Equation of a plane through 2 points, parallel to a line.

[cjthibeault]

Homework Statement

I'm trying to understand my professor's solution to a quiz problem. I can follow it up to a point, but then I'm getting confused as to how he continues. The question is:

Find an equation of the plane $$p$$ through $$A(2,1,-1)$$ and $$B(1,5,-4)$$ parallel to the line $$L: x=1 + 2t, y=-1+t, z=-2-t$$.

Homework Equations

n/a

The Attempt at a Solution

This is the solution that was given:

$$p: A(x-2)+B(y-1)+C(z+1)=0$$

$$B \in p: A(1-2)+B(5-1)+C(-4+1)=0$$

$$L \|p \Leftrightarrow \vec{d}\bot \vec{n}\Leftrightarrow 2A+B-C=0$$

He then solves for $$C$$, getting $$C=2A+B$$.

The next step is where I get lost:

$$9B=7C \Rightarrow B=\frac{7}{9}C$$

$$A=4B-3C=4 \cdot \frac{7}{9}C - 3C=\frac{1}{9}C$$

$$A=1, B=7, C=9$$

$$\vec{n}=(1,7,9)$$

$$x-2+7(y-1)+9(z+1)=0 \Rightarrow x+7y+9z=0$$

It is the $$9B=7C$$ part that I don't get. Where are those coefficients coming from? Thanks for any assistance anyone can provide.

 

[Mark44]

I think what your prof is doing is using the idea that the cross product of a vector in the plane with the direction of the line will be parallel to the normal to the plane.

I.e., <-1, 4, -3> X <2, 1, -1> = k<A, B, C>

The first vector in the equation above is AB, the second is the direction of the line.

It is unfortunate that A and B are used both for points in the plane and for the coordinates of the plane's normal. This duplication can lead to confusion.

 

[cjthibeault]

Thank you so much. Taking the cross product does lead to the normal vector he eventually arrived at. I still don't understand exactly what algebraic manipulations he was performing to get there his way, but as long as I have some way of arriving at the correct answer, I'll be happy.

Thanks again.

 

[Mark44]

The cross product works in this problem because the vector from A to B is not parallel to the direction vector of the line.