दो दिखाएँ जिनके सदिश समीकरण निम्नलिखित हैं, के बीच कि न्यूनतम दुरी ज्ञात कीजिये|
(i) `vecr =hati + 2hatj +3hatk + lambda(2hati + 3hatj + 4hatk)`, तथा (and)
`vecr = 2hati + 4hatj + 5hatk +mu(3hati + 4hatj + 5hatk)`

Explanation

दो रेखाओं `vecr = veca_(1) +lambda vecb_(1)` तथा `vecr = veca_(2) + muvecb_(2)` के बीच कि न्यूनतम दुरी,
`d=|((veca_(2)-veca_(1))(vecb_(1) xx vecb_(2)))/(|vecb_(1)-vecb_(2)|)|`............(1)
(i) यहाँ `veca_(1) = hati + 2hatj + 3hatk, vecb_(1) = 2hati + 3hatj + 4hatk`
`veca_(2) = 2hati + 4hatj + 5hatk` तथा `vecb_(2) = 3hati + 4hatj + 5hatk`
अब `veca_(2) -veca_(1) = (2hati + 4hatj + 5hatk)- (hati + 2hatj + 3hatk)`
`=hati + 2hatj + 2hatk`.............(2)
`vecb_(1) xx vecb_(2) = |:(hati, hatj,hatk),(2,3,4),(3,4,5):|= (15-16)hati - (10-12)hatj + (8-9)hatk`
`=-hati + 2hatj - hatk`.........(3)
`|vecb_(1) xx vecb_(2)|= sqrt((-1)^(2) + 2^(2) + (-1)^(2))=sqrt(6)`...........(4)
`(veca_(2)-veca_(1)).(vecb_(1) xx vecb_(2)) =(hati + 2hatj + 2hatk).(-hati + 2hatj - hatk)`
`=1 xx (-1) +2 xx 2 + 2 xx (-1) =1`........(5)
(4) तथा (5) में प्राप्त मानों को (1) में रखने पर हमें मिलता हैं,
`d=|1/sqrt(6)|=1/sqrt(6)`