# Equation of plane through two points and parallel to line

• May 18th 2011, 03:45 PM
blackbear
Equation of plane through two points and parallel to line
Quote:

Find the equation of the plane passing through the points (3, 2, -1) and (1, -1, 2) that is parallel to the line r(t) = <1, -1, 0> + t<3, 2, -1>
Let P = (3, 2, -1) , Q = (1, -1, 2) and R = <3, 2, -1> = direction vector from r(t)

a = PQ = <1, -1, 2> - <3, 2, -1> = <-2, 3, -3>
b = QR = <3, 2, -1> - <1, -1, 2> = <2, 3, -3>

N = a x b = <0, -12, -12>

Now I'm going to choose a point on the plane, P = (3, 2, -1)

d = n dot OP = <0, -12, -12> dot <3, 2, -1> = -12(2) + 12 = -12

So it follows that the plane has equation -12y + -12z = -12 or y + z = 1 So it appears that using the direction vector, as I did, from r(t) and generating the plane using those three points is completely wrong. Any guidance?
• May 19th 2011, 03:32 AM
HallsofIvy
Quote:

Originally Posted by blackbear
Let P = (3, 2, -1) , Q = (1, -1, 2) and R = <3, 2, -1> = direction vector from r(t)

a = PQ = <1, -1, 2> - <3, 2, -1> = <-2, 3, -3>

This isn't "PR". It would be RQ, which is the negative of QR, except that you have subtracted wrong. For the second component, -1- 2= -3, not 3 and for the third component 2- (-1)= 3, not -3.

Quote:

b = QR = <3, 2, -1> - <1, -1, 2> = <2, 3, -3>

N = a x b = <0, -12, -12>

Now I'm going to choose a point on the plane, P = (3, 2, -1)

d = n dot OP = <0, -12, -12> dot <3, 2, -1> = -12(2) + 12 = -12

So it follows that the plane has equation -12y + -12z = -12 or y + z = 1 So it appears that using the direction vector, as I did, from r(t) and generating the plane using those three points is completely wrong. Any guidance?
Yes. Repeat what you did but this time do it correctly.

You don't actually need to find "PR" or "QR" (technically, you can't- R is a vector not a point). Just use PQ, a vector in the surface, and R. R is not necessarily in the surface but it is parallel to it so the cross product of PQ and R will be perpendicular to the surface.
• Jan 9th 2014, 03:56 PM
OVOXO
Re: Equation of plane through two points and parallel to line
Hi, could you please explain this question, i am completely stuck on it. Thanks!!
• Jan 9th 2014, 03:57 PM
OVOXO
Re: Equation of plane through two points and parallel to line
Hi, would you be able to re-explain this question from the beginning, i have to understand it for school tomorrow and Im so confused.
• Jan 9th 2014, 04:52 PM
Prove It
Re: Equation of plane through two points and parallel to line
A plane with equation \displaystyle \begin{align*} a\,x + b\,y + c\,z = d \end{align*} has the same coefficients as the normal vector \displaystyle \begin{align*} \mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \end{align*}. To get a normal vector to the plane, you take the cross product of two vectors in the plane. You already have one vector \displaystyle \begin{align*} \mathbf{r} \end{align*}. You can use the two points you are given to get another vector in the plane.

Go from here.
• Jan 9th 2014, 06:05 PM
OVOXO
Re: Equation of plane through two points and parallel to line
So, I take PQ X R as my coefficients of the plane?
(-2,-3,3)x(3,2,-1)=(-3,7,5)
-3x+7y+5z=0 is parallel to the line and includes p and q?
• Jan 9th 2014, 06:06 PM
OVOXO
Re: Equation of plane through two points and parallel to line
and lol at the song lyrics
• Jan 10th 2014, 07:32 AM
Hartlw
Re: Equation of plane through two points and parallel to line
P, Q, and X are in a plane perpendicular to r(t) in dir d

(PQ)X(PX) is perpendicular to plane and has to be perpendicular to dir d of r(t)

(PQ)X(PX)•d = 0, or, in determinant form: |(PQ);(PX);d|=0
(PQ)=(3,2,-1) - (1,-1,2)=(2,3,-1)
(PX)=(3,2,-1) – (x,y,z)=[(3-x),(2-y),(-1-z)]
d=(3,2,-1)

3x-7y-5z=0
• Jan 11th 2014, 02:31 AM
Prove It
Re: Equation of plane through two points and parallel to line
Quote:

Originally Posted by OVOXO
So, I take PQ X R as my coefficients of the plane?
(-2,-3,3)x(3,2,-1)=(-3,7,5)
-3x+7y+5z=0 is parallel to the line and includes p and q?

How do you know that d = 0? Use a point you know lies on the plane as (x, y, z) in the equation of your plane to evaluate d.
• Jan 11th 2014, 08:58 AM
Hartlw
Re: Equation of plane through two points and parallel to line
Thanks Prove It, you are right:

l = (3.2,-1) dir of line
PQXl = (2,3,-3)X(3,2,-1) = (3,-7,-5) is normal to plane par to line
3x-7y-5x=d is equation of plane.
Substitute P = (3,2,-1) to determine d.
• Jan 11th 2014, 05:18 PM
OVOXO
Re: Equation of plane through two points and parallel to line
I had subbed it in and d was equal to zero.
• Jan 11th 2014, 05:22 PM
HallsofIvy
Re: Equation of plane through two points and parallel to line
Good. That was what Prove It was asking.