Find both the parametric and the vector equation of the line.
-The line through (0, 1, 0) in the direction of the vector v = (3, 0, -1)
A) x = 3, y = t, z = -1; r = (3, 0, -1) + t(0, 1, 0)
B) x = 3, y = t, z = -1; r = (0, 1, 0) + t(3, 0, -1)
C) x = 3t, y = 1, z = -t; r = (3, 0, -1) + t(0, 1, 0)
D) x = 3t, y = 1, z = -t; r = (0, 1, 0) + t(3, 0, -1)

Question

Quizplus · Accepted Answer

The parametric equations of a line can be found using the point and direction vector given. For a line through point $P_0(x_0, y_0, z_0)$ in the direction of vector $\mathbf{v} = (a, b, c)$, the parametric equations are $x = x_0 + at$, $y = y_0 + bt$, and $z = z_0 + ct$. Here, $P_0 = (0, 1, 0)$ and $\mathbf{v} = (3, 0, -1)$, so the parametric equations are $x = 0 + 3t = 3t$, $y = 1 + 0t = 1$, and $z = 0 - 1t = -t$. The vector equation of a line is $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$, where $\mathbf{r}_0$ is the position vector of a point on the line and $\mathbf{v}$ is the direction vector. Thus, the vector equation is $\mathbf{r} = (0, 1, 0) + t(3, 0, -1)$.

Find Both the Parametric and the Vector Equation of the Line