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I am not really able to understand the question. I didn’t get the part ‘contains the projection of the line on the plane....’

How can a plane contain the projection of a line on another plane? Wouldn’t that just mean that required plane is coincident with $2x+3y-z=5$?

I know there is already such a question on MSE but I am not convinced with that answer.

Aditya
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  • The projection of a line on a plane is another line. So you need to calculate the equation of the projection, and find a plane containing two different lines. – MikeG Feb 22 '21 at 05:12
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    @MikeG according to the answer, it’s much simpler than that. The plane required contains the vectors $2i+2j+2k$ and $2i+3j-k$. That’s the part where I am confused – Aditya Feb 22 '21 at 05:15
  • Oh you are right. It is indeed much easier to think in that way. Maybe I can write a short answer to better explain it than in the comment. – MikeG Feb 22 '21 at 05:18
  • and the first should be +3k. – MikeG Feb 22 '21 at 05:19
  • While in general we cannot ask for the plane that "contains two lines" in three dimensions (because in general two skew lines are not contained in any one plane), it is reasonable to use the parametric form of the line $(x-3)/2 = (y+2)/2 = (z-1)/3$ to work out the projection of each of its points on the other plane. We'd hope that those projected points form a line that is parallel to or intersecting with the first line. – hardmath Feb 22 '21 at 05:20
  • Additional comment to @hardmath: the line is not parallel to the plane so we can expect an intersection point. Then the line and its projection is of course in the same plane(which means this problem is well-formulated) since they all pass the same point. – MikeG Feb 22 '21 at 05:27

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The first vector you provide, $2i+2j+3k$, is the direction of the line $\frac{x-3}{2}=\frac{y+2}{2}=\frac{z-1}{3}$. Surely the plane containing the line should contain this.

The second vector is the "direction" of the plane $2x+3y-z=5$, which is perpendicular to it. And if you can draw a picture on the draft, you know that the plane we want should contain this, since the projected line, the original line, and the perpendicular vector $2i+3j-k$ should be on the same plane.

enter image description here

Finally you need the plane to contain $(3,-2,1)$ which is on the original line. One point and two different vectors is now sufficient to determine a plane--

$a(x-3)+b(y+2)+c(z-1)=0$ is what you want. Here $(a,b,c)\perp(2,2,3)$ and $(a,b,c)\perp(2,3,-1)$ as we discussed above. You can then either solve the equation directly or use outer product.

MikeG
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  • Here is my problem. The required plane contains the projection of the the given line, which lies on the plane $2x+3y-z=5$ and is thus perpendicular to the $2i+3j-k$. So how can the required plane contain this vector? – Aditya Feb 22 '21 at 05:52
  • The required plane is NOT perpendicular to that--it actually contains it. The vector is perpendicular to 2x+3y-z=5. – MikeG Feb 22 '21 at 05:54
  • And it contains it because it contains the original line and the projected line, since the vector 2i+3j-k should be on the same plane as them. – MikeG Feb 22 '21 at 05:55
  • @Aditya Just add a picture. It is ugly but tells the spirit. – MikeG Feb 22 '21 at 05:59
  • Oh that does make sense, but my point of view was different. What if the plane we wanted was parallel to the given plane, or at any other angle instead of being perpendicular. Isn’t that a valid case? – Aditya Feb 22 '21 at 07:18
  • Also the given line should be parallel to the required plane – Aditya Feb 22 '21 at 07:19
  • But the plane we want contains the projection line, and projection without any classification is considered to be perpendicular projection, which gives the fact that the plane we want is perpendicular to the given one, even if the given line is parallel to the given plane(but in your explicit example it is not, so I draw it that way). – MikeG Feb 22 '21 at 07:33
  • So in short can we consider the problem to have insufficient data? – Aditya Feb 22 '21 at 07:40
  • What do you mean by "insufficient data"? If you mean the conditions are not enough for us to find the plane,then it's not true. – MikeG Feb 22 '21 at 07:47
  • No I meant we don’t know whether the planes are perpendicular to each other. As I said, they might not be perpendicular and still satisfy all the conditions – Aditya Feb 22 '21 at 08:26
  • Let us continue this discussion in chat. – MikeG Feb 22 '21 at 08:39