यदि दो परस्पर लम्ब रेखाओ की दिक्-कोजनाएँ `l _(1),m_(1),n_(1)`और `l _(2),m_(2),n_(2)`
हों तो बताओ कि एक दोनों पर लम्ब रेखा कि दिक्-कोसाइन
`m_(1)n_(2)-m_(2)n_(1),n_(1)l_(2)-n_(1)l_(1),l_(1)m_(2)-l_(2)m_(1)` है .

Explanation

दी गई रेखाओ के direction cosines क्रमशः `l_(1),m_(1),n_(1)" तथा "l_(2),m_(2),n_(2)` है . इसलिए इनके, अनुदिश इकाई सदिशा है,
`hata=l_(1)hati+m_(1)hatj+n_(1)hatk" "...(1)`
`hatb=l_(2)hati+m_(2)hatj+n_(2)hatk" "...(2)`
अब, `hataxxhatb=(l_(1)hati+m_(1)hatj+n_(1)hatj)xx(l_(1)+m_(2)hatj+b_(2)hatk)`
`=.{:(hati, hatj, hatk),(l_(1),m_(1), n_(1)),(l_(2), m_(2), n_(2)):}.`
`=(m_(1)n_(2)-m_(2)n_(1))hati+(n_(1)l_(2)-n_(2)l_(1))hatj+(l_(1)m_(2)-l_(2)m_(1))hatk`
`because.hataxxhatb.=.hata..hatb .sin 90^(@)=1`
अतः अभीष्ट दिक्-कोज्याएँ (direction cosines ) है,
`=(m_(1)n_(2)-m_(2)n_(1)),(n_(1)l_(2)-n_(2)l_(1)), (l_(1)m_(2)-l_(2)m_(1)).`
Seconed method:
माना कि अभीष्ट रेखा के दिक्-अनुपात `l_(3),m_(3),n_(3)` है .
चूँकि यह दो रेखाओ जिनकी दिक्-कोज्याएँ (direction cosines ) `l_(1), m_(1), n_(1)` तथा ` l_(2),m_(2), n_(2)` है पर लम्ब है,
`thereforel_(1)l_(3)+m_(1)+n_(1)n_(3)=0" "...(1)`
तथा `l_(2)l_(3)+m_(2)m_(3)++n_(2)n_(3)=0" "...(2)`
(1) तथा (2) को हल करने पर हमें मिलता है,
`(l_(3))/(m_(1)n_(2)-m_(2)n_(1))=(m_(3))/(n_(1)l_(2)-n_(2)l_(1))=(n_(3))/(l_(1)m_(3)-l_(2)m_(1)).`
`thereforem _(1)n_(2)-m_(2)n_(1),n_(1)l_(2)-n_(2)l_(1),l_(1)m_(2)-l_(2)m_(1)` अभीष्ट रेखा के दिक्-अनुपात है.
`therefore` रेखा के दिक्-कोज्याएँ (direction cosines ) है,
`(m_(1)n_(2)-m_(2)n_(1))/(sqrt(sum(m_(1)n_(2)-m_(2)n_(1))^(2))), (n_(1)l_(2)-n_(2)l_(1))/(sqrt(sum(m_(1)n_(2)-m_(1)n_(2))^(2))),(l_(1)m_(2)-l_(2)m_(1))/(sqrt(sum(m_(1)n_(2)-m_(2)n_(1))^(2)))`
अब `sum (m_(1)n_(2)-m_(2)n_(1))^(2)`
`=(m_(1)n_(2)-m_(2)n_(1))^(2)+(n_(1)l_(2)-n_(2)l_(1))^(2)+(l_(1)m_(2)-l_(2)m_(1))^(2)`

`=(l_(1)^(2)+m_(1)^(2)+n_(1)^(2))(l_(2)^(2)+m_(2)^(2)+n_(2)^(2))-(l_(1)l_(2)+m_(1)+n_(1)n_(2))^(2)`
`=1.1-0^(2)=1.`
`therefore ` अभीष्ट दिक्-कोज्याएँ `(m_(1)n_(2)-m_(2)n_(1)),(n_(1)l_(2)-n_(2)l_(2)), (l_(1)m_(2)-l_(2)m_(1))` है .