Solving 2 Line Equations: Find Point of Intersection

AI Thread Summary
To determine the intersection of two lines represented by the equations r1=a+lp and r2=b+mq, the condition for intersection is given by (a-b)·(p×q)=0, where p and q are the direction vectors of the lines. This condition indicates that the lines are either parallel or intersecting, but does not guarantee intersection if p and q are multiples of each other. For finding the point of intersection, it is necessary to express the lines in vector form and solve for parameters l and m that satisfy both equations simultaneously. If p and q are not multiples, the lines can either intersect at a single point or not at all. Understanding these conditions is crucial for solving problems involving the intersection of lines in three dimensions.
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Homework Statement


One of those annoying questions that should be simple, but that I've forgotten how to do:

Two lines are given by the equations r1=a+lp and r2=b+mq. Find the condition for the lines to cross, and find there position of intersection.

Homework Equations



The Attempt at a Solution


I've done the first bit: (a-b).(p*q)=0 where * means the cross product. But I can't find a way of doing the second bit (find the point of intersection), I'd know how to do it if the actual vectors were given, but how do you write in in a nice vector form?
Thanks
 
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Where did you get that "first bit". If p and q are perpendicular, then pXq= 0 so that condition is satisfied for all a and b but the line do not necessarily intersect. Or are you working in 2 dimensions?
 
I'm working in three dimensions.
To get the first part I used the result that the minimum distance between two skew lines= (a-b).(p*q)/|p*q| where || means modulus which comes from the fact that the vector across the minimum distance will be perpendicular to both lines, and using the dot product to find the cosine of an angle. Then I set this equal to zero to find when the intersect.

As you pointed out (and I didn't realize) this is only going to work if p and q are not a multiple of one another, so I either need to add that onto the end as a condition with an explanation there could in this case either be infinite intersetions or no intersection. Alternatively do you know a better way of finding a condition?
Thanks for your help.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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