Normal line to a tangent plane

1. Jul 12, 2008

tsw303

1. The problem statement, all variables and given/known data
I have two problems that don't seem to match. The task is to find the parametric equation for the normal line to the tangent plane. I'm seeing that the vector value of "z" is (-1) with one formula, but not in the other. These two seem to be the same, but the first was solved implicitly in the book (?) and arrived at (-1). The second was solved easily with the second formula, but arrived at a unique value for z.

The normal line to (x^2)+(y^2)+(z^2)=25 at p(-3,0,4)
is x=-3+(3t/4) y=0 z=4-t

The normal line to (x^2)+(y^2)+4(z^2)=12 at p(2,2,1)
is x=2+t y=2+t z= 1+2t

2. Relevant equations

when z=f(x.,y.)+fx(x.,y)(x-x.)+f(x.,y.)(y-y.)
it makes sense that z= f(x.,y.)-t

when fx(x.,y.,z.)(x-x.)+fy(x.,y.,z.)(y-y.)+fz(x.,y.,z.)(z-z.)=0
different values for z seem ok, such as z= z.+fz(x.,y.,z.)

3. The attempt at a solution

The two problems seem like they are nearly the same and should both be solved with the second formula. I'm completly lost trying to figure out what I'm missing.

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

2. Relevant equations

3. The attempt at a solution

2. Jul 13, 2008

HallsofIvy

Staff Emeritus
You don't say what formulas you are using! Also I have no idea what you mean by a "vector value of -1".

In my opinion, the simplest way to do such a problem is to treat the surface as a "level surface". We can thinkg of the first as a level surface for $\ph(x,y,z)= x^2+ y^2+ z^2$ and then we know that the $\nabla\phi= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}$ is normal to the surface and so normal to the tangent plane. At (-3, 0, 4) that is $-6/vec{i}+ 8\vec{k}$ and so has parametric equations x= -3- 3t, y= 0, z= 4+ 8t. Of course, the "length" of the vector, and whether it is pointing "up" or "down" is irrelevant so you could divide that normal vector by -8, getting $3/4\vec{i}+ \vec{k}$ which gives the parametric equations x= -3+ (3/4)t, y= 0, z= 4- t, the equations you have. If, in the second set of equations, you replace t by 4s, you get the first set of equations.

Similarly, the normal vector to $\phi(x,y,z)= x^2+ y^2+ 4z^2= 12$ is $\nabla \phi= 2x\vec{i}+ 2y\vec{j}+ 8z\vec{k}$. At (2, 2, 1) that is [itex]4\vec{i}+ 4\vec{j}+ 8\vec{k} so the parametric equations can be written x= 2+ 4t, y= 2+ 4t, z= 1+ 8t. Or we could divide the normal vector by 2 getting a normal vector of [itex]2\vec{i}+ 2\vec{j}+ 4\vec{k} and parametric equations x= 2+ 2t, y= 2+ 2t, z= 1+ 4t, as you give.

Remember that parametric equations are not unique. A single line (or curve) can be parameterized in many different ways.

3. Jul 13, 2008

tsw303

Thank You.
Now that you mentioned the length and direction of the vector, I can understand the answers in my text-book (such as dividing by 4 or -8). I initially got the same answers as you (but in the first, isn't it x=-3 -6t ?). Is "veci" the unit vector i ?

However, where I'm really stumped (I need to take better notes in class): I thought my teacher said that, in the equation for the normal line, "z" defaults to z=f(x.,y.)-t instead of z= z.+fz(x.,y.,z.)t . Maybe that is only for f(x,y) and not for f(x,y,z)?

4. Jul 14, 2008

tsw303

... I may have just figured it out.
I am confused about the normal line to the tangent plane because when the equation is f(x,y), x=x.+fx(x.,y.)t , y=y.+fy(x.y.)t , yet z=f(x.,y.)-t . When the equation is in the form of f(x,y,z), I guess "z" must be the same as "x" and "y" such that z=z.+fz(x.,y.,z.)t . It must be the structure of the equation and what happens when everything is moved to one side.