Question

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the point}\mathit{\text{P}}\text{(1, 1, 2) in the plane}2x-2y+4z+5=0 \\ \text{Then,}\mathit{\text{PM}}\text{\hspace{0.17em}is normal to the plane. So, the direction ratios of}\mathit{\text{PM}}\text{are proportional to 2, -2, 4.} \\ \text{Since}\mathit{\text{PM}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}(1,\; 1,\; 2)\text{and has direction ratios proportional to}\mathrm{2,\; -2}and4,\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=r\text{(say)} \\ \text{Let the coordiantes of}\mathit{\text{M}}\text{be}\left(2r+1,-2r+1,4r+2\right)\text{.} \\ \text{Since}\mathit{\text{M}}\text{lies in the plane}2x-2y+4z+5=0, \\ 2\left(2r+1\right)-2\left(-2r+1\right)+4\left(4r+2\right)+5=0 \\ \Rightarrow 4r+2+4r-2+16r+8+5=0 \\ \Rightarrow 24r+13=0 \\ \Rightarrow r=\frac{-13}{24} \\ \text{Substituting this in the coordinates of}\mathit{\text{M}}\text{, we get} \\ M=\left(2r+1,-2r+1,4r+2\right)=\left(2\left(\frac{-13}{24}\right)+1,-2\left(\frac{-13}{24}\right)+1,4\left(\frac{-13}{24}\right)+2\right)=\left(\frac{-1}{12},\frac{25}{12},\frac{-1}{6}\right) \\ \text{Now, the length of the perpendicular from}\mathit{\text{P}}\text{onto the given plane} \\ =\frac{\left|2\left(1\right)-2\left(1\right)+4\left(2\right)+5\right|}{\sqrt{4+4+16}} \\ =\frac{13}{\sqrt{24}}\text{units} \end{gathered}$$