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Projecting two non-parallel lines until they meet

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Projecting two non-parallel lines until they meet


this ostensibly seems to be a simple problem, but i'm not sure how to go about solving it.

I'll present a simplified, hypothetical example:

I have two non-parallel lines, lets say of length 3 and 5 (in whatever units). I also know the distances between all four end points of the lines - and therefore the angles of the polygon that could be made by joining the line ends together. What I want to do is project both lines until they meet, then calculate the angle between the lines at the point at which they meet. Ideally, I would also know the length of the projected sections along each line. Is possible? Any suggestions would be greatly appreciated.



Hi Steve,
Please take paper & pencil and draw the figure. Set PQ=L=5 horizontal (you don't loss generality with this). Above P and to the right put R, so PR=D (known). Above Q and to the left, but below R, put S, so RS=l=3 and SQ=d (known). Thus we have the counter-clockwise "quadrilater" PQSRP. Prolongations of RS and PQ meet at O where the unknown "meet angle M" is formed. You also said diagonals PS=p and QR=q are known (as you know all the distances between the end points). So we apply cosine formula to the triangles PQR and SRQ, i.e.

p^2 = L^2 + D^2 - 2LD\cos P, so angle at P is obtained,
q^2 = L^2 + d^2 - 2Ld\cos Q, so angle at Q is obtained.

From R trace down a vertical line intersecting PQ at F. From S, and to the left, trace a horizontal line intersecting the vertical RF at H. So you have formed the right triangle SHR; here the internal angle at S is also the meet angle M (two parallels cutted by the same secant line, i.e. the line supporting the segment OS). Trace from S a vertical line intersecting PQ at G (which will be to the right of F). Now you will easily see how to obtain the meet angle M. That is,

D\cos P + l\cos M + d\cos Q = L, so meet angle M is obtained.

But you also want to find out distances OS=a and OQ=A, which are easily computed by the equations (for instance)

a\sin M = d\sin Q, so "a" is obtained,
(A + d\cos Q)\tan M = d\sin Q, so A is obtained.

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