# Linear Algebra Lines vs. Vectors

• Sep 24th 2008, 03:15 PM
Hellreaver
Linear Algebra Lines vs. Vectors
Hi, I'm new to the forum, and to linear algebra. Not having much fun with that latter, I'm afraid. So I hope that this helps me understand the concepts that it deals with.
I guess I'll start with a question, try to work my way through it, and hopefully get it right, but if not, please correct me, and tell me where I went wrong and why.

Find the equation of the line parallel to v=(5,-2,1) passing through the point (1,6,2), and determine whether the point (5,4,3) is on this line.

I think that the equation of the line would be x=(1,6,2) + t (5,-2,1), because this would mean that (1,6,2) is on the line, and that the line has a direction of (5,-2,1). I'm not exactly sure how to figure out if the point is on the line, or if I'm even going in the right direction with this problem...
• Sep 24th 2008, 03:29 PM
Plato
Quote:

Originally Posted by Hellreaver
Find the equation of the line parallel to v=(5,-2,1) passing through the point (1,6,2), and determine whether the point (5,4,3) is on this line.
I think that the equation of the line would be x=(1,6,2) + t (5,-2,1)

You are correct!
Now, is there a value of t for which x=(5,4,3)?
• Sep 24th 2008, 03:33 PM
Hellreaver
So would it then become:

(5,4,3)= (1,6,2) + t (5,-2,1)
(5,4,3)-(1,6,2)=t (5,-2,1)
(4,-2,1)= t (5,-2,1)

and since the x values are different, and changing t from 1 to any other number would change the values of y and z, the point is not on the line?

Edit: Wow, dumb question, of course it isn't. Alright, thanks a bunch for that. I have another one, kinda similar to the first, but I'm not sure how to go about it...

Find parametric equations of the line that is perpendicular to the plane x+2y+3z=4 and passes through the point (1,1,-1).

I'm not even sure where to being on this one... Should I solve for x in the equation of the plane, or what?