Equation of a line that passes through a point.

-Dragoon-

1. The problem statement, all variables and given/known data
Find the equation of the line that passes through A(1,-4, 2) and is parallel to the intersection line of the two planes x - 2y + 3z - 1 = 0 and x - 4y+ 2z - 8 = 0


2. Relevant equations
N/A


3. The attempt at a solution
First I set the first and second equations to [1] and [2]:
x - 2y + 3z - 1 = 0 [1]
x - 4y+ 2z - 8 = 0 [2]
I then multiply [1] by 2 and use elimination to get rid of the y variable for now:
2x - 4y + 6z - 2 = 0
x - 4y + 2z - 8 = 0
________________________
x + 4z + 6 = 0 [3]

I'll then let z = t to solve for x in equation [3]:
x + 4t + 6 = 0
x = -4t - 6
Now I substitute z = t and x = -4t - 6 into equation [1] to solve for y:
-4t - 6 - 2y + 3t - 1 = 0
y = (-1/2)t - 7/2

Now that I have the values of all the unknowns, I first express it in parametric form:
x = -4t - 6
y = (-1/2)t - 7/2
z = t
Knowing this, finally, the direction vector for the line that passes through A(1, -4, 2) can be expressed:
(x,y,z) = (1, -4, 2) + t(-4, 1/2, 1)

I just wanted to know, did I do this correctly? I feel as if I did something wrong. If I did, can you point where I went wrong? Thank you in advance.

jbunniii

Your solution looks fine to me.

LCKurtz

Another possibly easier way to work this kind of problem is to note that the cross product of the two normal vectors to the planes gives a direction vector for the line. Do you see why?