Question

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

Answer 1

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