Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

We know that the length of the perpendicular from the point $\left(x_1,y_1,z_1\right)$ to the plane $ax+by+cz+d=0$ is $\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$ and the co-ordinate $(\alpha ,\beta ,\gamma )$ of the foot of the $\perp$ are given by $\frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{\gamma -z_1}{c}$ $=-\left(\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}\right)$ ...(1) In the given ques, $x_1=7,y_1=14,z_1=5$, $a=2b=4,c=-1,d=-2$ By putting these values in (1), we get $\frac{\alpha -7}{2}=\frac{\beta -14}{4}=\frac{\gamma -5}{-1}=-\frac{63}{21}$ $\Rightarrow \alpha =1,\beta =2$ and $\gamma =8$ Hence, foot of $\perp$ is (1,2,8)