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The plane containing the line $\frac{x-3}{2}= \frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2x + 3y – z = 5$, contains which one of the following points? $(A) (2, 2, 0) (B) (–2, 2, 2) (C) (0, –2, 2) (D) (2, 0, –2)$

I think that plane would be $a(x-3)+b(y+2)+c(z-1)=0$, where $a,b,c$ are direction ratios of the normal of the plane. So, $2a-b+3c=0$. I don't understand how the projection of the line on another plane could lie on the first plane. Does that mean the two planes are intersecting? Do they have to perpendicular for this? Because, I guess this is what the solution is implying as it has written $2a+3b-c=0$.

aarbee
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  • The projection is nothing but another line. This is also the line of intersection of the required plane and the given plane. They’re not necessarily perpendicular. – Vishu Apr 24 '20 at 10:07
  • @Tavish Thanks. Should the planes be perpendicular? – aarbee Apr 24 '20 at 10:09

1 Answers1

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See that since the plane contains the line and and it's projection onto the plane $2x+3y–z=5$, it means that the plane would contain the normal to the plane $2x+3y–z=5$ and yes, would be perpendicular to it.

To find the direction ratios of the plane , you just need to find the cross product of $(2,3,-1)$ and $(2,-1,3)$ as both the directions are parallel to the plane.

udit narayan pandey
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  • narayan ji, a big fan. But not able to imagine how the normal to the second plane would be contained in the first i.e. not able to imagine why the two planes should be perpendicular. sur nahi pakad pa raha. – aarbee Apr 24 '20 at 10:17
  • see that the projection line is formed by projecting each point of the line on to the plane , take a point $P_1$ on the line,let $P_2$ be it's projection, the way we project you would see that the line segment $P_1P_2$ is perpendicular to the plane and since $P_1$ and $P_2$ both lie on the required plane, it means the segment $P_1P_2$ lies on the required plane .... – udit narayan pandey Apr 24 '20 at 10:25