Find parametric equations for the line described below.
-The line through the point P(6, 0, 5) and parallel to the line x = 2t + 5, y = 3t + 4, z = 3t - 6
A) x = 2t + 6, y = 3t, z = 3t + 5
B) x = 2 - 6, y = 3t, z = 3t - 5
C) x = 6, y = 3t, z = -3t + 5
D) x = 3t + 6, y = -2t, z = 5

Question

Quizplus · Accepted Answer

The direction vector of the given line is determined by the coefficients of $t$ in the parametric equations, which are $2, 3,$ and $3$ for $x, y,$ and $z$ respectively. To find a line parallel to the given line and passing through point $P(6, 0, 5)$, we use the same direction vector and the point $P$ to write the new parametric equations: $x = 2t + 6, y = 3t + 0, z = 3t + 5$, which simplifies to $x = 2t + 6, y = 3t, z = 3t + 5$.

Find Parametric Equations for the Line Described Below