Vectors Questions

Solution : We know that the vector equation of line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is, $\vec{r}$ = $\lambda$ $(\vec{b} – \vec{a})$ Here $\vec{a}$ = $3\hat{i} + 4\hat{j} – 7\hat{k}$ and $\vec{b}$ = $\hat{i} – \hat{j} + 6\hat{k}$. So, the vector equation of the required line is $\vec{r}$ = (\(3\hat{i} […]

Solution : We have, $\vec{a}$ = $4\hat{i} – 3\hat{j} + 5\hat{k}$ and $\vec{b}$ = $3\hat{i} + 4\hat{j} + 5\hat{k}$ Let $\theta$ is the angle between the given vectors. Then, cos$\theta$ = $\vec{a}.\vec{b}\over |\vec{a}||\vec{b}|$ $\implies$ cos$\theta$ = $12 – 12 + 25\over \sqrt{16 + 9 + 25} \sqrt{16 + 9 + 25}$ = $1\over 2$ $\implies$

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

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}$ + $\vec{a}$) = ($\vec{a}\times\vec{b}$).$\vec{c}$ + ($\vec{a}\times\vec{c}$).$\vec{c}$ + ($\vec{b}\times\vec{c}$).$\vec{c}$ + ($\vec{a}\times\vec{b}$).$\vec{a}$ + ($\vec{a}\times\vec{c}$).$\vec{a}$ + ($\vec{b}\times\vec{c}$).$\vec{a}$ =

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 $|\vec{b}||\vec{c}|sin{\pi\over 2}$ = |$\vec{a}$| 

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