Question

Find an equation of the plane that passes through the points (3,2,1) and (3,1,5) and is perpendicular to the plane with the equation 6x+7y+2z=10

Answer 1

Okay... to get the equation of a plane, you need a point on that plane (no problem, you already have two of them) and a normal vector... any ideas how to get one?

Answer 2

just any vector that has a dot product of zero with the vector that we can make with the two points rite?

Answer 3

No. Okay, let's clear one thing up....
The vector with the two points can be gotten by subtracting components... like so...
\[\huge \vec{v_1}= <0,-1,4>\]

Answer 4

so we said that the vector between the two points is (0,-1,4) and a normal vector is (0,4,1)

Answer 5

Afraid it's not that simple, if it were, where's the need for it to be perpendicular to the other plane you just gave? :)

Answer 6

Beside, you can't use just any vector perpendicular to this v1, because you don't know if it will be normal to the plane you're looking for... to illustrate, suppose we were looking for this plane, and it contained these two points..

Answer 7

Sure, we can get the vector between these two points...

Answer 8

And this vector here, is normal to the vector I just drew...
However, it lies in the plane, ergo, it is NOT a normal vector of the plane...

do you get what I'm saying? :)

Answer 9

so that black vecotr is perpendicular to the brown one correct but not parallel to the other plane?

Answer 10

Yes. The vector you got (out of nowhere, apparently, LOL)
<0 , 4 , 1>
IS perpendicular to the vector in the plane, but is NOT perpendicular to the plane :)

Okay?

Answer 11

ok i get what you are saying but I still cant figure out what we need to do. I have an idea that the normal vector of the plane that we are trying to create must be parallel to the plane that we are given rite?

Answer 12

The relationship between two planes is tied to the relationship between their normal vectors
In fact, the relationship is similar...
So, if two planes are perpendicular, it really means that their two normal vectors are perpendicular (orthogonal) to each other

Answer 13

ok so then we just need to find a vestor that is orthogonal to both the norm of the given plane and the given vector?

Answer 14

Yes. Exactly :D
And when you have two vectors and are tasked to find a vector that's orthogonal to BOTH of them, what's the ideal way to do it? :)

Answer 15

I would think that the best way would be to take the cross product of both vectors

Answer 16

That is correct :)
Get to it, champ :D

Answer 17

Thanks alot :D now I get it

Answer 18

Awesome :)

Answer 19

did we find the 2 vectors we need to cross?

Answer 20

He got me at "normal vector to the plane" and "vector that the two points make"

Hopefully yes :D

Answer 21

the normal vector of the given plane is not (0,4,1) unless im reading that post wrong

Answer 22

Oh, that (0,4,1) is a result of a faulty assumption that any vector that is normal to the vector made by the two points (which is <0,-1,4>) would also be a normal vector to the plane.

(0,4,1) is such a vector, but it is not normal to the plane...

I think I rectified the situation through my drawings (not nearly as good as yours though, @amistre64 :D )

Answer 23

the glass eye helps :)

so to be sure, we want the normal vector of the given plane (since its already perp and should be in the unknown plane):

6x+7y+2z=10 --> (6,7,2)


and the vector between the points:

( 3, 2, 1) and (3, 1, 5)
-3-2-1 -3 -2 -1
----------------------
(0,0,0) (0,-1,4) <-- that one