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Let ` vec adot vec b=0,w h e r e vec aa n d vec b` are unit vectors and the unit vector ` vec c` is inclined at an angle `theta` to both ` vec aa n d vec bdot` If ` vec c=m vec a+n vec b+p( vec axx vec b),(m ,n , p in R),` then `pi/4lt=thetalt=pi/4` b. `pi/4lt=thetalt=(3pi)/4` c. `0lt=thetalt=pi/4` d. `0lt=thetalt=(3pi)/4`

A

`pi/4lethetalepi/4`

B

`pi/4lethetale(3pi)/4`

C

`0 le thetale pi/4`

D

`0 le thetale(3pi)/4`

Text Solution

Verified by Experts

The correct Answer is:
a

`vecc = mveca + nvecb + p(veca xx vecb)`
taking dot product with `veca and vecb` we have
`m =n= cos theta`
`Rightarrow |vecc|= |cos theta veca + cos theta veca + cos theta vecb + p (veca xx vecb)|=1`
squaring both sides, we get
`cos^(2)theta + cos ^(2) theta + p^(2) =1 `
`or cos theta= +-sqrt(1-p^(2))/sqrt2`
Now ` - 1/sqrt2 le cos theta ge 1/sqrt2` ( for real value of `theta`)
`pi/4 le cos theta le (3pi) / 4`
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