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`

Options

A. `alpha = beta `
B. `gamma^(2) = 1- 2alpha^(2)`
C. `gamma^(2) =-cos 2 theta`
D. `beta^(2) = (1+ cos 2theta)/2`

Answer

a,b,c,d

Explanation

Since, `veca, vecb, and vecc` are unit vectors inclined at an angle `theta` we have
`|veca|=|vecb|=1 and cos theta = veca.vecc = vecb.vecc`
Now, ` vecc = alpha veca + beta vecb + gamma (veca xx vecb)`
` Rightarrow veca.vecc=alpha(veca. veca)+beta(veca.vecb) +gamma{veca.(vecaxx vecb)}`
`Rightarrow cos theta=alpha|veca|^(2) " " (therefore veca. vecb=0, veca. (vecaxxvecb) =0) = alpha`
similarly, by taking dot product on both sides of (i)
by `vecb " we get " beta =cos tehta`
Again ` vecc= alpha veca + beta vecb + gamma (veca xx vecb) `
` Rightarrow |vecc|^(2) = |alphaveca + beta vecb + gamma (veca xx vecb)`
`=alpha^(2) |veca|^(2) +beta^(2) |vecb|^(2) +gamma^(2)|vecaxxvecb|^(2)`
` + 2alphabeta(veca .vecb) + 2alphagamma{veca. (vecaxxvecb)}`
`+ 2 betagamma(vecb.{vecaxxvecb})`
` Rightarrow 1 = alpha^(2) +beta^(2) +gamma^(2) |vecaxxvecb|^(2)`
` = 2alpha^(2) + gamma^(2) {|veca|^(2) |vecb|^(2) sin^(2) pi//2}`
`= 2alpha^(2) +gamma^(2) or alpha^(2)(1-gamma^(2))/2`
But `alpha = beta = cos theta`
`1=2alpha^(2) +gamma^(2) Rightarrow1-2cos^(2) theta-cos 2theta`
` beta^(2) (1-gamma^(2))/2 = (1+cos2theta)/2`