108k views
3 votes
Given the plane 3x + 5y - z = 9, give an equation of a plane parallel to the given plane, and a plane orthogonal to the given plane.

User Danny Guo
by
8.2k points

1 Answer

2 votes

Answer:

  • parallel: 3x +5y -z = 0
  • perpendicular: 5x -3y = 0

Explanation:

A parallel plane will have the same normal vector (the coefficients of x, y, z), so will differ only in the constant. Changing the constant from 9 to anything else gives the equation of a parallel plane.

A perpendicular plane will have a normal vector that is perpendicular to the normal vector of the given plane. That is, the dot-product of the normal vectors will be zero. There are an infinite number of possible solutions. One of them is ...

5x -3y = 0

Its normal vector is <5, -3, 0> and the dot-product of that with the normal vector of the given plane is ...

<5, -3, 0> · <3, 5, -1> = (5)(3) +(-3)(5) +(0)(-1) = 15 -15 +0 = 0

Any plane whose coefficients a, b, c satisfy 3a+5b-c = 0 will be a normal plane.

User YogevSitton
by
7.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories