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.

Step-by-step 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.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.