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Let veca= 2hati+hatj -2hatk and vecb=ha...

Let `veca= 2hati+hatj -2hatk and vecb=hati+hatj`. If `vec c` is a vector such that `veca*vec c=|vec c|, |vec c-veca|=2sqrt(2)` and the angle between `vec a xx vec b and vec c` is `30^(@)` then `|(veca xx vec b) xx vec c|` is equal to :

A

`(1)/(2)`

B

`(3)/(2)`

C

3

D

`(3sqrt(3))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`|(vec(a) xx vec(b) )xxvec(c) | = |vec(a) xx vec(b)| |vec(c)| sin 30^(@)`
Now , `|vec(a) xx vec(b)| = sqrt(|a|^(2)|vecb|^(2) - (vec(a). vec(b))^(2)) = sqrt((9)(2) - (3)^(2)) = 3`
Also , `|vec(c) - vec(a)| = 2sqrt(2) rArr |vec(c)|^(2) + |a|^(2) - 2|vec(c)| = 8`
`rArr |vec(c)|^(2) - 2|vec(c)|^(2) + 1 = 0 rArr |vec(c) | = 1 (As vec(a) . vec(c) = |vec(c)|.)`
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