Question

Find the image of the point with position vector $3\hat{i}+\hat{j}+2\hat{k}$ in the plane $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+\hat{k}\right)=4.$ Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3\hat{i}+\hat{j}+2\hat{k}.$

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{Q}}\text{be the image of the point}\mathit{\text{P}}\text{(3}\hat{i}\text{+}\hat{j}\text{+2}\hat{k}\text{) in the plane}\overrightarrow{r.}\left(2\hat{i}-\hat{j}+\hat{k}\right)\text{= 4} \\ \text{Since}\mathit{\text{PQ}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}\text{and is normal to the given plane, it is parallel to the normal vector}2\hat{i}-\hat{j}+\hat{k}.\mathrm{So},\mathrm{the}\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \overrightarrow{r}=\left(3\hat{i}\text{+}\hat{j}\text{+2}\hat{k}\right)+\lambda \left(2\hat{i}-\hat{j}+\hat{k}\right) \\ \text{As Q lies on PQ, let the position vector of Q be}\left(3+2\lambda \right)\hat{i}\text{+}\left(1-\lambda \right)\hat{j}\text{+}\left(2+\lambda \right)\hat{k}\text{.} \\ \text{Let}\mathit{\text{R}}\text{be the mid-point of}\mathit{\text{PQ}}\text{. Then, the position vector of R is} \\ \frac{\left[\left(3+2\lambda \right)\hat{i}\text{+}\left(1-\lambda \right)\hat{j}\text{+}\left(2+\lambda \right)\hat{k}\right]\text{+}\left[3\hat{i}\text{+}\hat{j}\text{+2}\hat{k}\right]}{2} \\ \text{=}\frac{\left(6+2\lambda \right)\hat{i}+\left(2-\lambda \right)\hat{j}+\left(4+\lambda \right)\hat{k}}{2} \\ =\left(3+\lambda \right)\hat{i}+\left(1-\frac{\lambda }{2}\right)\hat{j}+\left(2+\frac{\lambda }{2}\right)\hat{k} \\ \text{Since}\mathit{\text{R}}\text{lies in the plane}\overrightarrow{r.}\left(2\hat{i}-\hat{j}+\hat{k}\right)\text{= 4,} \\ \left[\left(3+\lambda \right)\hat{i}+\left(1-\frac{\lambda }{2}\right)\hat{j}+\left(2+\frac{\lambda }{2}\right)\hat{k}\right].\left(2\hat{i}-\hat{j}+\hat{k}\right)\text{= 4} \\ \Rightarrow 6+2\lambda -1+\frac{\lambda }{2}+2+\frac{\lambda }{2}=4 \\ \Rightarrow 7+2\lambda +\frac{\lambda }{2}+\frac{\lambda }{2}=4 \\ \Rightarrow 14+6\lambda =8 \\ \Rightarrow 6\lambda =8-14 \\ \Rightarrow \lambda =-1 \\ \mathrm{Putting}\mathrm{\lambda }=-1\mathrm{in}\mathrm{Q},\mathrm{we}\text{get} \\ Q=\left(3+2(-1)\right)\hat{i}\text{+}\left(1-(-1)\right)\hat{j}\text{+}\left(2+(-1)\right)\hat{k} \\ =\hat{i}\text{+ 2}\hat{j}+\hat{k}or(1,2,1) \\ \text{Therefore, by putting}\mathrm{\lambda }=-1\text{in}R,\text{we get} \\ R=\left(3+(-1)\right)\hat{i}+\left(1-\frac{(-1)}{2}\right)\hat{j}+\left(2+\frac{(-1)}{2}\right)\hat{k} \\ =2\hat{i}+\frac{3}{2}\hat{j}+\frac{3}{2}\hat{k} \end{gathered}$$