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

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Homework Help Overview

The discussion revolves around determining the conditions under which two sets of lines, defined by parameters m and n, are concurrent. The lines are expressed in parametric form, and participants are exploring the implications of specific values for m and n on the intersection of these lines.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the conditions for concurrency, specifically focusing on the equation m + n = 5. There are attempts to identify exceptions to this condition, such as the case where m = 10 and n = 0. Some participants question the validity of certain assumptions regarding parallelism and intersection.

Discussion Status

The discussion is ongoing, with participants sharing their findings and questioning the implications of their results. There is a recognition of the complexity of the problem, particularly regarding the conditions under which the lines may or may not intersect. Some guidance has been offered regarding the nature of the lines based on the values of n.

Contextual Notes

Participants note that for n = -1 or n = 2, the direction vectors of the lines become parallel, which raises questions about the nature of their intersection. There is also mention of potential arithmetic errors in the equations being used, which may affect the conclusions drawn.

0kelvin
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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.
 
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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?
 
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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?
 
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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.
 

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