# Math Help - Point on a surface

1. ## Point on a surface

Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.

I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.

2. Originally Posted by tdat1979
Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.

I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.
I think I can help you but first you have to tell how will you get tangent plane and what is the equation of tangent plane you got.

3. Quote:
Originally Posted by tdat1979
Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.

I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.

I think I can help you but first you have to tell how will you get tangent plane and what is the equation of tangent plane you got.
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I don't know if if doing it right but i got the point (2,7,5) and vector
(-3,8,-1) from the equation of the line. So the tangent plane would be
-3(x-2) + 8(y-7) - (z-5) = 0 which would give me -3x + 8y - z = 45. If I'm wrong please tell me what I did wrong.

4. Originally Posted by tdat1979
Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.

I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.
So we want to know when the gradient of the surface is parallel to the tangent of the line.

$F(x,y,z)=3x^2+2y^2+z-8$

$\nabla F(x,y,z)=6x \vec i +4y \vec j +\vec k$

The tangent vector to the line is $-3 \vec i + 8\vec j -\vec k$

We know that 2 vectors are parallel if and only if $v_1=cv_2$ for some scalar (or if their cross product is 0)

Now we need to solve the vector equation

$6x \vec i +4y \vec j +\vec k =c(-3 \vec i + 8\vec j -\vec k)$

Note that our only choice for c is $c=-1$ from the z component.

so this implies that $6x=3 \iff x=\frac{1}{2}$ and
$4y=-8 \iff y=-2$

Now just plug these back into the function to find z.