Parametric equations for a line

In summary, the surface 25x^2+25y^2+4z^2=54 has a parametric equation for a line going through point P=(1,1,1). The equation for the tangent plane through P can be found by using the coefficients of the surface equation as the perpendicular vector to the plane. Another way to find the perpendicular vector is by taking the partial derivatives of the surface equation and using them to calculate the normal vector. This normal vector can then be used to find the equation of the tangent plane.
  • #1
UrbanXrisis
1,196
1
considering the surface [tex]25x^2+25y^2+4z^2=54[/tex]

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

[tex]x=1+50t[/tex]
[tex]y=1+50t[/tex]
[tex]z=1+8t[/tex]

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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
Here's the way I like to do it:
Define [itex]F(x,y,z)= 25x^2+25y^2+4z^2[/itex] 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 [itex]\nabla F\cdot v[/itex]. In particular, if u is a vector tangent to the level surface, since F does not change in that direction, [itex]\nabla F\cdot v= 0[/itex]. That is [itex]\nabla F[/itex], at a point, is perpendicular to any level surface of containing that point.

In this case, [itex]\nabla F= 50x i+ 50y j+ 8z[/itex], evaluated at (1, 1, 1) is [itex]\nabla F(1,1,1)= 50 i+ 50 j+ 8z[/itex]. 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:
[tex]A(x- a)+ B(y- b)+ C(z- c)= (Ai+ Bj+ Ck)\cdot((x-a)i+ (y-b)j+ (z-c)k)[/tex].
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.
 
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What are parametric equations for a line?

Parametric equations for a line are a set of equations that describe the coordinates of points on a line in terms of one or more parameters.

How do you write parametric equations for a line?

Parametric equations for a line can be written in the form x = x0 + at and y = y0 + bt, where x0 and y0 are the coordinates of a point on the line, and a and b are constants.

What is the significance of parametric equations for a line?

Parametric equations for a line allow us to describe the motion or position of a point on the line, and can also be used to graph the line in a coordinate plane.

Can parametric equations for a line be used in 3D space?

Yes, parametric equations for a line can also be used to describe lines in three-dimensional space. In this case, the equations would be written in the form x = x0 + at, y = y0 + bt, and z = z0 + ct, where x0, y0, and z0 are the coordinates of a point on the line, and a, b, and c are constants.

What are some real-world applications of parametric equations for a line?

Parametric equations for a line are commonly used in physics, engineering, and computer graphics to describe the motion of objects, trajectories of projectiles, and the shape of curves and surfaces.

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