# Intersection of lines

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1. Mar 25, 2017

### 0kelvin

1. The problem statement, all variables and given/known data
For which m and n the lines are concurrent?
$r: \begin{cases} x & - & y & = & 1 \\ nx & - & y & - & 2z & + & m & + & 1 & = & 0\end{cases}$

$s: \begin{cases} x & - & nz & + & m & + & n & = & 0 \\ x & + & y & - & 2nz & + & 11 & = & 0 \end{cases}$

Solving r gives me: $\left(1,0,\frac{m + 1 + n}{2}\right) + y\left(1,1,\frac{n - 1}{2}\right)$

Solving s gives me: $(-m - n, n - 11, 0) + z(n, n, 1)$

For n = -1 or n = 2 the direction vectors are parallel.

The answer in the book is that for $n \ne 2$ and $n \ne -1$ and $n + m = 5$ the lines are concurrent. However, I've found that for m = 10 and n = 0 the lines intersect at a single point.

2. Mar 25, 2017

### LCKurtz

I haven't checked your work, but assuming it's correct, I don't see what is bothering you. You know if $n \ne 2$ and $n \ne -1$ the lines are skew in 3D. So they may not intersect or may intersect at a point. So what's the problem?

3. Mar 25, 2017

### 0kelvin

I'm trying to reach the condition m + n = 5 for the lines to intersect at a single point. I plugged in some values for m and n such that m + n ≠ 5 and it's true, the lines don't intersect if m + n ≠ 5. However, it seems that m = 10 and n = 0 is an exception.

4. Mar 25, 2017

### LCKurtz

But the problem says if $m+n=5$ the lines are concurrent, i.e., the same line. So why are you trying to show they only intersect at a single point?

Last edited: Mar 25, 2017
5. Mar 25, 2017

### LCKurtz

Also, in your equation for $s$: $s = (-m - n, n - 11, 0) + z(n, n, 1)$, that point $(-m - n, n - 11, 0)$ doesn't satisfy the second equation for $s$ so you must have an arithmetic error.

6. Mar 25, 2017

### 0kelvin

I'm trying to find for which m the lines intersect and for which m they don't.