# Giving the equation of a line

1. Oct 31, 2013

### fogvajarash

1. The problem statement, all variables and given/known data
Find the scalar equations of the line passing through p(6, 0, 3), intersecting the line (x y z) = (1 2 -3) + t(1 -2 0) and perpendicular to it

2. Relevant equations
-

3. The attempt at a solution
I don't know where to start with the problem, I tried using projections and then giving out equations for the scalar product of the two lines (which is 0) and the equations of the points of intersection, but there are so many variables I can't do anything with them. I'm stuck with this problem.

2. Oct 31, 2013

### Dick

If t0 is the value of t where they intersect then (6,0,3)-[(1,2,-3)+t0(1,-2,0)] must be perpendicular to (1,-2,0), yes? Find t0. There aren't that many variables. There's really only one.

Last edited: Oct 31, 2013
3. Nov 2, 2013

### fogvajarash

Got it, I can't believe i was so blind. Just asking, does the direction vector of a line always have to be given in the lowest integers as possible? (for example, we can have a direction vector (8 -4 2) and can we express it as (4 -2 1)?

4. Nov 3, 2013

### Dick

If the two direction vectors point in the same direction then it's the same line. No, there's no requirement to do it like that. You could equally well say (400, -200, 100) or (.4, -.2, .1).

5. Nov 3, 2013

### Jose.Guerrero

A solution to the problem.

in order to find an equation of a line you need two things: a direction (V) and a point (Po). In order to find the direction we need to find another point on the wanted line. To do this, we use the perpendicular line to our advantage. I set x=6 since Z always = -3 then I found the parameter t and plugged in the value to get the y... Now you have a second point on the wanted line (P1) the direction vector can now be found (vector from Po to P1) then by taking this vector and one to those 2 points (doesn't matter, Po or P1) we can put into vector form r=<ro>+t<v> and then put it into parametric from... Walah! you're done. :)