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Ex 11.2, 14 Find the shortest distance between the lines whose vector equations are 𝑟 ⃗ = (𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 (𝑖 ̂ – 3𝑗 ̂ + 2𝑘 ̂) and 𝑟 ⃗ = (4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂) + 𝜇 (2𝑖 ̂ + 3𝑗 ̂ + 𝑘 ̂)Shortest distance between the lines with vector equations 𝑟 ⃗ = (𝑎_1 ) ⃗ + 𝜆 (𝑏"1" ) ⃗ and 𝑟 ⃗ = (𝑎"2" ) ⃗ + 𝜇(𝑏"2" ) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | Given, 𝒓 ⃗ = (𝒊 ̂ + 2𝒋 ̂ + 3𝒌 ̂) + 𝜆 (𝒊 ̂ − 3𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗, (𝑎1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ & (𝑏1) ⃗ = 1𝑖 ̂ − 3𝑗 ̂ + 2𝑘 ̂ 𝒓 ⃗ = (4𝒊 ̂ + 5𝒋 ̂ + 6𝒌 ̂) + 𝝁(2𝒊 ̂ + 3𝒋 ̂ + 𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗, (𝑎2) ⃗ = 4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂ & (𝑏2) ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 1𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂) − (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) = (4 − 1)𝑖 ̂ + (5 − 2) 𝑗 ̂ + (6 − 3) 𝑘 = 3𝒊 ̂ + 3𝒋 ̂ + 3𝒌 ̂ ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1& −3&2@2&3&1)| = 𝑖 ̂ [(−3×1)−(3×2)] − 𝑗 ̂ [(1×1)−(2×2)] + 𝑘 ̂ [(1×3)−(2×−3)] = 𝑖 ̂ [−3−6] − 𝑗 ̂ [1−4] + 𝑘 ̂ [3+6] = 𝑖 ̂ (−9) − 𝑗 ̂ (−3) + 𝑘 ̂(9) = − 9𝒊 ̂ + 3𝒋 ̂ + 9𝒌 ̂ Magnitude of ((𝑏1) ⃗ × (𝑏2) ⃗) = √((− 9)2+32+92) |(𝒃𝟏) ⃗" × " (𝒃𝟐) ⃗ | = √(81+9+81) = √171 = √(9×19 ) = 3√𝟏𝟗 Also, ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) . ((𝒂𝟐) ⃗ – (𝒂𝟏) ⃗) = (−9𝑖 ̂ + 3𝑗 ̂ + 9𝑘 ̂).(3𝑖 ̂ + 3𝑗 ̂ + 3𝑘 ̂) = (−9 × 3) + (3 × 3) + (9 × 3) = −27 + 9 + 27 = 9 So, shortest distance = |("(" (𝑏1) ⃗×" " (𝑏2) ⃗")" ."(" (𝑎2) ⃗ −" " (𝑎1) ⃗")" )/|(𝑏1) ⃗× (𝑏2) ⃗ | | = |9/(3√19)| = 𝟑/√𝟏𝟗 Therefore, shortest distance between the given two lines is 3/√19.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo