Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
So here’s a problem; Given line l with the equation xyz equals 5, 3, 0 plus t times 1 -4, 6. Find an equation of the line through 2, 1, 4 that’s parallel to l.
Any line that’s parallel to l will have a direction vector that’s a scalar multiple of this one. But we could use exactly the same direction vector and get a line parallel. So let’s try x, y, z equals, and you use this as our initial point. Vector 2, 1, 4 is the position vector of that point plus, t times the same direction vector 1, -4, 6. That’s a vector equation of the line through 2, 1, 4 that’s parallel to our given line.
How do we find an equation of a line through point 5, 3, 0 that’s perpendicular to l? Now here are actually lots of lines perpendicular to l, but we want to find just one of them. So the trick here is going to be to find a direction vector that’s perpendicular to this one. Remember the test for that is all their dot products equal to 0.
So you have to think about a vector that’s going to have the dot product of 0 with this one. So I’ll write dot 1, -4, 6. Now I’m noticing that if I multiply this, the 1 by -42, and the -4 by 1, I get -46. If I add that to 6 I’ll get 0. So this will work. -2, 1, 1. I’ll get minus 2 minus 4 plus 6 with the dot product. So this is a vector that I could use. Remember there are infinitely many other vector that you could use to find a perpendicular line.
So I’ll use this is my direction vector, and this will be my initial point. x, y, z equals 5, 3 ,0 plus t times my direction vector -2, 1, 1. So this could be a line that passes through the point 5, 3, 0 that’s perpendicular to our given line.