Question

If the distance of the point $P(1,-2,1)$ from the plane $x+2y-2z=\alpha$, where $\alpha >0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

• A. $(\frac{8}{3},\frac{4}{3}–\frac{7}{3})$
• B. $(\frac{4}{3},\frac{4}{3}–\frac{1}{3})$
• C. $(\frac{1}{3},\frac{2}{3},\frac{10}{3})$
• D. $(\frac{2}{3},-\frac{1}{3},\frac{5}{2})$

Answer 1

The correct option is A

$(\frac{8}{3},\frac{4}{3}–\frac{7}{3})$

Explanation for correct option

Step 1. Find the value of $\alpha$

We know that distance of a point $\left(p,q,r\right)$ fro plane $ax+by+cz+d=0$is

$d=\frac{\left|ap+bq+cr+d\right|}{\sqrt{a^2+b^2+c^2}}$

$\therefore$Distance of the point $\left(1,-2,1\right)$ from the plane $x+2y-2z=\alpha$is

$$\begin{gathered} \Rightarrow \frac{\left|1+2\left(-2\right)-2\left(1\right)-\alpha \right|}{\sqrt{1^2+2^2+{\left(-2\right)}^2}}=5 \\ \Rightarrow \frac{\left|-5-\alpha \right|}{\sqrt{9}}=5 \\ \Rightarrow \frac{\left|-5-\alpha \right|}{3}=5 \\ \Rightarrow \left|\alpha +5\right|=15 \\ \Rightarrow \alpha +5=\pm 15 \\ \Rightarrow \alpha =10,-20 \end{gathered}$$

Step 2. Find the foot of perpendicular

The equation of the line $PQ$ is $\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}$

We know that

The equation of line $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ can be represented as $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=\lambda$

Then any point on the line is $\left(\lambda a+x_0,\lambda b+y_0,\lambda c+z_0\right)$

Therefore point $Q=\left(\lambda +1,2\lambda -2,-2\lambda +1\right)$

The point $Q$ lie on the plane

$\therefore$For $\alpha =10$

$$\begin{gathered} \therefore x+2y-2z=10 \\ \Rightarrow \left(\lambda +1\right)+2\left(2\lambda -2\right)-2\left(-2\lambda +1\right)=10 \\ \Rightarrow 9\lambda =5+10 \\ \Rightarrow \lambda =\frac{5}{3} \\ \therefore Q=\left(\frac{8}{3}\frac{4}{3},-\frac{7}{3}\right) \end{gathered}$$

$\therefore$For $\alpha =-20$

$$\begin{gathered} \therefore x+2y-2z=-20 \\ \Rightarrow \left(\lambda +1\right)+2\left(2\lambda -2\right)-2\left(-2\lambda +1\right)=-20 \\ \Rightarrow 9\lambda =5-20 \\ \Rightarrow \lambda =-\frac{5}{3} \\ \therefore Q=\left(-\frac{2}{3},-\frac{16}{3},\frac{13}{3}\right) \end{gathered}$$

Hence, option A is correct