Question

Finding the parametric equations of the line... ?

Answer 1

...that passes through the point A(6, 4, 0) and is parallel to the line passing through B(-2, 0, 4) and C(3, -2, 1)?

Answer 2

...if (-4, m, n) is a point on "l", find m and n?

Answer 3

It really REALLY helps to draw a sketch, here I'll give you one you can print out and work on :-)

The parametric equations for this should be easy if you think of how they would be project on each face of the cube's walls:
\[x(t) = 5 t-2\]
\[y(t) = -2 t\]
\[z(t) = 4-3 t\]

Answer 4

so...
x=6+5t
y=4-2t
z=-3t

?

Answer 5

Find a point on this 3D line that shares some coordinates in common with A would be the next step

The last step is after you have your new equation you'll want to simply solve for the other two points using the knows. I couldn't do this with out visualizing it myself, that's just what kind of learner I am I guess.

I'm going to try your equations here in just a moment...

Answer 6

\[ r(t) = (x_1, y_1, z_1) + t(x_2 - x_1, y_2 - y_1, z_2 - z_1)\]
or
\[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} = t \]
from here you can work out

Answer 7

For the two points B & C:
x(0) = -2
y(0) = 0
z(0) = 4
x(1) = 3
y(1) = -2
z(1) =1

(this is simple plug & chug)

Then this formula (which hopefully you covered in class):
\[\frac{x−x1}{x2−x1}=\frac{y−y1}{y2−y1}=\frac{z−z1}{z2−z1}=t\]

If you're a bit rusty, that's ok and here's a pretty good refresher: http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx

Answer 8

And of course @experimentX beats me to the final explanation :P hehe