1

How to find the the canonical equation of the projection of the line $$\frac{1}{3} \left(x - 4\right) = -\frac{1}{2} \left(y + 1\right) = \frac{z}{4}$$

on the plane $$x-3y-z+8=0.$$

Gibbs
  • 8,230
Elvin
  • 220
  • 1
  • 9
  • 1
    A line is defined by two points. Find two points on the line’s projection. – amd Apr 18 '18 at 22:11
  • @amd,can you show me how ? – Elvin Apr 18 '18 at 22:14
  • Another possibility is to convert the line into parametric form, project that onto the plane, then convert back. The other method is probably simpler. – amd Apr 18 '18 at 22:30

1 Answers1

1

Following the hint given by amd in the comment, two points on the line are given by

  • $P=(4,-1,0)$
  • $Q=(7,-3,4)$

to find the projections of the points onto the plane we can consider the lines through the two points and orthogonal to the plane (i.e. direction vector = normal vector for the plane)

  • $P(t)=(4,-1,0)+t(1,-3,-1)$
  • $Q(t)=(7,-3,4)+t(1,-3,-1)$

and then find by the intersections the projection $P_0$ and $Q_0$ onto the plane.

user
  • 154,566