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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))` thenA. `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` |
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Answer» Correct Answer - A 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)` |
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