by: