Tangent line, to a surface, through a point, parallel to a plane

[crc1559]

Hey y'all, this is my first post. I am currently stuck on a multivariable question. Please let me know if you can help.

Homework Statement

The point, P = (1, 2, 2) lies on the surface z = x^2 + y^2 -3x. Find parametric equations for the tangent line to the surface through the point P parallel to the plane x = 1.

Homework Equations

Gradient vector ∇F(x,y) = < dF/dx, dF/dy>
Normal vector n = < dF/dx, dF/dy, -1>

General form of tangent vector:
dF/dx(x-x0) + dF/dy(y-y0) + dF/dz(z-z0)

The Attempt at a Solution

∇F(x,y) = < 2x -3, 2y >
n = < 2x-3, 2y, -1>
n(1, 2, 2) = < -1, 4, -1>

-1(x-1) + 4(y-2) -1(z-2) = 0
-x + 1 +4y - 8 -z + 2 = 0
-x + 4y -z = 5

This is where I am stuck.

In order to be parallel to the plane x=1, should the parametric equations be
x = 1, y = 2 + 4t, z = 2 - t
or would it still be the tangent equation
x = 1 - t, y = 2 + 4t, z = 2-t?

 

[HallsofIvy]

Hey y'all, this is my first post. I am currently stuck on a multivariable question. Please let me know if you can help.

Homework Statement

The point, P = (1, 2, 2) lies on the surface z = x^2 + y^2 -3x. Find parametric equations for the tangent line to the surface through the point P parallel to the plane x = 1.

Homework Equations

Gradient vector ∇F(x,y) = < dF/dx, dF/dy>
Normal vector n = < dF/dx, dF/dy, -1>

General form of tangent vector:
dF/dx(x-x0) + dF/dy(y-y0) + dF/dz(z-z0)

This is not a tangent vector (or any vector at all). You cannot simply get an equation for "the" tangent vector to a surface at a point because there exist an infinite number of tangent vectors at a given point. I think what you mean is that the tangent plane satisfies the equation $(\partial F/\partial)(x- x_0)+ (\partial F/\partial y)(y- y_0)+ (\partial F/\partial z)(z- z_0)= 0$. For this exercise that is -(x- 1)+ 4(y- 2)- (z- 2)= 0 or x- 4y- z= 5. You want a line in that plane and in the plane x=1. That is, the line of intersection of the two planes.

The Attempt at a Solution

∇F(x,y) = < 2x -3, 2y >
n = < 2x-3, 2y, -1>
n(1, 2, 2) = < -1, 4, -1>

-1(x-1) + 4(y-2) -1(z-2) = 0
-x + 1 +4y - 8 -z + 2 = 0
-x + 4y -z = 5

This is where I am stuck.

In order to be parallel to the plane x=1, should the parametric equations be
x = 1, y = 2 + 4t, z = 2 - t
or would it still be the tangent equation
x = 1 - t, y = 2 + 4t, z = 2-t?

Where did you get these equations from? The first set, x= 1, y= 2+4t, z= 2- t does not satisfy -x+ 4y- z=5: -1+4(2+ 4t)- (2- t)= -1+ 8+ 16t- 2+ t= 5+ 17t, not 5 so does not lie in the tangent plane. The second does not have x always equal to 1 so does not lie in the plane x= 1 nor is it parallel to it.

Since x= 1 at the given point, the plane x=1 is through the given point and "parallel to the plane x= 1" means "lies in the plane x= 1" so the x coordinate is the constant, 1. Since you have correctly determined that the tangent plane is given by -x+ 4y- z= 5, the intersection of the two planes gives -1+ 4y- z= 5 or z= 4y- 6. Taking y itself as parameter, the line of intersection is x= 1, y= t, z= 4t- 6.