Equation for a plane through 2 points

[Punkyc7]

find an equation of a plane through the points (,1,2,3) and (3,2,1,) and is perpendicular to the plane 4x-y+2z=-7

My answer is x+6y+z=16, I am just wondering if its right

 

[Mark44]

find an equation of a plane through the points (,1,2,3) and (3,2,1,) and is perpendicular to the plane 4x-y+2z=-7

My answer is x+6y+z=16, I am just wondering if its right

Yes, it's right. There are an infinite number of planes that work in this problem, however. Another one is 4x - y + 2z = 8.

 

[asdofindia]

There are an infinite number of planes that work in this problem, however. Another one is 4x - y + 2z = 8.

Isn't that parallel to the plane (to which it's supposed to be perpendicular)

(Something tells me there can only be one solution for this question. But I don't get it in the form of equations)

 

[Mentallic]

Mark must've missed that detail. There is in fact just one distinct solution.

 

[pongo38]

punkyc7: Please show your working. There is more than one way to do this.

 

[Metaleer]

I believe the thread's title is the misleading factor. Title claims it's through two points - actual post also mentions and additional plane.

 

[Mentallic]

I believe the thread's title is the misleading factor. Title claims it's through two points - actual post also mentions and additional plane.

Well he obviously couldn't put the whole post into his title. It's a brief summary of what his problem is

 

[HallsofIvy]

Any plane through (1, 2, 3) is of the form A(x- 1)+ B(y- 2)+ C(z- 3)= 0. If that plane also contains (3, 2, 1) then we must have A(3- 1)+ B(2- 2)+ c(1- 3)= 2A- 2C= 0 or C= A.

In order that it be perpendicular to 4x-y+2z=-7, we must have 4A- B+ 2C= 0. Since C= A, that gives B= 6A.

Choosing any value for A gives a solution. Punkyc7's answer corresponds to A= 1, but all choices for A give the same plane.

 

[Punkyc7]

Wait so is my answer right?

 

[Mentallic]

Wait so is my answer right?

Punkyc7's answer corresponds to A= 1, but all choices for A give the same plane.

Yes. All Hallsofivy is getting at is that it's like saying the solution is y=2x², but another solution is 2y=4x² and in general, Ay=2Ax².

 

[asdofindia]

So 1(x-a)+1(y-b)+1(z-c) = 0
is the same plane as 2(x-a)+2(y-b)+2(z-c) = 0
That's an interesting insight.
Yeah, after all 1i + 1j + 1k is parallel to 2i + 2j + 2k

 

[asdofindia]

Wait so is my answer right?

Yes. you're right, you can easily verify that.
a plane through the points (1,2,3) and (3,2,1,)
and perpendicular to the plane 4x-y+2z=-7
answer is x+6y+z=16

1+6*2+3=16
3+6*2+1=16
4*1+-1*6+2*1=0

Correct!

 

[Mark44]

Mark must've missed that detail. There is in fact just one distinct solution.

Yes, I wasn't thinking about this the right way. I was thinking about the plane parallel to the given plane, not perpendicular to it. Sorry for any misleading advice.