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.
Let’s do another problem. Given plane M with equation; 5x minus 2y plus 4z equals 10, find an equation of a plane parallel to M and then we’ll find one perpendicular to M.
First let’s do parallel to M. You have to find the normal vector of the plane, and the normal vector just comes from the coefficients. You have normal vector <5, -2, 4>. In order for a plane to be parallel to M, it’s got to have a normal vector that is a scalar multiple of this, or you could just use this normal vector. That’s what I’m going to do. Let’s use this normal vector and that means the coefficients are going to be exactly the same, 5x minus 2y plus 4z. You just can’t have it equal 10, because that would be the same plane. So any other number; -20, that’s a plane parallel to M.
Now let’s try perpendicular. That’s a little trickier but again, the normal vector is <5, -2, 4>. Here, to get a plane perpendicular to M, we have to find one whose normal vector is perpendicular to this one. So we have to find a vector that has a dot product with this one of zero.
Let me write components dot <5, -2, 4> equals 0. You could just start playing around with numbers because there are infinitely many vectors that will satisfy this. Why don’t we make a equal 1, b equal 2 and I’ll leave it as c, dot <5, -2, 4>? What does that equal? On the left we have 1 times 5, 5 plus 2 times -2, -4, plus 4c equals 0. We have 1 plus 4c equals 0, c equals -1/4. So if I make c equals to -1/4, this vector will be perpendicular to this vector. Let me do that.
So I'll use, I’ll call it m, is going to be <1, 2, -1/4>. My plane will be 1 times x, plus 2 times y minus 1/4z equals, and then any number will do. Any number because the perpendicularities are determined but these coefficients, not by the number over here. So, 100. That plane is perpendicular to M and that’s all there is to it.
If you want to find a plane parallel to a plane, then make the normal vector a scalar multiple of the original normal vector. To find one perpendicular, make the normal vector perpendicular by making sure that its dot product with the original normal vector is zero.