If the acute angle that the vector `alphahati+betahatj+gammahatk` makes with the plane of the two vectors `2hati+3hatj-hatk` and `hati-hatj+2hatk` is `tan^(-1)1/(sqrt(2))` then

Options

A. `alpha(beta+gamma)=beta gamma`
B. `beta(gamma+alpha)=gamma alpha`
C. `gamma(alpha+beta)=alpha beta`
D. `alpha beta =beta gamma +gamma alpha =0`

Answer

A

Explanation

Let `theta` be the angle between `vecr=alpha hati+beta hatj+gammahatk` and the plane containig the vectors `veca=2hati+hatj-hatk` and `vecb=hati-hatj+2hatk`. Then `(pi//2-theta)` is the angle between `vecr` and `vecaxxvecb`.
`:.cos((pi)/2-theta)=(vecr.(vecaxxvecb))/(|vecr||vecaxxvecb|)`
`impliessin theta=([(vecr, veca, vecb)])/(|vecr||vecaxxvecb|)`
`implies1/(sqrt(3))=([(vecr, veca, vecb)])/(|vecr|(5sqrt(3))) [ :. tan theta=1/(sqrt(2))impliessin theta=1/(sqrt(3), and vecaxxvecb=5hati-5hatj-5hatk]`
`implies5|vecr|=[(vecr, veca, vecb)]`
`implies5|vecr|=5(alpha-beta-gamma)`
`implies|vecr|=alpha-beta-gamma`
`impliessqrt(alpha^(2)+beta^(2)+gamma^(2))=alpha-beta-gamma`
`impliesbeta gamma=alpha gamma +alpha betaimpliesbeta gamma=alpha (beta+gamma)`