Question

Find the equation of the plane through the origin which is parallel to z=4x-3y+8

Answer 1

I'm assuming that you know that a plane in 3D space is defined by its normal vector and a point in the plane. In this case the "point in the plane" is conveniently defined as (0,0,0). Where...and how...are we going to derive / find a vector normal to the given plane z=4x-3y+8? Since the new plane will be parallel to this given plane, we need only find the normal vector to z=4x-3y+8.

Answer 2

Hint: in your shoes I'd rewrite z=4x-3y+8 in the form ax+by+cz=d

Answer 3

Then the equation of the plane would become \[a(x-x _{0})+b(y-y _{0})+c(z-z _{0})=0\]... can you identify a, b and c? Hint: the constants come from our knowing that the new plane goes through the origin.

Answer 4

@eylult What is the normal vector of the plane that you had given ?

Answer 5

Hint #2: a, b and c for the "new" plane will be the same as a, b and c for the given plane, and all the info you need to specify the normal vector are right there in z=4x-3y+8.

Answer 6

-4i+3j+k=8 i the normal vector of it right?

Answer 7

so a=-4, b=+3, c=1 ?

Answer 8

That's right @eylult

Answer 9

O'h c=-1

Answer 10

Ok thank you so much

Answer 11

\(z=4x-3y+8=>4x-3y-x+8=0\)

Answer 12

\(n(4,-3,-1)\)