Parametric equations for a line

[UrbanXrisis]

considering the surface 25x^2+25y^2+4z^2=54

The parametric equation for a line going thought point P=(1,1,1) is

x=1+50t
y=1+50t
z=1+8t

A plane an equation for the tangent plane through P.

Here's what I know:
the equation for a plane needs a perpendicular vector to the plane and a point on the plane. I have a point on the place P=(1,1,1) but how do I find the perpendicular vector? I have a line ON the plane, what can I do to get an equation that is perpendicular to the plane?

 

[DieCommie]

Im struggling with these concepts as well.. but I think your perp. vector is <25,25,4>. Simply the coefficients of teh surface equation.

I wonder why these coefficients is the perp. and I asked my teacher. All he could tell me was its a way to describe the plane. So I have no idea why the coefficients are the perp. vector, they simply are.

 

[0rthodontist]

Commie, you can only say that when the equation is already actually a plane, which is not what his equation is. the reason for it is that given a point P on the plane and normal vector N, the plane is the set of points Q such that (Q - P).N = 0, or in other words the set of points Q such that the line between Q and P is perpendicular to the normal vector (makes sense?). The coefficients are then derived from Q.N: the coefficients of q1, q2, q3 are n1, n2, n3, which is just N.

Urban, you can find two vectors parallel to your plane--think partial derivatives. Now given two vectors, what operation can you do to get a vector perpendicular to those two vectors?

 

[HallsofIvy]

Here's the way I like to do it:
Define F(x,y,z)= 25x^2+25y^2+4z^2 so your surface is a level surface of F: F(x,y,z)= 54. The derivative of F in the direction of any unit vector, u is \nabla F\cdot v. In particular, if u is a vector tangent to the level surface, since F does not change in that direction, \nabla F\cdot v= 0. That is \nabla F, at a point, is perpendicular to any level surface of containing that point.

In this case, \nabla F= 50x i+ 50y j+ 8z, evaluated at (1, 1, 1) is \nabla F(1,1,1)= 50 i+ 50 j+ 8z. Parametric equations for the normal line to the surface at that point (not just "a" line) are
x= 1+ 50t, y= 1+ 50t, z= 1+ 8t.

Since that vector must also be perpendicular to the tangent plane at that point, an equation for the tangent plane is 50(x-1)+ 50(y-1)+ 8(z-1)= 0 or 50x+ 50y+ 8z= 108.

I wonder why these coefficients is the perp. and I asked my teacher. All he could tell me was its a way to describe the plane. So I have no idea why the coefficients are the perp. vector, they simply are.

Commie, an equation of a plane containing the point (a, b, c) and having normal vector Ai+ Bj+ Ck is A(x- a)+ B(y- b)+ C(z- c)= 0. To see why, think of it as a dot product:
A(x- a)+ B(y- b)+ C(z- c)= (Ai+ Bj+ Ck)\cdot((x-a)i+ (y-b)j+ (z-c)k).
Since (a,b,c) and (x,y,z) are both in the plane, the vector (x-a)i+ (y-b)j+ (z-c)k is in the plane. It's dot product, for any (x,y,z) in the plane, with Ai+ Bj+ Ck is 0 if and only if Ai+ Bj+ Ck is perpendicular to the plane.