Let L1, and L2 denotes the lines →r=^i+λ(−^i+2^j+2^k),λ∈R and →r=μ(2^i−^j+2^k),μ∈R respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following option describe(s) L3?
A
→r=29(4^i+^j+^k)+t(2^i+2^j−^k),t∈R
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B
→r=29(2^i−^j+2^k)+t(2^i+2^j−^k),t∈R
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C
→r=13(2^i+^k)+t(2^i+2^j−^k),t∈R
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D
→r=t(2^i+2^j−^k),t∈R
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Solution
The correct options are A→r=29(4^i+^j+^k)+t(2^i+2^j−^k),t∈R B→r=29(2^i−^j+2^k)+t(2^i+2^j−^k),t∈R C→r=13(2^i+^k)+t(2^i+2^j−^k),t∈R Equation of line L1:→r=^i+λ(−^i+2^j+2^k),λ∈R Equation of line L2:→r=μ(2^i−^j+2^k),μ∈R Therefore, Point on lines L1 and L2 are A(1−λ,2λ,2λ) and B(2μ,−μ,2μ) ∴−−→AB=(2μ+λ−1)^i+(−μ−2λ)^j+(2μ−2λ)^k L3 is perpendicular to both L1 and L2 ∴L3∥(L1×L2) L3=m∣∣
∣
∣∣^i^j^k−1222−12∣∣
∣
∣∣=3m(2^i+2^j−^k)=t(2^i+2^j−^km,t∈R) Direction ratios of AB will be (2t,2t,−t) which is propotional to (2,2,−1) Hence, 2μ+λ−12=−μ−2λ2=2μ−2λ−1=c(Let) 2μ+λ−1=2c...(i) −μ−2λ=2c...(ii) 2μ−2λ=−c...(iii) On solving equations we get, λ=19,μ=29 ∴A=(89,29,29), B=(49,−29,49) Mid Point of −−→AB=(23,0,13) ∴ Equation of line L3 can be →r=89^i+29^j+29^k+t(2^i+2^j−^k)t∈R →r=29(4^i+^j+^k)+t(2^i+2^j−^k)t∈R or →r=49^i−29^j+29^k+t(2^i+2^j−^k)t∈R →r=29(2^i−^j+^k)+t(2^i+2^j−^k)t∈Ror →r=23^i+13^k+t(2^i+2^j−^k)t∈R →r=13(2^i+^k)+t(2^i+2^j−^k)t∈R