Finding the equation of a plane by its parametric equations

Question

How can I find the equation of a plane by its parametric equations?

Like this one:

p: x = h + t y = -1 + 2h - 3t z = -3 + h - t

The exercise is to see if a given straight (r in this case) is contained in the plan (Pi), as the following photo

Could you guys please help me out?

Dr. Sonnhard Graubner · Accepted Answer · 2019-11-19T17:09:42.850

1

You can write $$x-2=t,\frac{y+2}{2}=t,z+3=t$$ where $t$ is a real number. So we have $$[x,y,z]=[2,-2,-3]+t[1,2,1]$$ That is true, since we have $$t+2-2=\frac{2t-2}{2}=t-3+3=t$$.True!

edited Nov 19 '19 at 17:09

answered Nov 19 '19 at 16:57

Dr. Sonnhard Graubner

95,283

I did that, then I found that: x = t + 2; y = 2t - 2; z = t -3; – Matheus Minguini Nov 19 '19 at 17:03
Now, to see if the straight is contained in the given plan, I would have to replace that in the plan's equation, wouldn't I? – Matheus Minguini Nov 19 '19 at 17:05
Where is your given plane? – Dr. Sonnhard Graubner Nov 19 '19 at 17:07
It is π (It is in the photo) – Matheus Minguini Nov 19 '19 at 17:10
I got it, see the lines from me! – Dr. Sonnhard Graubner Nov 19 '19 at 17:11
I see, thanks for the reply, I really appreciate it! Lemme ask a silly question: Let's say that I got the plan's equation, in order to a straight be contained in the plan, I select two points in this straight and I replace those points in the plan's equation, if the equation evaluates to zero in the two given points coordinates, it is true to be said that the straight is contained in the plan, right? It would also be a possible solution, wouldn't it? – Matheus Minguini Nov 19 '19 at 17:16

Finding the equation of a plane by its parametric equations

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