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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2010 · Paper 2 · Mathematics · Three Dimensional Geometry

If the distance of the point P(1,-2,1) from the plane x+2 y-2 z=α ,...

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

Options

  1. A(83,43,−73)\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
  2. B(43,−43,13)\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)
  3. C(13,23,103)\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)
  4. D(23,−13,52)\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)

Correct answer

A. (83,43,−73)\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)

Step-by-step solution

 Distance of point \text{ Distance of point } P  from plane =5∴=∣1−4−2−α3∣α=10\text{ from plane }=5 \begin{gathered} \therefore=\left|\frac{1-4-2-\alpha}{3}\right|\\ \alpha=10 \end{gathered}If the distance of the point P(1,-2,1) from the plane x+2 y-2 z=α , where α>0 , is 5 , then the foot of theFootofperpendicular\\ Foot of perpendicularx−11=y+22=z−1−2=53⇒x=83,y=43,z=−73\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}$Thus,thefootoftheperpendicularisThus, the foot of the perpendicular isA\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$
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