InterviewSolution
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InterviewSolution
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In a four-dimensional spacewhere unit vectors along the axes are ` hat i , hat j , hat ka n d hat l ,a n d vec a_1, vec a_2, vec a_3, vec a_4`are four non-zero vectors such that novector can be expressed as a linear combination of others and `(lambda-1)( vec a_1- vec a_2)+mu( vec a_2+ vec a_3)+gamma( vec a_3+ vec a_4-2 vec a_2)+ vec a_3+delta vec a_4=0,`thena. `lambda=1`b. `mu=-2//3`c. `gamma=2//3`d. `delta=1//3`A. `lamda =1`B. `mu = -2//3`C. `gamma = 2//3`D. `delta = 1//3` |
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Answer» Correct Answer - A::B::D `(lamda -1) (veca_1 - veca_2) + mu(veca_2 + veca_3)+ gamma (veca_3 + veca_4- 2veca_2) + veca_3 + deltaveca_4 = vec0` i.e., `(lamda -1) veca_1 + (1-lamda +mu- 2gamma)veca_2 + (mu + gamma +1)veca_3 + ( gamma + delta) veca_4=vec0` Since `veca_1, veca_2, veca_3 and veca_4` are linearly inependent, we have `lamda-1 =0, 1-lamda + mu - 2gamma =0, mu+gamma +1=0 and gamma +delta =0` i.e., `lamda =1, mu=2gamma, mu +gamma+1 =0, gamma +delta =0` Hence, `lamda =1, mu = - (2)/(3), gamma = - (1)/(3), delta = (1)/(3)` |
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