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

Click For Summary
The discussion focuses on determining the conditions under which two sets of lines, defined by equations r and s, are concurrent. The consensus is that the lines are concurrent when n is not equal to 2 or -1, and when the sum of m and n equals 5. However, an exception is noted where m equals 10 and n equals 0, which results in the lines intersecting at a single point despite not meeting the general condition. Participants highlight the importance of verifying calculations, particularly in the equations for s, to ensure accurate results. The conversation emphasizes the nuances in the relationships between m and n for line concurrency.
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 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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
907
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
883
Replies
5
Views
2K
Replies
6
Views
2K