Answer:
[tex]r=(0,11,-8)+(4,-3,7)t[/tex]
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:
[tex]r=r_{0} +tv[/tex]
where v is the direction vector, and [tex]r_{0}[/tex] is a point of the line.
So, for this case, the line pass for the point (0, 11, −8):
[tex]r_{0}=(0,11,-8)[/tex]
With the direction, the vector equation is:
[tex]r=(0,11,-8)+(4,-3,7)t[/tex]
The parametric equations are just the simplification of the vector equation:
x=4t
y=11-3t
z=-8+7t
The vector parametric equation of the line parallel to the given one and that passes through the point (0, 11, −8) is:
<x = 0 + 4t, y = 11 - 3t, z = -8 + 7t>
If you recall to 2-dimensional lines, two lines are parallel if the lines have the same rate of change.
The same happens here, if each variable has the same rate of change, then the lines will be parallel.
So if we want parametric equation of a line parallel to:
x = −1 + 4t, y = 6 − 3t, z = 3 + 7t
We must have:
x = a + 4t, y = b - 3t, z = c + 7t
Now, if we want this line to pass through the point (0, 11, -8), then we can replace the correspondent values in the constant term for each equation:
<x = 0 + 4t, y = 11 - 3t, z = -8 + 7t>
This is the parametric equation of the line parallel to the given one and that passes through the point (0, 11, −8)
If you want to learn more about parametric equations, you can read:
https://brainly.com/question/12695467
