Question

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the point}\mathit{\text{P}}\text{(3, 2, 1) in the plane}2x-y+z+1=0. \\ \text{Then,}\mathit{\text{PM}}\text{\hspace{0.17em}is the normal to the plane. So, the direction ratios of}\mathit{\text{PM}}\text{are proportional to 2, -1, 1.} \\ \text{Since}\mathit{\text{PM}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}(3,\; 2,\; 1)\text{and has direction ratios proportional to}\mathrm{2,\; -}1,1, \\ \text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}=r\text{(say)} \\ \text{Let the coordinates of}\mathit{\text{M}}\text{be}\left(2r+3,-r+2,r+1\right)\text{.} \\ \text{Since}\mathit{\text{M}}\text{lies in the plane}2x-y+z+1=0, \\ 2\left(2r+3\right)-\left(-r+2\right)+r+1+1=0 \\ \Rightarrow 4r+6+r-2+r+2=0 \\ \Rightarrow 6r=-6 \\ \Rightarrow r=-1 \\ \text{Substituting the value of r in the coordinates of}\mathit{\text{M}}\text{, we get} \\ M=\left(2r+3,-r+2,r+1\right)=\left(2\left(-1\right)+3,-\left(-1\right)+2,-1+1\right)=\left(1,3,0\right) \\ \text{Now, the length of the perpendicular from}\mathit{\text{P}}\text{onto the given plane} \\ =\frac{\left|2\left(3\right)-\left(2\right)+\left(1\right)+1\right|}{\sqrt{4+1+1}} \\ =\frac{6}{\sqrt{6}} \\ =\sqrt{6}\text{units} \end{gathered}$$