# Vectors Questions

## Find dot product of vectors $$\vec{a}$$ = $$2\hat{i}+2\hat{j}-\hat{k}$$ and $$\vec{b}$$ = $$6\hat{i}-3\hat{j}+2\hat{k}$$

Solution : We have $$\vec{a}$$ = $$2\hat{i}+2\hat{j}-\hat{k}$$ and $$\vec{b}$$ = $$6\hat{i}-3\hat{j}+2\hat{k}$$ $$\vec{a}$$.$$\vec{b}$$ = ($$2\hat{i}+2\hat{j}-\hat{k}$$).($$6\hat{i}-3\hat{j}+2\hat{k}$$) = (2)(6) + (2)(-3) + (-1)(2) = 12 – 6 – 2 = 4 Similar Questions Find the angle between the vectors with the direction ratios proportional to 4, -3, 5 and 3, 4, 5. Find the vector equation of a …

## For any three vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ prove that [$$\vec{a}$$ + $$\vec{b}$$ $$\vec{b}$$ + $$\vec{c}$$ $$\vec{c}$$ + $$\vec{a}$$] = 2[$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$]

Solution : We have [$$\vec{a}$$ + $$\vec{b}$$ $$\vec{b}$$ + $$\vec{c}$$ $$\vec{c}$$ + $$\vec{a}$$] = {($$\vec{a}$$ + $$\vec{b}$$)$$\times$$($$\vec{b}$$ + $$\vec{c}$$)}.($$\vec{c}$$ + $$\vec{a}$$) = {$$\vec{a}$$$$\times$$$$\vec{b}$$ + $$\vec{a}$$$$\times$$$$\vec{c}$$ + $$\vec{b}$$$$\times$$$$\vec{b}$$ + $$\vec{b}$$$$\times$$$$\vec{c}$$}.($$\vec{c}$$ + $$\vec{a}$$)  {$$\vec{b}$$$$\times$$$$\vec{b}$$ = 0} = {$$\vec{a}$$$$\times$$$$\vec{b}$$ + $$\vec{a}$$$$\times$$$$\vec{c}$$ + $$\vec{b}$$$$\times$$$$\vec{c}$$}.($$\vec{c}$$ + $$\vec{a}$$) = ($$\vec{a}\times\vec{b}$$).$$\vec{c}$$ + ($$\vec{a}\times\vec{c}$$).$$\vec{c}$$ + ($$\vec{b}\times\vec{c}$$).$$\vec{c}$$ + ($$\vec{a}\times\vec{b}$$).$$\vec{a}$$ + ($$\vec{a}\times\vec{c}$$).$$\vec{a}$$ + ($$\vec{b}\times\vec{c}$$).$$\vec{a}$$ = …

## If $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are three non zero vectors such that $$\vec{a}\times\vec{b}$$ = $$\vec{c}$$ and $$\vec{b}\times\vec{c}$$ = $$\vec{a}$$, prove that $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are mutually at right angles and |$$\vec{b}$$| = 1 and |$$\vec{c}$$| = |$$\vec{a}$$|

Solution : $$\vec{a}\times\vec{b}$$ = $$\vec{c}$$ and $$\vec{b}\times\vec{c}$$ = $$\vec{a}$$ $$\implies$$  $$\vec{c}\perp\vec{a}$$ , $$\vec{c}\perp\vec{b}$$ and $$\vec{a}\perp\vec{b}$$, $$\vec{a}\perp\vec{c}$$ $$\implies$$  $$\vec{a}\perp\vec{b}$$, $$\vec{b}\perp\vec{c}$$ and $$\vec{c}\perp\vec{a}$$ $$\implies$$  $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are mutually perpendicular vectors. Again, $$\vec{a}\times\vec{b}$$ = $$\vec{c}$$ and $$\vec{b}\times\vec{c}$$ = $$\vec{a}$$ $$\implies$$ |$$\vec{a}\times\vec{b}$$| = |$$\vec{c}$$| and |$$\vec{b}\times\vec{c}$$| = |$$\vec{a}$$| $$\implies$$  $$|\vec{a}||\vec{b}|sin{\pi\over 2}$$ = |$$\vec{c}$$| and $$|\vec{b}||\vec{c}|sin{\pi\over 2}$$ = |$$\vec{a}$$|  …

## Find the vector of magnitude 5 which are perpendicular to the vectors $$\vec{a}$$ = $$2\hat{i} + \hat{j} – 3\hat{k}$$ and $$\vec{b}$$ = $$\hat{i} – 2\hat{j} + \hat{k}$$

Solution : Unit vectors perpendicular to $$\vec{a}$$ & $$\vec{b}$$ = $$\pm$$$$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$$ $$\therefore$$  $$\vec{a}\times\vec{b}$$ = $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}$$ = $$-5\hat{i} – 5\hat{j} – 5\hat{k}$$ $$\therefore$$ Unit Vectors = $$\pm$$ $$-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}$$ Hence the required …