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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.
Reply 1
Avatar for nuodai
nuodai
17
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.


When you see an equation of the form ax+by+cz=kax+by+cz=k, the normal to the plane is always (a,b,c)(a,b,c).

Why? Well let r=(x,y,z)\mathbf{r}=(x,y,z) and n=(a,b,c)\mathbf{n}=(a,b,c). Choose numbers u,v,wu,v,w such that au+bv+cw=kau+bv+cw=k (that is, we can choose (u,v,w)(u,v,w) to be any point in the plane). Then if we write u=(u,v,w)\mathbf{u}=(u,v,w), we have

ax+by+cz=au+bv+cwax+by+cz=au+bv+cw

which is precisely the statement that r⋅n=u⋅n\mathbf{r} \cdot \mathbf{n} = \mathbf{u} \cdot \mathbf{n}

Equivalently, (r−u)⋅n=0(\mathbf{r} - \mathbf{u}) \cdot \mathbf{n} = 0

That is, n\mathbf{n} is perpendicular to (x−u,y−v,z−w)(x-u,y-v,z-w) for every choice of x,y,zx,y,z.

But any direction vector in the plane can be written in the form (x−u,y−v,z−w)(x-u,y-v,z-w), since (u,v,w)(u,v,w) is a fixed point in the plane and (x,y,z)(x,y,z) is the coordinates (position vector) of any given point. So what the equation (r−u)⋅n=0(\mathbf{r}-\mathbf{u}) \cdot \mathbf{n} = 0 tells us is that n\mathbf{n} is perpendicular to all directions in the plane. That is, n\mathbf{n} 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.
(edited 11 years ago)
Reply 2
Avatar for anonstudent1
anonstudent1
OP
13
Original post by nuodai
When you see an equation of the form ax+by+cz=kax+by+cz=k, the normal to the plane is always (a,b,c)(a,b,c).


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. :smile:
Reply 3
Avatar for nuodai
nuodai
17
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. :smile:


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 :p:)
Reply 4
Avatar for ztibor
ztibor
10
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.


Yes, the normal wolud be (7,-4,2) in your example.
And why?
For the equation of a plane we need a point on the plane
P0(x0,y0,z0)P_0(x_0,y_0,z_0) and a vector being perpendicular to
the plane.This is the normal vector, and let n⃗=Ai⃗+Bj⃗+Ck⃗\vec n=A\vec i+B\vec j+C\vec k
THe position vector pointing to P0 is
Unparseable latex formula:

\vec r_0=x_0\cdot \vec i+y_0\cdot \vec j+z_0\cdot \veck

.
Let a point in the plane P(x,y,z)P(x,y,z), sor the position vector
pointing this point is r⃗=x⋅i⃗+y⋅j⃗+z⋅k⃗\vec r=x\cdot \vec i+y\cdot \vec j+z\cdot \vec k.
It is clear that the vector r⃗−r⃗0\vec r-\vec r_0 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:
(r⃗−r⃗0)⋅n⃗=0\left (\vec r -\vec r_0\right )\cdot \vec n=0
Calculating the dot product
A(x−x0)+B(y−y0)+C(z−z0)=0A(x-x_0)+B(y-y_0)+C(z-z_0)=0 gives the scalar equation.
Arranging
Ax+By+Cz=Ax0+By0+Cz0=DAx+By+Cz=Ax_0+By_0+Cz_0=D
whre D is a constant value.
From this you can read down the coordinates of the normal vector (A,B,C)
Reply 5
Avatar for shiinkii
shiinkii
8
Original post by nuodai
When you see an equation of the form ax+by+cz=kax+by+cz=k, the normal to the plane is always (a,b,c)(a,b,c).

Why? Well let r=(x,y,z)\mathbf{r}=(x,y,z) and n=(a,b,c)\mathbf{n}=(a,b,c). Choose numbers u,v,wu,v,w such that au+bv+cw=kau+bv+cw=k (that is, we can choose (u,v,w)(u,v,w) to be any point in the plane). Then if we write u=(u,v,w)\mathbf{u}=(u,v,w), we have

ax+by+cz=au+bv+cwax+by+cz=au+bv+cw

which is precisely the statement that r⋅n=u⋅n\mathbf{r} \cdot \mathbf{n} = \mathbf{u} \cdot \mathbf{n}

Equivalently, (r−u)⋅n=0(\mathbf{r} - \mathbf{u}) \cdot \mathbf{n} = 0

That is, n\mathbf{n} is perpendicular to (x−u,y−v,z−w)(x-u,y-v,z-w) for every choice of x,y,zx,y,z.

But any direction vector in the plane can be written in the form (x−u,y−v,z−w)(x-u,y-v,z-w), since (u,v,w)(u,v,w) is a fixed point in the plane and (x,y,z)(x,y,z) is the coordinates (position vector) of any given point. So what the equation (r−u)⋅n=0(\mathbf{r}-\mathbf{u}) \cdot \mathbf{n} = 0 tells us is that n\mathbf{n} is perpendicular to all directions in the plane. That is, n\mathbf{n} 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.


Are you a wizard?
n=2i +1j -2k is already the normal vector.

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