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Options

A. 0
B. 1
C. `1/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2) +b_(2)^(2)+b_(2)^(2))`
D. `3/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2) +b_(2)^(2)+b_(2)^(2)) (c_(1)^(2) + c_(2)^(2)+c_(2)^(2))`

Answer

c

Explanation

We are given that `veca = a_(1)hati+a_(2)hatj +a_(3)hatk`
`vecb = b_(1)hati +b_(2)hatj +b_(3)hatk`
`vecc =c_(1)hati +c_(2)hatj +c_(3)hatk`
`"then"|{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|^(2)=[veca vecbvecc]^(2)`
` (veca xx vecb.vecc)^(2)`
`(|veca xx vecb|.1cos)^(@2)`
(since `vecc` is `bot "to" veca and vecb, vecc "is " bot "to" vecaxx vecb)`
`(|veca xx vecb|)^(2)`
`(|veca||vecb|.sin""pi/6)^(2)`
`(1/2sqrt(a_(1)^(2)+a_(2)^(2)+a_(3)^(2))sqrt(b_(1)^(2)+b_(2)^(2)+b_(3)^(2)))^(2)`
`1/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2)+b_(2)^(2)+b_(3)^(2))`