Let the shortest distance between the lines L:x−5−2=y−λ0=z+λ1, λ≥0 and L1:x+1=y−1=4−z be 36 If (α,β,γ) lies on L then which of the following is Not possible?

r1=(5,λ,−λ)+t(−2,0,1)=a¯+tb¯ r2=(−1,1,−4)+s(1,1,−1)=c¯+sd¯ b¯×d¯=−i−J−2k a¯−c¯=−6i+(1−λ)J+(4+λ)k¯ S.D=|(a¯−c¯),(b¯×d¯)|b¯×d¯||=36 λ=15 ∴α=−2t+5, β=15, γ=t−15 α+2γ=−25 2α+γ=−4t+10+t−15=−3t−5 =−5t+25

Let the shortest distance between the lines L:x−5−2=y−λ0=z+λ1, λ≥0 and L1:x+1=y−1=4−z be 36 If (α,β,γ) lies on L then which of the following is Not possible?

r1=(5,λ,−λ)+t(−2,0,1)=a¯+tb¯ r2=(−1,1,−4)+s(1,1,−1)=c¯+sd¯ b¯×d¯=−i−J−2k a¯−c¯=−6i+(1−λ)J+(4+λ)k¯ S.D=|(a¯−c¯),(b¯×d¯)|b¯×d¯||=36 λ=15 ∴α=−2t+5, β=15, γ=t−15 α+2γ=−25 2α+γ=−4t+10+t−15=−3t−5 =−5t+25

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Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z b e 3 \sqrt{6} I f (α, β, γ)$ lies on L then which of the following is Not possible?

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$2 α + γ = 4$

$α - 2 γ = 5$

$2 α - γ = 5$

$α + 2 γ = 20$

answer is B.

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Detailed Solution

$r_{1} = (5, λ, - λ) + t (- 2, 0, 1) = \bar{a} + t \bar{b} r_{2} = (- 1, 1, - 4) + s (1, 1, - 1) = \bar{c} + s \bar{d} \bar{b} \times \bar{d} = - i - J - 2 k \bar{a} - \bar{c} = - 6 i + (1 - λ) J + (4 + λ) \bar{k} S . D = | \frac{(\bar{a} - \bar{c}), (\bar{b} \times \bar{d})}{| \bar{b} \times \bar{d} |} | = 3 \sqrt{6} λ = 15 ∴ α = - 2 t + 5, β = 15, γ = t - 15 α + 2 γ = - 25 2 α + γ = - 4 t + 10 + t - 15 = - 3 t - 5 = - 5 t + 25$