इकाई मापांक वाले उस सदिश को ज्ञात कीजिए जो , सदिशों ` veca = 3hati + hatj – 4hatk ” तथा” vecb = 6hati + 5hatj – 2hatk ` से लम्बवत हो |
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माना सदिश `vecc= vecc_(1) hati + c_(2) hatj + c_(3) hatk ` है तथा मापांक = 3
इकाई , तब
`|vecc| = 3 `
` rArr sqrt(c_(1)^(2) + c_(2)^(2) + c_(3)^(2)) = 3 `
`rArr c_(1)^(2)+ c_(2)^(2) + c_(2)^(2) = 9` …(1)
तथा यह भी दिया है की `vecc , veca ” तथा” vecb` से लम्बवत है तब
`vecc bot veca`
` rArr vecc. veca= 0 `
`rArr (c_(1)hati + c_(2)hatj + c_(3)hatk).(3hati + hatj – 4hatk) = 0 `
` rArr 3c_(1) + c_(2) – 4c_(3) = 0` … (2)
तथा `vecc bot vecb`
`rArr (c_(1)hati + c_(2)hatj + c_(3) hatk). (6hati + 5hatj – 2hatk) = `
`rArr 6c_(1) + 5c_(2) – 2c_(3) = 0 ` …(3)
समी (2) तथा (3) से
`(c_(1))/((-2+ 20))= (c_(2))/((-24 + 6)) = (c_(3))/((15-6)) = lambda`
`rArr (c_(1))/(18) = (c_(2))/(-18) = (c_(3))/(9) = lambda`
`rArr (c_(1))/(2) = (c_(2))/(-2) = (c_(3))/(1) = lambda`
`rArr c_(1) = 2lambda, c_(2) = – 2lambda ” तथा” c_(2)= lambda `
यह मान समी (1) में रखने पर
`4lambda^(2) + 4lambda + lambda^(2) = 9`
`rArr lambda^(2) = 1`
`rArr lambda = 1`
`therefore c_(1) = 2, c_(2) = – 2 “तथा ” c_(3) = 1`
अतः अभीष्ट सदिश `= (2hati – 2hati + hatk)` .