# Finding the scalar equation for a plane

## Homework Statement

Find the scalar equation for the plane containing L1 and L2.
Consider the lines:
L1 : x = t + 1, y = 2t, z = 3t - 1
L2 : x = s - 1, y = 2s + 1, z = 3s - 1

## Homework Equations

Scalar equation for a plane:
a(x - x0) + b(y - y0) + c(z - z0) = 0

## The Attempt at a Solution

These two lines have the same vector <1, 2, 3>, so they are parallel. I let t = 0 and s = 0 so that I found a point (1, 0, -1) for L1 and a point (-1, 1, -1) for L2. Where do I go from here? Thanks and I appreciate it.

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ehild
Homework Helper
Find the vector which connects those points. It will not be parallel with the original lines.

ehild

Ok so I found the displacement vector between the two points to be (-1 - 1, 1 - 0, -1 -(-1)), which comes up to (-2, 1, 0). Now that I have the vector, does it matter which point I use to plug into the formula to find the scalar equation?

ehild
Homework Helper
You have two vectors lying in the plane, but you need a normal vector of the plane to get the scalar equation. How do you get a vector that is normal to the vectors of the plane?

What does the scalar equation of the plane mean? What are the parameters a,b,c?

ehild

Okay, I took the cross product with the displacement vector <-2, 1, 0> and the other vector <1, 2, 3> which ended up as <-3, -6, 5>. So now I would use that vector and pick one of those points to come up with the scalar equation. I end up with -3x - 6y + 5z = -8

HallsofIvy
Homework Helper
Now, it is easy to check if that is correct, isn't it? The line L1 is given by x = t + 1, y = 2t, z = 3t - 1. Putting that into your equation for the plane, -3(t+1)- 6(2t)+ 5(3t-1)= -3t- 3- 12t+ 15t- 5= (-3- 12+ 15)t+ (-3- 5)= -8 which is correct for all t. Every point of L1 is in that plane. The line L2 is given by x = s - 1, y = 2s + 1, z = 3s - 1. Putting that into your equation, -3(s- 1)- 6(2s+ 1)+ 5(3x- 1)= -3s+ 3- 12s- 6+ 15x- 5= (-3-12+15)s+ (3- 6- 5)= -8 which is correct for all s. Every point of L2 is in that plane.

Thanks a lot for the help! To follow up on this problem. If I wanted to find the distance between the lines, would I just use the distance formula between the two points?

ehild
Homework Helper
The distance between two parallel lines is the length of the segment of line, perpendicular to them. You know already a vector which connects the two lines: find its normal component.

ehild

Using the displacement vector <-2, 1, 0>, the component would be ( (-2)^2 + 1^2 + 0^2 ) ^ (1/2), which ends up as square root of 5. That would then be the distance correct?

ehild
Homework Helper
As the displacement vector happens to be perpendicular to the lines, the distance is correct.

ehild