How do you find parametric equations for the line which passes through the point (1,−2,3) and is parallel to both of the planes 3x + y + 5z = 4 and z = 1 − 2x?

1 Answer
Ultrilliam
Jun 25, 2018
  • #bb r(t) = (:1,−2,3:) + t(:1 , 7,-2:)#

Explanation:

Line passes through the point (1,−2,3) so:

  • #bbr(t) = (: 1,−2,3:) + t(:a,b,c:)#

For direction vector #(:a,b,c:)#, line is parallel to both of the planes:

  • # 3x + y + 5z = 4#, which has normal vector #(:3,1,5:)#

  • #2x + z = 1 #, which has normal vector #(:2,0,1:)#

Which means it runs in the same direction as the line of intersection of those planes. It will have direction:

#(:a,b,c:) = (:3,1,5:) times (:2,0,1:) #

#= det [(hat x, hat y, hat z),(3,1,5),(2,0,1)]#

#= hat x(1- 0) - hat y (3 - 10) + hat z (0 - 2)#

#= (:1 , 7,-2:)#

#:. bbr(t) = (:1,−2,3:) + t(:1 , 7,-2:)#

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