Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Use foot of perpendicula $(x,y,z)$ of a point $\left(x_1,y_1,z_1\right)$ in a plane $ax+by+cz$ $+d=0$ is given by $\frac{x-x_1}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{c}}=\underline{-(\mathrm{ax}}$ Given equation of plane is $2x-y+5z-3$ $=0$ $\therefore$ Foot of perpendicular drawn from origin to the given plane $\begin{array}{rl}& \Rightarrow \frac{x-0}{2}=\frac{y-0}{-1}=\frac{z-0}{5}\\ & =\frac{-(0+0+0-3)}{(2{)}^2+(-1{)}^2+(5{)}^2}\\ & \Rightarrow \frac{x}{2}=\frac{y}{-1}=\frac{z}{5}=\frac{3}{30}=\frac{1}{10}\\ & \therefore \frac{x}{2}=\frac{1}{10},\frac{y}{-1}=\frac{1}{10},\frac{z}{5}=\frac{1}{10}\\ & \Rightarrow x=\frac{1}{5},y=-\frac{1}{10},z=\frac{1}{2}\end{array}$