# Point on a surface where the tangent plane is perpendicular to another plane?

Discussion in 'Undergraduate Math' started by lstormont, Apr 22, 2007.

1. ### lstormontGuest

I have no trouble finding the equations of the tangent plane and
normal line to a surface at a given point. However, how do I find
another point on the surface where its tangent plane is perpendicular
to the first plane I found?

lstormont, Apr 22, 2007

2. ### [Mr.] Lynn KurtzGuest

You have a normal vector to the surface at the given point, call it N.
If your surface is f(x,y,z)=0, let V = grad(f), which is normal to the
surface at any point (x,y,z) on the surface. You need V(x,y,z) dot N =
0 and f(x,y,z) = 0. Here's an example:

Say your surface is the unit sphere:

f(x,y,z) =x^2 + y^2 + z^2 - 1 = 0 and the given point is (1,0,0).
The normal vector N at that point is < 1, 0, 0 >
V = grad(f) = < 2x, 2y, 2z >, might as well use < x, y, z >.

So we need < x, y, z > dot < 1, 0, 0 > = 0 and
x^2 + y^2 + z^2 = 1.

The first equation gives x = 0 so the second equation gives y^2 + z^2
= 1.

These equations describe the intersection of the sphere with the yz
plane and, sure enough, every point on that circle is a point on the
sphere where the tangent plane is perpendicul to the original.

--Lynn

[Mr.] Lynn Kurtz, Apr 22, 2007

3. ### LisaGuest

Thank you. With a normal vector like, say <2,1,3>, you then have two
equations: (2x+y+3z = 0) and x^2 + y^2 +z^2=1, and three variables.
How does one proceed from there?

Lisa, Apr 23, 2007
4. ### LisaGuest

Thanks. Since my equations/values resulted in 2 equations of 3
variables, I simply let one variable equal zero and found values that
would work for the other two. I appreciate your help.

Lisa, Apr 23, 2007
5. ### [Mr.] Lynn KurtzGuest

Each of those equations represents a surface. Where they intersect
gives the values (x,y,z) that you are looking for. In this case, the
plane and sphere intersect in a circle. Generally, two surfaces
intersect in a curve, although they might not intersect at all.

--Lynn

[Mr.] Lynn Kurtz, Apr 23, 2007
6. ### [Mr.] Lynn KurtzGuest

Why let one variable be zero. Why not 1/2, or 5 or whatever? If you
arbitrarily give it a value you may find a point or two that work but
you may miss others. Generally when you have more equations than
unknowns, you can let the extra variable(s) be anything. So let that
variable = t instead of zero. Then get the other two variables in
terms of t. So you will wind up with something like this, assuming for
illustration that x is the variable you set equal to t:

x = t, y = g(t), z = h(t)

This would give you the parametric equation of the curve of
intersection of the two surfaces.

--Lynn

[Mr.] Lynn Kurtz, Apr 23, 2007
7. ### [Mr.] Lynn KurtzGuest

I meant more unknowns than equations, which is what you have.

--Lynn

[Mr.] Lynn Kurtz, Apr 23, 2007
8. ### LisaGuest

Lynn, I only needed to find A point, not every point, that would
satisfy the problem. Was what I did not alright in that case?

Lisa, Apr 23, 2007
9. ### The Qurqirish DragonGuest

What you did MAY have been fine, but what if, for example, the
variable you set to 0 didn't yield any points? then you would have to
guess another value. What if that one didn't work? and so on.
By getting the parametric form, you can see if there are any
restrictions on the parameter first. Now you can choose any value of
the parameter in its domain to get a point.

Simple example: let the surface be the unit sphere, centered at
(2,2,2), and let any point on the sphere determine your first plane.
Since no point on the sphere has any coordinate of 0, choosing one to
be zero will yield no solution, since one of the two equations would
be:
(x - 2) ^2 + (y - 2) ^2 + (z - 2) ^2 = 1,
and this has no solution if you force one coordinate to 0. If you
paramatrize (say, x=t), then you will have a domain of 1<=t<=3. The
second equation may further restrict the domain, but if it didn't, you
could then choose any value in this range to get a specific point.

The Qurqirish Dragon, Apr 23, 2007
10. ### LisaGuest

Ok, I guess this is something I do not know how to do, nor have I had
to at this point. I've got 2 equations like 3a + 4b + 5c = 0, and
2a^2 + 3b^2 + 4c^2 = 75. I'm not sure how to restrict the domain here
- I see how it would work in your example but do no tknow how to apply
it to mine (note: the equations I gave were NOT the problem I'm
working on so I have no idea what solution, if any, they'd have).

Lisa, Apr 23, 2007
11. ### LisaGuest

Ok, I guess this is something I do not know how to do, nor have I had
to at this point. I've got 2 equations like 3a + 4b + 5c = 0, and
2a^2 + 3b^2 + 4c^2 = 75. I'm not sure how to restrict the domain
here
- I see how it would work in your example but do not know how to
apply
it to mine (note: the equations I gave were NOT the problem I'm
working on so I have no idea what solution, if any, they'd have).

Edited to add: I used software to solve the system and got a
parametrization, which confirmed that the answer I did get, worked. I
still do not know though how to do it the old-fashioned way.

Lisa, Apr 23, 2007
12. ### The Qurqirish DragonGuest

(i) 3a + 4b + 5c = 0
(ii) 2a^2 + 3b^2 + 4c^2 = 75

Now, assuming that you are looking for the plane tangent to (ii) and
orthogonal to (i), we need "grad ((ii))" (please pardon the abuse of
notation), which gives (choosing V for this to conform to Lynn Kurtz's
notation above):
V=<4a, 3b, 8c>
The normal to the plane (i) is N=<3,4,5>

So, you need
(ii) 2a^2 + 3b^2 + 4c^2 = 75
and
(iii) V dot N = 12a + 12b + 40c = 0

You can choose any variable as the parameter. I will choose b, but in
general you want to choose the variable that is hardest to isolate
from the other 2. In this problem all three choices are equivelantly
hard.

Let b = t
rewriting (iii) by dividing by 4 gives:
(iii*) 3a + 3t + 10c = 0
and so
(iv) c = -0.3 (a + t)
You could also choose to solve for a, but you do not want to solve for
t, as that is your parameter.
Substituting (iv) into (ii) gives you a quadratic in a, which you
solve to get a as a function of t.
Use a(t) in (iv) to get c as a function of t.

You solution set is thus ( a(t) , t , (c(t) ). When you solve the
quadratic to get a(t), you will find the domain for acceptible t
values. (set the discriminent of the quadratic to 0 for the bounds).

This example doesn't appear to give nice values, so I won't put the
rest of the solution here. (It is all simple algebra anyway)

The Qurqirish Dragon, Apr 24, 2007