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Let veca=2hati+hatj-2hatk, vecb=hati+hat...

Let `veca=2hati+hatj-2hatk`, `vecb=hati+hatj`. If `vecc` is a vector such that `veca.vecc=|vecc|` and angle between vectors `vecaxxvecb and vecc` is `30^@` , then `|(vecaxxvecb)xxvecc|`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))xx vec(c )|=|vec(a)xx vec(b)||vec(c )| sin 30^(@)`
Now, `|vec(a)xx vec(b)|=sqrt(|a|^(2)|vec(b)|^(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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