Question

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the point}\mathit{\text{P}}\text{(2, 3, 7) in the plane}3x-y-z=7. \\ \text{Then,}\mathit{\text{PM}}\text{\hspace{0.17em}is normal to the plane. So, the direction ratios of}\mathit{\text{PM}}\text{are proportional to 3, -1, -1.} \\ \text{Since}\mathit{\text{PM}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}\text{(2, 3, 7) and has direction ratios proportional to}\mathrm{3,\; -1}and-1,\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z-7}{-1}=r\text{(say)} \\ \text{Let the coordinates of}\mathit{\text{M}}\text{be}\left(3r+2,-r+3,-r+7\right)\text{.} \\ \text{Since}\mathit{\text{M}}\text{lies in the plane}3x-y-z=7, \\ 3\left(3r+2\right)-\left(-r+3\right)-\left(-r+7\right)=7 \\ \Rightarrow 9r+6+r-3+r-7=7 \\ \Rightarrow 11r=11 \\ \Rightarrow r=1 \\ \text{Substituting this in the coordinates of}\mathit{\text{M}}\text{, we get} \\ M=\left(3r+2,-r+3,-r+7\right)=\left(3\left(1\right)+2,-1+3,-1+7\right)=\left(5,2,6\right) \\ \text{Now, the length of the perpendicular from}\mathit{\text{P}}\text{onto the given plane} \\ =\frac{\left|3\left(2\right)-3-7-7\right|}{\sqrt{9+1+1}} \\ =\frac{11}{\sqrt{11}} \\ =\sqrt{11}\text{units} \end{gathered}$$