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## 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?