Intersection of Lines: Solving for m and n to Determine Concurrent Lines

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
0kelvin
Messages
50
Reaction score
5

Homework Statement


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.
 
Physics news on Phys.org
0kelvin said:

Homework Statement


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.
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?
 
  • Like
Likes   Reactions: Floydd
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.
 
0kelvin said:
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.
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:
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.
 
I'm trying to find for which m the lines intersect and for which m they don't.