Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The perpendicular distance of the point $P(1,-2,1)$ from the plane $x+2y-2z-\alpha =0$ is given by:

$d=\frac{|1(1)+2(-2)-2(1)-\alpha |}{\sqrt{1^2+2^2+(-2{)}^2}}=5$

$\frac{|1-4-2-\alpha |}{\sqrt{9}}=5$

$\frac{|-5-\alpha |}{3}=5$

$|-5-\alpha |=15$

$-5-\alpha =15\text{or}-5-\alpha =-15$

$\alpha =-20\text{or}\alpha =10$

Since it is given that $\alpha >0$, we have $\alpha =10$.

The equation of the plane is $x+2y-2z-10=0$.

Let the foot of the perpendicular from $P(1,-2,1)$ to the plane be $(x_2,y_2,z_2)$. The coordinates of the foot of the perpendicular are given by the formula:

$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}=-\frac{a{x}_1+b{y}_1+c{z}_1-d}{a^2+b^2+c^2}$

Substituting the values, we get:

$\frac{x_2-1}{1}=\frac{y_2-(-2)}{2}=\frac{z_2-1}{-2}=-\frac{1(1)+2(-2)-2(1)-10}{1^2+2^2+(-2{)}^2}$

$\frac{x_2-1}{1}=\frac{y_2+2}{2}=\frac{z_2-1}{-2}=-\frac{-15}{9}=\frac{5}{3}$

Equating each term to $\frac{5}{3}$:

$x_2-1=\frac{5}{3}\Rightarrow x_2=\frac{8}{3}$

$y_2+2=\frac{10}{3}\Rightarrow y_2=\frac{4}{3}$

$z_2-1=-\frac{10}{3}\Rightarrow z_2=-\frac{7}{3}$

Therefore, the foot of the perpendicular is $\left(\frac{8}{3},\frac{4}{3},\frac{-7}{3}\right)$.

Answer: $\left(\frac{8}{3},\frac{4}{3},\frac{-7}{3}\right)$