# Intersection of Lines (Vectors)

1. Aug 26, 2013

### FeDeX_LaTeX

1. The problem statement, all variables and given/known data
Show that if the two lines

$\frac{x - c_1}{d_1} = \frac{y - c_2}{d_2} = \frac{z - c_3}{d_3}$ and

$\frac{x - d_1}{c_1} = \frac{x - d_2}{c_2} = \frac{x - d_3}{c_3}$

intersect, they lie in the plane $r.(c \times d) = 0$

where c = c1i + c2j + c3k and d = d1i + d2j + d3k

3. The attempt at a solution

I've shown that they intersect at $x = c_1 + d_1, y = c_2 + d_2, z = c_3 + d_3$. It seems intuitively obvious that the lines lie in the plane r.(c x d) = 0, but I just can't seem to properly show this. Any advice?

2. Aug 26, 2013

### verty

It is possible for the lines to intersect at infinitely many points, for example let $c_i = d_i = i$. In this case, $r . (c × d) = 0$ isn't a plane. So I think one must assume we are talking about lines that intersect at exactly one point, if at all.

Given this, and assuming your proof is correct that they intersect at c+d if they intersect at all, showing that the points (c, d, c+d) are in that plane will suffice because each line goes through two of those points.

3. Aug 26, 2013

### FeDeX_LaTeX

Thanks -- I assume that the lines are non-parallel. All three points lie in the plane.

4. Aug 26, 2013

### haruspex

The question is a little odd. Those two lines necessarily intersect (at x = c+d).
Given an equation like $\frac{x - c_1}{d_1} = \frac{y - c_2}{d_2} = \frac{z - c_3}{d_3}$, you can set each term to equal some scalar parameter t, then express the line as a vector equation. It should then be obvious that the line lies in the subspace (be it a line or a plane) generated by c and d.