यदि ` veca = hati + hatj + hatk " तथा" vecb = hatj - hatk` , तब एक सदिश ` vecc `
इस प्रकार ज्ञात कीजिए ` veca xx vecc = vecb " तथा " veca . vecc = 3`

Explanation

दिया है , ` veca = hati + hatj + hatk " तथा " vecb = hatj - hatk `
माना ` vecc = c_(1) hati + c_(2)hatj + c_(3) hatk ` तब ,
` veca . vecc = 3 "व् " veca xx vecc = vecb `
` rArr (hati + hatj + hatk) . (c_(1) hati + c_(2) hatj + c_(3) hatk) = 3 `
तथा ` (hati + hatj hatk ) xx (c_(1) hati + c_(2) hatj + c_(3) hatk) = (hatj - hatk)`
` rArr c_(1) c_(2) + c_(3) = 3 ` ...(1)
तथा `|{:(hati , hatj , hatk ),(1,1,1),(c_(1), c_(2), c_(3) ):}|= (hatj - hatk)` ...(2)
समी (2) से
` (c_(3) - c_(2)) hati - (c_(3) - c_(1)) + (c_(2) - c_(1)) hatk = (hatj - hatk)`
`rArr c_(3) - c_(2) = 0 , c_(1) - c_(3) = 1 " तथा " c_(2) - c_(1) = - 1`
`rArr c_(3) = c_(2) " तथा " c_(1) - c_(2) = 1` ...(3)
ये मान समी (1) में रखने पर
` c_(1) + 2c_(2) = 3` ...(4)
समी (3) तथा (4) से
` c_(2) = (2)/(3) " तथा " c_(1) = (5)/(3)`
ये मान समी (1) में रखने पर
` c_(3) = (2)/(3)`
अतः ` c = (5)/(3) hati + (2)/(3) hatj + (2)/(3) hatk = (1)/(3) (5hati + 2hatj + 2hatk)` .