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निम्न रेखाओं के मध्य लघुत्तम दूरी ज्ञात कीजिए \( \frac{x-1}{2}=\frac{y+1}{3}=z\) और \(\frac{x + 1}{3} =\frac{y - 2}{1}= z\) तथा लघुतम दूरी वाली रेखा का समीकरण भी ज्ञात कीजिए।

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अतः रेखा (1) पर कोई बिंदु P(2r1 + 1, 3r1 – 1, r1) तथा

रेखा (2) पर कोई बिंदु Q(3r2 – 1, r2 + 2, r2 + 2)

तब रेखा PQ के दिक्-अनुपात

= 3r2 – 2r1 – 2, r2 – 3r1 + 3, 2 – r1

PQ रेखा (1) के लम्बवत् हैं, इसलिए

a1a2 + b1b2 + c1c2 = 0

2(3r2 – 2r1 – 2) + 3(r2 – 3r1 + 3) + 1(2 – r1) = 0

9r2 – 14r1 = 7 …(3)

PQ रेखा (2) के लम्बवत् है।।

इसलिए a1a2 + b1b2 + c1c2 = 0

3(3r2 – 2r1 – 2) + 1(r2 – 3r1 + 3) + 0(2 – r1) = 0

10r2 – 9r1 – 3 = 0

समी. (3) व (4) को हल करने पर

r1 = \(\frac{97}{59}\) , r2 = \(\frac{105}{59}\)

r1 व r2 के ये मान बिन्दु P व Q में रखने पर

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