# Getting vector normal from parametric equations

• Oct 25th 2008, 03:12 PM
Craka
Getting vector normal from parametric equations
Question states "The Cartesian equation of the plane containing the line x=3t, y=1+t, z=2-t and passing through the point (-1,2,1) is?"

I know I need to get the vector normal from the parametric equations, but I don't know how to do this, and than apply the relationship

n.(r-r0)= 0

ie <n1,n2,n3> . <x-(-1), y-2, z-1 > = 0

So I think the biggest problem I have is trying to work out the vector normal from the parametric
• Oct 25th 2008, 03:16 PM
Jhevon
Quote:

Originally Posted by Craka
Question states "The Cartesian equation of the plane containing the line x=3t, y=1+t, z=2-t and passing through the point (-1,2,1) is?"

I know I need to get the vector normal from the parametric equations, but I don't know how to do this, and than apply the relationship

n.(r-r0)= 0

ie <n1,n2,n3> . <x-(-1), y-2, z-1 > = 0

So I think the biggest problem I have is trying to work out the vector normal from the parametric

the line is given by: (0, 1, 2) + t<3, 1, -1>

thus you want to find a vector <a,b,c> such that

<a, b, c> . <3, 1, -1> = 0
• Oct 25th 2008, 03:42 PM
Craka
Sorry I can see what you have done. But I still do not understand how this helps. Could you give me some more information?
• Oct 25th 2008, 04:04 PM
Jhevon
Quote:

Originally Posted by Craka
Sorry I can see what you have done. But I still do not understand how this helps. Could you give me some more information?

i wrote the line in its vector form. then i took the direction vector. since the line is in the plane, the direction vector is parallel to the plane. thus, we need to find a vector orthogonal to that vector. so it is a vector where its dot product with the direction vector gives zero. since the dot product of orthogonal vectors gives 0

now find a, b, and c for which that works and you are done
• Oct 25th 2008, 05:11 PM
Craka
I follow what you are saying but I don't understand how to solve for a, b, c.

Could you show me some working, I sometimes find it to understand from examples
• Oct 25th 2008, 05:17 PM
Jhevon
Quote:

Originally Posted by Craka
I follow what you are saying but I don't understand how to solve for a, b, c.

Could you show me some working, I sometimes find it to understand from examples

for 2-dimensional vectors, there is a nice way to do it. the vector <-b,a> is always orthogonal to <a,b>. but in 3-dimensions, i don't know of a nice formula. i usually just pick to random values for a and b, usually a = b = 1, and then i solve for what c has to be
• Oct 25th 2008, 06:29 PM
Craka
Yes I knew how to find a orthogonal vector in 2 dimension by that method , however I didn't know how to find one easily in 3 dimensions.

From what I understood I did this to find a orthogonal vector.
<1,1,c> . <3,1,-1> =0

giving C = 4
so vector normal is < 1,1,4>

thus <1,1,4> . < x-(-1), y-2, z-1 > =0

giving x+y+4z = 5 but this still is not the answer.

Am I interpreting this incorrectly?
• Oct 25th 2008, 08:09 PM
Jhevon
Quote:

Originally Posted by Craka
Yes I knew how to find a orthogonal vector in 2 dimension by that method , however I didn't know how to find one easily in 3 dimensions.

From what I understood I did this to find a orthogonal vector.
<1,1,c> . <3,1,-1> =0

giving C = 4
so vector normal is < 1,1,4>

thus <1,1,4> . < x-(-1), y-2, z-1 > =0

giving x+y+4z = 5 but this still is not the answer.

Am I interpreting this incorrectly?

i would think that both answers are correct. but anyway, new plan.

find the normal vector the hard way, that is, using cross-products.

we know the direction vector of the line, and that is one vector in the plane. if we find another vector in the plane (not parallel to the first), we can take the cross product of that vector and the said direction vector, and get the vector orthogonal to the plane.

first, note that the point (-1,2,1) is not on the line. so pick a point on the line, say when t = 0, we have the point (0, 1, 2). so another vector in the plane we want is <0 - (-1), 1 - 2, 2 - 1> = <1,-1,1>

now the normal vector is $\left< 3,1,-1 \right> \times \left<1,-1,1 \right> = \left< 0,4,4 \right> = \left<0,1,1 \right>$

now continue, you will get the same answer

use this procedure from now on, forget that picking random numbers foolishness
• Oct 25th 2008, 09:42 PM
Craka
Thankyou very much Jhevon.

Only thing isn't the cross product of <3,1,-1> x <1,-1,1> equal to <0,-1,-1> ?
• Oct 25th 2008, 09:47 PM
Jhevon
Quote:

Originally Posted by Craka
Thankyou very much Jhevon.

Only thing isn't the cross product of <3,1,-1> x <1,-1,1> equal to <0,-1,-1> ?

<0,-1,-1> is parallel to the vectors I gave, so it is the same thing for our purposes

but i took the form that will give you the same answer as the book--for the sake of your piece of mind
• Oct 26th 2008, 04:37 AM
HallsofIvy
Quote:

Originally Posted by Craka
Question states "The Cartesian equation of the plane containing the line x=3t, y=1+t, z=2-t and passing through the point (-1,2,1) is?"

I know I need to get the vector normal from the parametric equations, but I don't know how to do this, and than apply the relationship

n.(r-r0)= 0

ie <n1,n2,n3> . <x-(-1), y-2, z-1 > = 0

So I think the biggest problem I have is trying to work out the vector normal from the parametric

When t= 0, x= 3(0)= 0, y= 1+0= 1, and z= 2-0= 2 so A= (0, 1, 2) is a point in the plane. When t= 1, x= 3(1)= 3, y= 1+1= 2, and z= 2-1= 1 so B= (3, 2, 1) is also in the plane. You are told that C= (-1, 2, 1) is in the plane. Find the vectors AB and AC and take their cross product to find the normal vector.