What would the vector, parametric, and if possible, symmetric equations of the line through C(2, -2, 1) and parallel to the line with parametric equations x = -1 + 5t, y = 2 – t, z = 3 – 4t.

Answer 1

Here's my attempt.
Parametric equation: x = -1 + 5t, y = 2 - t, z = 3 - 4t since the lines are parallel.
Vector equation: (-1, 2, 3) + t(5, -1, -4)
Symmetric equation: (x + 1)/5 = (y - 2)/1 = (z - 3)/4

Can someone double check this for me please?

Answer 2

you have ignored the new point and are simply using the original given equation
Since the new line is parallel to the old one, simply edit your equations using the
new point.
Your given equation goes through the point (-1, 2, 3) , replace it with (2, -2, 1)

Answer 3

I also noted that in your answer for the symmetric equation you had the
direction vector incorrect.

Using the new point, it should have been
(x - 2)/5 = (y + 2)/-1 = (z - 1)/-4

Answer 4

Okay so the parametric eq'n would be: x = 2 + 5t, y = -2 - t, z = 1 - 4t and the vector would then be: (2, -2, -4) + t (5, -1, -4)

Answer 5

Made a mistake w/ the vector one. It should be (2, -2, 1)