The symmetric equations of the normal line to the surface $x ^ { 2 } - 3 y ^ { 2 } - 4 z ^ { 2 } = 2$ at (3, 1, 1) are

A) $\frac { x - 3 } { - 3 } = \frac { y - 1 } { 3 } = \frac { z - 1 } { - 4 }$
B) $\frac { x - 3 } { 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { 4 }$
C) $\frac { x - 3 } { 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { - 4 }$
D) $\frac { x - 3 } { - 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { 4 }$
E) $\frac { x - 3 } { 3 } = \frac { y - 1 } { 3 } = \frac { z - 1 } { 4 }$

Question

The symmetric equations of the normal line to the surface $x ^ { 2 } - 3 y ^ { 2 } - 4 z ^ { 2 } = 2$ at (3, 1, 1) are

A) $\frac { x - 3 } { - 3 } = \frac { y - 1 } { 3 } = \frac { z - 1 } { - 4 }$
B) $\frac { x - 3 } { 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { 4 }$
C) $\frac { x - 3 } { 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { - 4 }$
D) $\frac { x - 3 } { - 3 } = \frac { y - 1 } { - 3 } = \frac { z - 1 } { 4 }$
E) $\frac { x - 3 } { 3 } = \frac { y - 1 } { 3 } = \frac { z - 1 } { 4 }$

Quizplus · Accepted Answer

The gradient vector of the surface at a point gives the direction of the normal line. For the given surface $x^2 - 3y^2 - 4z^2 = 2$, the gradient is $\nabla f = (2x, -6y, -8z)$. At the point (3, 1, 1), the gradient is $(6, -6, -8)$, which gives the direction ratios of the normal line. Thus, the symmetric equations of the normal line are $\frac{x-3}{6} = \frac{y-1}{-6} = \frac{z-1}{-8}$, which simplifies to choice C when dividing all components by -2 for simplicity.

The Symmetric Equations of the Normal Line to the Surface x2−3y2−4z2=2x ^ { 2 } - 3 y ^ { 2 } - 4 z ^ { 2 } = 2x2−3y2−4z2=2

The Symmetric Equations of the Normal Line to the Surface $x ^ { 2 } - 3 y ^ { 2 } - 4 z ^ { 2 } = 2$