The symmetric equations of the normal line to the surface $z x ^ { 2 } - x y ^ { 2 } - y z ^ { 2 } = 18$ at (0, -2, 3) are

A) $\frac { x } { - 4 } = \frac { y + 2 } { - 9 } = \frac { z - 3 } { 12 }$
B) $\frac { x } { - 4 } = \frac { y + 2 } { 9 } = \frac { z - 3 } { 12 }$
C) $\frac { x } { 4 } = \frac { y + 2 } { - 9 } = \frac { z - 3 } { 12 }$
D) $\frac { x } { 4 } = \frac { y + 2 } { 9 } = \frac { z - 3 } { 12 }$
E) $\frac { x } { 9 } = \frac { y + 2 } { - 4 } = \frac { z - 3 } { 12 }$

Question

The symmetric equations of the normal line to the surface $z x ^ { 2 } - x y ^ { 2 } - y z ^ { 2 } = 18$ at (0, -2, 3) are

A) $\frac { x } { - 4 } = \frac { y + 2 } { - 9 } = \frac { z - 3 } { 12 }$
B) $\frac { x } { - 4 } = \frac { y + 2 } { 9 } = \frac { z - 3 } { 12 }$
C) $\frac { x } { 4 } = \frac { y + 2 } { - 9 } = \frac { z - 3 } { 12 }$
D) $\frac { x } { 4 } = \frac { y + 2 } { 9 } = \frac { z - 3 } { 12 }$
E) $\frac { x } { 9 } = \frac { y + 2 } { - 4 } = \frac { z - 3 } { 12 }$

Quizplus · Accepted Answer

The gradient vector of the surface at a point gives the direction of the normal line. To find the gradient, we differentiate the given equation with respect to x, y, and z to get the components of the normal vector. The partial derivatives at (0, -2, 3) are: $\frac{\partial z}{\partial x} = 2xz - y^2 = -4$, $\frac{\partial z}{\partial y} = -2xy - 2yz = -9$, and $\frac{\partial z}{\partial z} = x^2 - 2yz = 12$. Thus, the symmetric equations of the normal line are $\frac{x}{-4} = \frac{y + 2}{-9} = \frac{z - 3}{12}$.

The Symmetric Equations of the Normal Line to the Surface zx2−xy2−yz2=18z x ^ { 2 } - x y ^ { 2 } - y z ^ { 2 } = 18zx2−xy2−yz2=18

The Symmetric Equations of the Normal Line to the Surface $z x ^ { 2 } - x y ^ { 2 } - y z ^ { 2 } = 18$