Question

Recall that if a line is parallel to the vector v and passes through the point P0, which is the tip of the position vector r0, then the vector equation of the line is given by r(t) = r0 + tv. For the given line, we have r0 = 7, −8, 3 and v = 1, 6, − 1 3 . So the vector equation for this line is r(t) = 7, −8, 3 + t 1, 6, − 1 3 = .

Answer 1

Answer:

The vector equation of the line is

                                $\overrightarrow{r}=<7,-8,3>+t<1,6,-13>$

Parametric equations for given line are

                                       $x=7+t\\y=-8+6t\\z=3-13t$

Explanation:

The vector equation of the line is given by

                                     $r(t) = r_{o} + tv$

r₀ = (7, -8, 3)

v = (1, 6, -13)

At these points the vector equation for this line is:

$\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=<7,-8,3>+t<1,6,-13>$

Parametric equations for given line are

                                       $x=7+t\\y=-8+6t\\z=3-13t$