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

Using vectors,prove that $\cos (A-B)=\cos A \cos B+\sin A \sin B$.

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\hat{i}+\hat{j}-2 \hat{k}$ be three vectors. $A$ vector in the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ whose projection on $\overrightarrow{a}$ is of magnitude $\sqrt{\frac{2}{3}}$,is

If vectors $\vec{a}$ and $\vec{b}$ have the same magnitude, the angle between them is $60^\circ$, and their scalar product is $1/2$, then $|\vec{a}|$ is:

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

If $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ are the position vectors of points $A, B, C$ and $D$ respectively,then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.