The symmetric equations of the line through (3,-2,5) in the direction of $\frac { x - 3 } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { - 1 }$ is

A) $x = 3 t + 3 , y = - t + 2 , z = - t + 5$
B) $x = - 3 t - 3 , y = - t - 2 , z = - t - 5$
C) $x = 3 t + 3 , y = - t - 2 , z = - t - 5$
D) $x = 3 t + 3 , y = - t - 2 , z = - t + 5$
E) $x = - 3 t + 3 , y = - t - 2 , z = - t + 5$

Question

The symmetric equations of the line through (3,-2,5) in the direction of $\frac { x - 3 } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { - 1 }$ is

A) $x = 3 t + 3 , y = - t + 2 , z = - t + 5$
B) $x = - 3 t - 3 , y = - t - 2 , z = - t - 5$
C) $x = 3 t + 3 , y = - t - 2 , z = - t - 5$
D) $x = 3 t + 3 , y = - t - 2 , z = - t + 5$
E) $x = - 3 t + 3 , y = - t - 2 , z = - t + 5$

Quizplus · Accepted Answer

The Answer of The symmetric equations of the line through (3,-2,5) in the direction of $\frac { x - 3 } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { - 1 }$ is

A) $x = 3 t + 3 , y = - t + 2 , z = - t + 5$
B) $x = - 3 t - 3 , y = - t - 2 , z = - t - 5$
C) $x = 3 t + 3 , y = - t - 2 , z = - t - 5$
D) $x = 3 t + 3 , y = - t - 2 , z = - t + 5$
E) $x = - 3 t + 3 , y = - t - 2 , z = - t + 5$

The Symmetric Equations of the Line Through (3,-2,5) in the Direction