Tangent Plane Perpendicular to Vector <1,0,2> on x^2+y^2-z^2=-1 Surface

[Loppyfoot]

Homework Statement

Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>.

I'm confused!

The Attempt at a Solution

So, the gradient would be: <2x,2y,-2z>, but what do I do with the gradient vector to find these points?

thanks in advance!

 

[Mark44]

Homework Statement

Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>.

I'm confused!

The Attempt at a Solution

So, the gradient would be: <2x,2y,-2z>, but what do I do with the gradient vector to find these points?

At a given point on the surface, the gradient is perpendicular to the tangent plane at that surface. Does that give you any ideas?

 

[Loppyfoot]

Yes I understand that point. I'm confused on how to computationally solve this problem.

Any thoughts?

 

[Office_Shredder]

What does that tell you about the relationship between (2x, 2y, -2z) and (1,0,2)

 

[Loppyfoot]

So <2x,2y,-2z> dotted with <1,0,2> = 0

Ok, So if I do the dot product, I get:

2x-4z = 0 . So , x=2z. If I plug that into the original equation, I get:

4z^2 + y^2 -z^2 = -1 . Should I solve for y in terms of z?

 

[HallsofIvy]

No, you have that completely backwards! \nabla f= < 2x, 2y, -2z> is itself normal to the tangent plane. You want <2x, 2y, -2z> to be parallel to < 1, 0, 2>. You are solving for a vector perpendicular to <1, 0, 2>.

Two vectors are parallel if and only if one is a multiple of the other: <2x, 2y, -2z>= <a, 0, 2a> for some number a.

 

[Loppyfoot]

Ok, Now I understand. So I should solve for k first:

2x= k ; 2y=0 ; -2z=2k .

So x = k/2 and z=-k. If I substitute those back into the original equation, I get

k^2/4+ y^2-k^2 = -1. But I have two variables. Should I have set the y=0 so I can solve for k?

 

[Loppyfoot]

any ideas?

 

[Mark44]

Ok, Now I understand. So I should solve for k first:

2x= k ; 2y=0 ; -2z=2k .

You're not solving for k; you're solving for x, y, and z.

So x = k/2 and z=-k.

Don't forget y = 0.

If I substitute those back into the original equation, I get

k^2/4+ y^2-k^2 = -1. But I have two variables. Should I have set the y=0 so I can solve for k?

Yes.