Question

Find a vector of magnitude $5$ units which is coplanar with vectors $3\hat{i}-\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}-2\hat{k}$ and is perpendicular to the vector $2\hat{i}+2\hat{j}+\hat{k}$.

Answer 1

Let $\overrightarrow{a}=3\hat{i}-\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}-2\hat{k}and\overrightarrow{c}=2\hat{i}+2\hat{j}+\hat{k}$

Any vector coplaner with vectors $\overrightarrow{a}and\overrightarrow{b}$ is

$\begin{array}{c}\overrightarrow{r}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}where\lambda ,\mu aresomescalars.\\ \overrightarrow{r}=\lambda \left(3\hat{i}-\hat{j}-\hat{k}\right)+\mu \left(\hat{i}+\hat{j}-2\hat{k}\right)\\ \overrightarrow{r}=\left(3\lambda +\mu \right)\hat{i}+\left(-\lambda +\mu \right)\hat{j}+\left(-\lambda -2\mu \right)\hat{k}.....\left(1\right)\\ As\overrightarrow{r}isperpendicularto\overrightarrow{c},\overrightarrow{r}\cdot \overrightarrow{c}=0\\ \Rightarrow \left(3\lambda +\mu \right)\cdot 2+\left(-\lambda +\mu \right)\cdot 2+\left(-\lambda -2\mu \right)\cdot 1=0\\ \Rightarrow 3\lambda +2\mu =0\\ \Rightarrow \mu =-\frac{3}{2}\lambda \\ \end{array}$

substituting this value of $\mu$ in (1), we get

$\begin{array}{c}\overrightarrow{r}=\frac{3}{2}\lambda \hat{i}-\frac{5}{2}\lambda \hat{j}+2\lambda \hat{k}.......\left(2\right)\\ As\overrightarrow{r}isofmagnitude5units,\\ {\left(\frac{3}{2}\lambda \right)}^2+{\left(\frac{-5}{2}\lambda \right)}^2+{\left(2\lambda \right)}^2=5^2\\ \left(\frac{9}{4}+\frac{25}{4}+4\right){\lambda }^2=25\\ \frac{25}{2}{\lambda }^2=25\\ {\lambda }^2=2\\ \lambda =\pm \sqrt{2}\end{array}$

Substituting these values in eq(2)we get required vectors

$\begin{array}{c}=\pm \sqrt{2}\left(\frac{3}{2}\hat{i}-\frac{5}{2}\hat{j}+2\hat{k}\right)\\ =\pm \frac{1}{\sqrt{2}}\left(3\hat{i}-5\hat{j}+4\hat{k}\right)\end{array}$