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# What is the normal form for this plane? watch

1. I have calculated the parametric form of a line as: L=P1+tP1P3=<2,2,0>+t<1,2,2>.

If I am given a point K=<1,−1,−1>, how would I show the normal form of plane E that has the line L as its normal and contains the point K?
2. By "normal form", I take it you mean in the form r.n = d. If so, then you first need a normal vector n. The wording of your questions should indicate that such a vector is staring you in the face! Once you've got this, how can you use your given point to find d?
3. (Original post by Pangol)
By "normal form", I take it you mean in the form r.n = d. If so, then you first need a normal vector n. The wording of your questions should indicate that such a vector is staring you in the face! Once you've got this, how can you use your given point to find d?
Is it just <1,2,2><-4,1,1>=0
4. Where has the <-4,1,1> vector come from?
5. (Original post by Pangol)
Where has the <-4,1,1> vector come from?
<1,2,2><?>=0

I found a vector that multiplied by <1,2,2> equaled 0. (1x-4)+(2x1)+(2x1)=0
6. Am I right in thinking that when you say "normal form", you mean the form r.n = d? If so, then the r is variable - you need to identify the n and the d.

I think you have found the right n. You now need to find d, by substituting any point on the plane into your equation as r.

Sorry if we're talking at cross purposes - do let me know if I've got the wrong end of the stick!
7. (Original post by mls123)
<1,2,2><?>=0

I found a vector that multiplied by <1,2,2> equaled 0. (1x-4)+(2x1)+(2x1)=0
Sorry but I have absolutely no idea what you are trying to do here. You need the equation for the plane. It should be of teh form r.n=d where n is a vector and r is a scalar.
8. (Original post by Pangol)
Am I right in thinking that when you say "normal form", you mean the form r.n = d? If so, then the r is variable - you need to identify the n and the d.

I think you have found the right n. You now need to find d, by substituting any point on the plane into your equation as r.

Sorry if we're talking at cross purposes - do let me know if I've got the wrong end of the stick!
I do mean the form r.n=d. After substituting the <1,-1,-1> given from the question, I get <3,0,-2> which is r.

Then using the cross product <3,0,-2><1,2,2> = -1
9. (Original post by mls123)
After substituting the <1,-1,-1> given from the question, I get <3,0,-2> which is r.
I don't know what you mean here. Where have you substituted <1,-1,-1> into?
10. (Original post by mls123)
I do mean the form r.n=d. After substituting the <1,-1,-1> given from the question, I get <3,0,-2> which is r.

Then using the cross product <3,0,-2><1,2,2> = -1
r isn't something you find, r=(x,y,z) and x,y and z are variables.
11. (Original post by mls123)
I have calculated the parametric form of a line as: L=P1+tP1P3=<2,2,0>+t<1,2,2>.

If I am given a point K=<1,−1,−1>, how would I show the normal form of plane E that has the line L as its normal and contains the point K?
THe vector equation of a plane is

where r_0=(x_0,y_0,z_0) is a given point on the plane. In this case this is point K nad r is a genaral running point on the plane r=(x,y,z)
n is the normal vector of the plane and here this is the direction vector
of the line (A,B,C)=(1,2,2), because that is perpendicular to the plane.
THe equation in scalar form

or arranging

or vith vectors

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Updated: April 12, 2013
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