Question

Let `veca =2hati +hatj -2hatk and vecb = hati +hatj . " Let " vecc` be vector such that ` |vecc -veca|=3, |(veca xx vecb) xx vecc|=3` and the angle between `vecc and veca xx vecb " be " 30^(@)` Then , ` veca . Vecc` is equal to
A. `25/8`
B. 2
C. 5
D. `1/8`

Answer 1

Correct Answer - B
we, have , `veca = 2hati +hatj -2hatk and vecb =hati +hatj`
` |veca|=3, |vecb|=sqrt2, veca xx vecb = 2hati -2hatj +hatk and |veca xx vecb =3`
Now,
`|(veca xx vecb) xx vecc|=3`
` Rightarrow |(veca xx vecb)||vecc| sin 30^(@) =3`
` Rightarrow 3/2 |vecc| -3 Rightarrow |vecc| =2`
Now,
`|vecc -veca| =3`
` Rightarrow |vecc|^(2) + |veca|^(2) -2 (veca.vecc)=9`
` Rightarrow 4+ 9-2 (veca.vecc) =9 Rightarrow veca. vecc =2`