Recent content by silver1x

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?

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?

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

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?

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...