JEE Advanced · Mathematics · 31. 3D Geometry
If the distance of the point \(P(1,-2,1)\) from the plane \(x+2 y-2 z=\alpha\), where \(\alpha>0\), is 5 , then the foot of the perpendicular from \(P\) to the plane is
- A
\(\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)\)
- B
\(\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)\)
- C
\(\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)\)
- D
\(\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)\)
Answer & Solution
Correct Answer
(A)
\(\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)\)
Step-by-step Solution
Detailed explanation
\[
\text { Distance of point } P \text { from plane }=5
\]
\[
\begin{gathered}
\therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\
\alpha=10
\end{gathered}
\]
Foot of perpendicular
\[
\begin{aligned}
\quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3}
\end{aligned}
\]
Thus, the foot of the perpendicular is
\[
A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
\]
\text { Distance of point } P \text { from plane }=5
\]
\[
\begin{gathered}
\therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\
\alpha=10
\end{gathered}
\]
Foot of perpendicular
\[
\begin{aligned}
\quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3}
\end{aligned}
\]
Thus, the foot of the perpendicular is
\[
A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
\]
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