Question

Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel to the line x = −1 + 4t, y = 6 − 3t, z = 3 + 7t

Answer 1

Answer:

$r=(0,11,-8)+(4,-3,7)t$

x=4t

y=11-3t

z=-8+7t

Step-by-step explanation:

The line is parallel to x=-1+4t, y=6-3t, z=3+7t.

Two lines are parallel if they have the same direction, and in the parametric form, the direction of a line is always the vector of constants that multiply t (or the parameter). So in this case, the direction is (4,-3,7); this is also the direction of the missing line because they are parallel.

The vector equation of a line is given by:

$r=r_{0} +tv$

where v is the direction vector, and $r_{0}$ is a point of the line.

So, for this case, the line pass for the point (0, 11, −8):

$r_{0}=(0,11,-8)$

With the direction, the vector equation is:

$r=(0,11,-8)+(4,-3,7)t$

The parametric equations are just the simplification of the vector equation:

x=4t

y=11-3t

z=-8+7t