How to find a vector perpendicular to a plane.
Maths and statistics discussion, revision, exam and homework help.
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How to find a vector perpendicular to a plane.If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
The proof I was given was that (x,y,z).(2,1,-2)=5
This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??
Any help greatly appreciated. -
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Re: How to find a vector perpendicular to a plane.When you see an equation of the form(Original post by anonstudent1)
If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
The proof I was given was that (x,y,z).(2,1,-2)=5
This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??
Any help greatly appreciated.
, the normal to the plane is always 
.
Why? Well let
and 
. Choose numbers 
such that 
(that is, we can choose 
to be any point in the plane). Then if we write 
, we have
which is precisely the statement that
Equivalently,
That is,
is perpendicular to 
for every choice of 
.
But any direction vector in the plane can be written in the form, since is a fixed point in the plane and 
is the coordinates (position vector) of any given point. So what the equation 
tells us is that is perpendicular to all directions in the plane. That is, is normal to the plane.
I hope I haven't obfuscated this too much with the heavy use of notation. Let me know if you want anything clarified.Last edited by nuodai; 05-05-2012 at 14:19. -
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Re: How to find a vector perpendicular to a plane.Thank you for your help(Original post by nuodai)
When you see an equation of the form, the normal to the plane is always .
I think for the level of exams i'm doing right now, the above will be sufficient. Wow the rest looks complicated, will give it a read through when i have more time.
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Re: How to find a vector perpendicular to a plane.The rest isn't something you need to know, as such, but it shows why it's true so it's worth trying to understand once even if you don't remember it. (But as you say, maybe it's too late for that(Original post by anonstudent1)
Thank you for your help
I think for the level of exams i'm doing right now, the above will be sufficient. Wow the rest looks complicated, will give it a read through when i have more time.
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Re: How to find a vector perpendicular to a plane.Yes, the normal wolud be (7,-4,2) in your example.(Original post by anonstudent1)
If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
The proof I was given was that (x,y,z).(2,1,-2)=5
This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??
Any help greatly appreciated.
And why?
For the equation of a plane we need a point on the plane
and a vector being perpendicular to
the plane.This is the normal vector, and let
THe position vector pointing to P0 is
.
Let a point in the plane
, sor the position vector
pointing this point is
.
It is clear that the vector
is in the plane, so
perpendicular to the normal vector, that their dot product is zero.
This fact gives the vector equation of the plane:
Calculating the dot product
gives the scalar equation.
Arranging
whre D is a constant value.
From this you can read down the coordinates of the normal vector (A,B,C)
