# Equation of a line that passes through a point.

1. May 21, 2011

### -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.

2. May 21, 2011

### jbunniii

Your solution looks fine to me.

3. May 22, 2011

### 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?