Answer:
x = -4 - 18t
y = 5 + 8t
z = 2 + 20t
Explanation:
To find the equation for the tangent plane to the given surface at point P(-4, 5, 2), we need to follow these steps:
(a) Equation of the Tangent Plane:
Calculate the partial derivatives of the surface equation with respect to x, y, and z.
Evaluate the partial derivatives at point P to obtain their values.
Use the values of the partial derivatives and the coordinates of point P to construct the equation of the tangent plane.
Let's calculate the partial derivatives:
∂/∂x (x² - xyz) = 2x - yz
∂/∂y (x² - xyz) = -xz
∂/∂z (x² - xyz) = -xy
Evaluate the partial derivatives at point P(-4, 5, 2):
∂/∂x (x² - xyz) = 2(-4) - 5(2) = -8 - 10 = -18
∂/∂y (x² - xyz) = -(-4)(2) = 8
∂/∂z (x² - xyz) = -(-4)(5) = 20
Now we have the values of the partial derivatives at point P, (-18, 8, 20). Using these values and the coordinates of point P(-4, 5, 2), we can construct the equation of the tangent plane:
-18(x - (-4)) + 8(y - 5) + 20(z - 2) = 0
-18(x + 4) + 8(y - 5) + 20(z - 2) = 0
-18x - 72 + 8y - 40 + 20z - 40 = 0
-18x + 8y + 20z - 152 = 0
So, the equation of the tangent plane to the given surface at point P is -18x + 8y + 20z - 152 = 0.
(b) Parametric Equations of the Normal Line:
The parametric equations of the normal line passing through point P on the surface can be expressed as:
x = -4 + at
y = 5 + bt
z = 2 + ct
Here, a, b, and c are the direction ratios of the normal vector to the surface at point P, which are the coefficients of x, y, and z in the equation of the tangent plane. So, a = -18, b = 8, and c = 20.
Therefore, the parametric equations of the normal line are:
x = -4 - 18t
y = 5 + 8t
z = 2 + 20t
These equations represent the line passing through point P(-4, 5, 2) and normal to the surface at that point.