# Homework Help: Calc 3-normal lines to surfaces

1. Sep 22, 2010

### somebodyelse5

1. The problem statement, all variables and given/known data

Find the equations of the normal lines to the surfaces at the given points.

z=(3/4)x^2+3y^2 @ pt. (2,1)

2. The attempt at a solution

I have already found the equation of the tangent plane and know it is correct.
Tangent plane => (z-6)=3(x-2)+6(y-1)

Now, I am confused here, when it says it wants the equation of the line normal to the surface, it really wants the equation of the line normal to the tangent plane correct?

I know that the answer is <3t+2, 6t+1, -t+6> but I have no idea how to solve for it.
I understand where these values come from <3t+2, 6t+1, -1t+6> but what step am I missing to switch the signs of the x$$_{}0$$ y$$_{}0$$ and z$$_{}0$$

I need to be able to do this again on 7 more problems, 4 of which are from parameterizations (i dont think this maters). For those, I took the partial derivatives of the position vector and crossed them, is the result of the cross product the equation of the line normal to the surface at that point?

I think I have a solid grasp on the actual image of what im looking for, but i dont understand how to actually get it.

2. Sep 22, 2010

### vela

Staff Emeritus
Rewrite the equation of the plane in the form nx = c. The vector n is normal to the plane.

3. Sep 22, 2010

### somebodyelse5

where x is the vector used to create the equation of the tangent plane? what do i set as c?

4. Sep 22, 2010

### vela

Staff Emeritus
x=(x,y,z). c is a constant; it's where everything else in your equation of the plane ends up. Just multiply out your equation for the plane, move the terms with variables all to one side and everything else to the other. You should be able to read off what n is.

5. Sep 22, 2010

### somebodyelse5

k thanks!

6. Sep 23, 2010

### HallsofIvy

Here's another way to do it, without first finding the tangent plane: think of the surface $z= (3/4)x^2+ 3y^2$, which is the same as $(3/4)x^2+ 3y^3- z= 0$ as a "level surface" for $F(x,y,z)= (3/4)x^2+ 3y^2- z$. We know that the $\nabla F\cdot \vec{v}$, with $\vec{v}$ a unit vector, is the derivative (rate of change) of F in the $\vec{v}$ direction. In particular, if $\vec{v}$ is a vector in the tangent plane to the surface, then, since F is constant on the surface, that directional derivative is 0: we must have $\nabla F\cdot \vec{v}= 0$ for any vector $\vec{v}$ in the tangent plane. Since $\nabla F$ is perpendicular to every vector in the tangent plane, it is normal to the tangent plane and so to the surface.

With $F(x, y, z)= (3/4)x^2+ 3y^2- z$, $\nabla F= (3/2)x\vec{i}+ 6y\vec{j}- \vec{k}$ normal to the surface. At x= 2, y= 1, that is $2\vec{i}+ 6\vec{j}- \vec{k}$. Further, z= (3/4)(4)+ 3(1)= 6 so the point on the surface is (2, 1, 6). The line through (2, 1, 6) in the direction of the vector $2\vec{i}+ 6\vec{j}- \vec{k}$ is given by x= 2t+ 2, y= 6t+ 1, z= -t+ 6.