Parallel Tangent Planes on a Given Surface

[wbluem88]

Homework Statement

A surface is given by the equation

2x²+y²-z²=64

Find all points p on the surface where the tangent plane is parallel to the plane 12x+y-3z=0.

Homework Equations

I know the equation of a plane tangent to a surface at point (x₀,y₀,z₀) is given by the equation

z-z₀= (partialz/partialx)(x₀-y₀)+(partialz/partialy)(x₀-y₀)

The Attempt at a Solution

I can use the above equation to find the tangent plane at any certain point p, but I don't know how to find which ones are parallel to a certain plane.

 

[wbluem88]

By the way, this is on a review for my final exam. It's not homework, I just know that something similar will be on the final.

 

[mutton]

Two planes are parallel iff their normal vectors are scalar multiples, and this happens just when the cross product of their normal vectors is 0. Combine this knowledge with that formula you have.

 

[wbluem88]

*sigh* I did all the work but it came out wrong :(

The normal vector of a plane ax+by+cz+d=0 is specified by [a b c], so the normal vector for 12x+y-3z = 0 is [12 1 -3].

The normal vector of a surface f(x,y,z) at point (x₀,y₀,z₀) is [f_x f_y f_z], so the normal vector for 2x²+y²-z²=64 is [4x₀ 2y₀ -2z₀]

[12 1 -3] X [4x₀ 2y₀ -2z₀] = 0

<((1*-2z₀)-(-3*2y₀)), ((-3*4x₀)-(-12*-2z₀)),((12*2y₀)-(1*4x₀))>=0

-2z₀+6y₀=0
-12x₀+24z₀=0
12y₀-4x₀=0

I don't know where I went wrong.

 

[Defennder]

You don't have to use cross product for this. You have $$k(12,1,-3)^T = (4x,2y,-2z)^T$$ The k is the scalar multiple. Just solve x,y,z in terms of k and put it into the equation of the surface to solve for k. Then you have the answer.