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If e(1) = hati + hatj + hatk and e(2) = ...

If `e_(1) = hati + hatj + hatk` and `e_(2) = hati + hatj - hatk` and `veca` and `vecb` are two vectors such that `e_(1) = 2veca + vecb` and `vece_(2) = veca + 2vecb`, then angle between `veca` and `vecb` is:

A

`cos^(-1) (7/9)`

B

` cos^(-1)(7/11)`

C

`cos^(-1) (- 7/11)`

D

` cos^(-1) ((6 sqrt2)/11)`

Text Solution

Verified by Experts

The correct Answer is:
C
we have,
` 2 veca + vecb = hati + hatj + hatk and veca + 2vecb = hati + hatj-hatk`
Solving these two equations, we get
` veca = 1/3 ( hati +hatj + 3hatk) and vecb = 1/3 ( hati +hatj -3hatk)`
`1 cos theta = ( veca.vecb)/(|veca||vecb|) Rightarro cos theta = - 7/11 Rightarrow theta = cos^(-1) ( - 7/11)`
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