# Thread: Line intersection

1. ## Line intersection

Let Line x be defined by $\bar {p_{1}}(t) = \bar {p_{1}}+t \bar {V_{1}}$ and
let Line y be defined by $\bar {p_{2} }(s) = \bar {p_{2}} + s \bar {V_{2}}$

Find conditions on line x,y such that they have unique point of intersection. Find this point.

Would that be $\bar {V_{1}} \dot \bar {V_{2}} \neq 0$? And how would I find the point?

2. Originally Posted by tttcomrader
Would that be $\bar {V_{1}} \dot \bar {V_{2}} \neq 0$?
If you're working in two dimensions, yes. If you're working in more dimensions, you need to make sure they're not skew; the easiest way to do that is to attempt to find your intersection.

And how would I find the point?
Set the two equations equal to each other and solve for t, then plug t into both your equations to make sure it gives you the same point in each. If it does, you've found your intersection. If it doesn't, the lines are skew.

3. Originally Posted by tttcomrader
Let Line x be defined by $\bar {p_{1}}(t) = \bar {p_{1}}+t \bar {V_{1}}$ and
let Line y be defined by $\bar {p_{2} }(s) = \bar {p_{2}} + s \bar {V_{2}}$

Find conditions on line x,y such that they have unique point of intersection. Find this point.

Would that be $\bar {V_{1}} \dot \bar {V_{2}} \neq 0$? And how would I find the point?
The 2 lines intersect even in 3-D if
$
(\overrightarrow{p_1} - \overrightarrow{p_2}) \cdot \frac{\overrightarrow{V_1} \times \overrightarrow{V_2}}{| \overrightarrow{V_1} \times \overrightarrow{V_2} |} = 0$

With this formula you calculate the distance between 2 lines. If the distance is zero then there must be a point of intersection.