Find (a) tangent plane and (b) normal line at the point P0 on the given surface: (a) x 2 + y 2 − 2xy − x + 3y − z = −4, at P0(2, −3, 18).

We are given the surface equation:

\[
x^2 + y^2 - 2xy - x + 3y - z = -4
\]

We are tasked with finding the tangent plane and normal line at the point \( P_0(2, -3, 18) \).

### Step 1: Implicit differentiation for tangent plane

The equation of the surface can be written as a level surface of a function \( F(x, y, z) \), where

\[
F(x, y, z) = x^2 + y^2 - 2xy - x + 3y - z + 4 = 0
\]

To find the tangent plane at \( P_0(2, -3, 18) \), we first compute the partial derivatives of \( F(x, y, z) \):

\[
F_x = \frac{\partial F}{\partial x} = 2x - 2y - 1
\]
\[
F_y = \frac{\partial F}{\partial y} = 2y - 2x + 3
\]
\[
F_z = \frac{\partial F}{\partial z} = -1
\]

### Step 2: Evaluate the partial derivatives at \( P_0(2, -3, 18) \)

Substitute \( x = 2 \), \( y = -3 \), and \( z = 18 \) into the partial derivatives:

\[
F_x(2, -3, 18) = 2(2) - 2(-3) - 1 = 4 + 6 - 1 = 9
\]
\[
F_y(2, -3, 18) = 2(-3) - 2(2) + 3 = -6 - 4 + 3 = -7
\]
\[
F_z(2, -3, 18) = -1
\]

So, the gradient of \( F(x, y, z) \) at \( P_0(2, -3, 18) \) is:

\[
\nabla F(2, -3, 18) = (9, -7, -1)
\]

### Step 3: Equation of the tangent plane

The equation of the tangent plane at \( P_0(2, -3, 18) \) is given by:

\[
F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0
\]

Substitute the values of the partial derivatives and the coordinates of \( P_0(2, -3, 18) \):

\[
9(x - 2) - 7(y + 3) - (z - 18) = 0
\]

Simplifying the equation:

\[
9x - 18 - 7y - 21 - z + 18 = 0
\]
\[
9x - 7y - z - 21 = 0
\]

Thus, the equation of the tangent plane is:

\[
9x - 7y - z = 21
\]

### Step 4: Equation of the normal line

The direction of the normal line is given by the gradient vector \( \nabla F(2, -3, 18) = (9, -7, -1) \). The parametric equation of the normal line through \( P_0(2, -3, 18) \) is:

\[
x = 2 + 9t
\]
\[
y = -3 - 7t
\]
\[
z = 18 - t
\]

### Final answers:

(a) The equation of the tangent plane is:

\[
9x - 7y - z = 21
\]

(b) The parametric equations of the normal line are:

\[
x = 2 + 9t, \quad y = -3 - 7t, \quad z = 18 - t
\]

Do you want further details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How do you compute the gradient vector for a function of three variables?
2. What is the geometric meaning of a tangent plane to a surface at a point?
3. Can the normal line and the tangent plane be perpendicular to each other?
4. How can we interpret the parametric equations of the normal line geometrically?
5. How do the coefficients in the tangent plane equation relate to the surface's geometry at that point?

**Tip:** The gradient vector gives both the direction and rate of the steepest increase of a function at a point!

Find (a) tangent plane and (b) normal line at the point P0 on the given surface: (a) x^2 + y^2 − 2xy − x + 3y − z = −4, at P0(2, −3, 18).

Find the tangent plane and normal line to the surface defined by x^2 + y^2 − 2xy − x + 3y − z = −4 at the point P(2, −3, 18). This problem uses partial derivatives and the gradient vector to solve for the tangent plane equation and the parametric equations of the normal line. Suitable for students studying multivariable calculus.

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