Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the origin}\mathit{\text{P}}\text{(0, 0, 0) in the plane}2x-3y+4z-6=0. \\ \text{Then,}\mathit{\text{PM}}\text{\hspace{0.17em}is normal to the plane. So, the direction ratios of}\mathit{\text{PM}}\text{are proportional to 2, -3, 4.} \\ \text{Since}\mathit{\text{PM}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}(0,\; 0,\; 0)\text{and has direction ratios proportional to}\mathrm{2,\; -}3,\; 4,\text{the}\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}=r\text{(say)} \\ \text{Let the coordiantes of}\mathit{\text{M}}\text{be}\left(2r,-3r,4r\right)\text{.} \\ \text{Since}\mathit{\text{M}}\text{lies in the plane}2x-3y+4z-6=0, \\ 2\left(2r\right)-3\left(-3r\right)+4\left(4r\right)-6=0 \\ \Rightarrow 4r+9r+16r-6=0 \\ \Rightarrow 29r-6=0 \\ \Rightarrow r=\frac{6}{29} \\ \text{Substituting the value of r in the coordinates of}\mathit{\text{M}}\text{, we get} \\ M=\left(2r,-3r,4r\right)=\left(2\left(\frac{6}{29}\right),-3\left(\frac{6}{29}\right),4\left(\frac{6}{29}\right)\right)=\left(\frac{12}{29},\frac{-18}{29},\frac{24}{29}\right) \end{gathered}$$