1
Question
If the vectors α^i+^j+^k,^i+β^j+^k,^i+^j+λ^k(α,β,γ≠1) are coplanar, then the value of 11−α+11+β+11−γ is.
If the vectors α^i+^j+^k,^i+β^j+^k,^i+^j+λ^k(α,β,γ≠1) are coplanar, then the value of 11−α+11+β+11−γ is.
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Solution
The correct option is C 1
Vectors α^i+^j+^k,^i+β^j+^k,^i+^j+γ^k are co-planars
Thus ∣∣
∣∣α111β111γ∣∣
∣∣=0
α[βγ−1]−1[γ−1]+1[1−β]=0
αβγ−α−γ+1+1−β=0
αβγ+2=α+β+γ
αβγ=α+β+γ−2⟶1
11−α+11−β+11−γ=(1−β)(1−γ)+(1−α)(1−γ)+(1−α)(1−β)(1−α)(1−β)(1−γ)
=1−β−γ+βγ+1−α−γ+αγ+1−α−β+αβ[1−α−β+αβ][1−γ]
=3−2α−2β−2γ+αβ+βγ+αγ1−α−β+αβ−γ+αγ+βγ−αβγ
=3−2α−2β−2γ+αβ+βγ+αγ1−α−β−γ+αβ+βγ+αγ−(α+β+γ−2) [from equation 1]
=3−2α−2β−2γ+αβ+βγ+αγ3−2α−2β−2γ+αβ+βγ+αγ
=1
Vectors α^i+^j+^k,^i+β^j+^k,^i+^j+γ^k are co-planars
Thus ∣∣ ∣∣α111β111γ∣∣ ∣∣=0
α[βγ−1]−1[γ−1]+1[1−β]=0
αβγ−α−γ+1+1−β=0
αβγ+2=α+β+γ
αβγ=α+β+γ−2⟶1
11−α+11−β+11−γ=(1−β)(1−γ)+(1−α)(1−γ)+(1−α)(1−β)(1−α)(1−β)(1−γ)
=1−β−γ+βγ+1−α−γ+αγ+1−α−β+αβ[1−α−β+αβ][1−γ]
=3−2α−2β−2γ+αβ+βγ+αγ1−α−β+αβ−γ+αγ+βγ−αβγ
=3−2α−2β−2γ+αβ+βγ+αγ1−α−β−γ+αβ+βγ+αγ−(α+β+γ−2) [from equation 1]
=3−2α−2β−2γ+αβ+βγ+αγ3−2α−2β−2γ+αβ+βγ+αγ
=1
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