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In a four-dimensional space where uni...

In a four-dimensional space where 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 no vector 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,` then a. `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`

Text Solution

Verified by Experts

The correct Answer is:
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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