Eq. of plane passing through points, parallel to line...

[jwpaine]

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>

I know that the direction vector of r(t) needs to be orthogonal to the plane's normal vector.

Let P1 = (3, 2, -1) and P2 = (1, -1, 2) I need a third point P3 such that the cross product of vectors P1P2 x P2P3 is our vector normal N such that the dot product N dot <3, 2, -1> = 0

That's correct, right? How would I setup for this problem to find the equation of the plane passing through P1 and P2 and be parallel to r(t)? Just a bit of advice would be great - would rather not have anyone set it up in entirety.

John

 

[galactus]

Let's take the two points A(3,2,-1) and B(1,-1,2).

Let's use the point A(3,2,-1) as the fixed point on the line. We need a direction vector that is parallel to the vector AB.

$\displaystyle V=AB=(1-3)i+(-1-2)j+(2-(-1))k=-2i-3j+3k$

The symmetric equations are $\displaystyle \frac{x-3}{-2}=\frac{y-2}{-3}=\frac{z+1}{3}$

The parametric equations are $\displaystyle x=3-2t, \;\ y=2-3t, \;\ -1+3t$

The plane contains the line:

$\displaystyle x=3-2t, \;\ y=2-3t, \;\ z=-1+3t$

The plane contains the point (3,2,-1) and the normal of the plane must be orthogonal to the direction vector

$\displaystyle V_{1}=-2i-3j+3k$

Since the plane is parallel to the line $\displaystyle x=1+3t, \;\ y=-1+2t, \;\ z=-t$ the normal to the plane must also be orthogonal to a direction vector of that line, $\displaystyle V_{2}=3i+2j-k$

The cross product $\displaystyle V_{1}\times V_{2}=-3i+7j+5k$ is the normal to the plane.

Find the equation of the plane that contains the point (3,2,-1) with normal vector -3i+7j+5k

So, the plane has equation $\displaystyle -3(x-3)+7(y-2)+5(z+1)=0$

$\displaystyle -3x+7y+5z=0$

Let's now check to see if the lines don't intersect.

We have parametric equations of the lines as:

$\displaystyle x=3-2t, \;\ y=2-3t, \;\ z=-1+3t$

They must be expressed in different parameters, so the other line can be expressed as:

$\displaystyle x=1+3s, \;\ y=-1+2s, \;\ z=-s$

Set corresponding coordinates equal on each line and we get the system:

\(\displaystyle \left\{ \begin{array} \text{3-2t}=1+3s\\ 2-3t=-1+2s\\ -1+3t=-s\end{array}\) (I do not know why that little 'c' is at the top. Disregard).

The first two equations have solution \(\displaystyle (t,s)=(1,0)\)

Since these values do not satisfy the third equation, the lines are parallel.

Check my 'figgers' to make sure I did not go down the primrose path. Easy to do in all that.

 

[jwpaine]

Galactus, you always did go above and beyond in your posts - thanks so much!

I did:

Let P = (3, 2, 1) and Q = (1, -1, 2), PQ = (1, -1, 2) - (3, 2, -1) = <-2, -3, 3> which is the vector in the surface
Letting R = <3, 2, -1> be the directional vector from r(t), taking the cross product of PQ x R will yield a normal vector to the plane:

N = PQ x R = [3-6]i - [2-9]j + [-4 + 9]k = <-3, 7, 5>

d = n dot OP = <-3, 7, 5> dot <3, 2, -1> = 0

So it follows that the eq of the plane is -3x + 7y + 5z = 0

Thanks Galactus! Online Summer calc III course. I may or may not have passed it last semester :roll: