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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 9 Find the shortest distance between the lines l1 and l2 whose vector equations are 𝑟 ⃗ = 𝑖 ̂ + 𝑗 ̂ + 𝜆(2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂ ) and 𝑟 ⃗ = 2𝑖 ̂ + 𝑗 ̂ – 𝑘 ̂ + 𝜇 (3𝑖 ̂ – 5𝑗 ̂ + 2𝑘 ̂ )Shortest distance between lines 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ and 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | 𝒓 ⃗ = (𝒊 ̂ + 𝒋 ̂) + 𝜆 (2𝒊 ̂ − 𝒋 ̂ + 𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ (𝒂𝟏) ⃗ = 1𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂ & (𝒃𝟏) ⃗ = 2𝑖 ̂ – 1𝑗 ̂ + 1𝑘 ̂ 𝒓 ⃗ = (2𝒊 ̂ + 𝒋 ̂ − 𝒌 ̂) + 𝝁 (3𝒊 ̂ − 5𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ (𝒂𝟐) ⃗ = 2𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂ & (𝒃𝟐) ⃗ = 3𝑖 ̂ − 5𝑗 ̂ + 2𝑘 ̂ Now (𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ = (2𝑖 ̂ + 1𝑗 ̂ − 1𝑘 ̂) − (1𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂) = (2 − 1) 𝑖 ̂ + (1 − 1)𝑗 ̂ + (−1 − 0) 𝑘 ̂ = 1𝒊 ̂ + 0𝒋 ̂ − 1𝒌 ̂ (𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2& −1&1@3& −5&2)| = 𝑖 ̂ [(−1×2)−(−5×1)] − 𝑗 ̂ [(2×2)−(3×1)] + 𝑘 ̂[(2×−5)−(3×−1)] = 𝑖 ̂ [−2+5] − 𝑗 ̂ [4−3] + 𝑘 ̂ [−10+3] = 𝑖 ̂ (3) − 𝑗 ̂ (1) + 𝑘 ̂(−7) = 3𝒊 ̂ − 𝒋 ̂ − 7𝒌 ̂ Magnitude of ((𝑏1) ⃗ × (𝑏2) ⃗) = √(32+(−1)2+(−7)^2 ) |(𝒃𝟏) ⃗× (𝒃𝟐) ⃗ | = √(9+1+49) = √𝟓𝟗 Also, ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) .((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (3𝑖 ̂ − 𝑗 ̂ − 7𝑘 ̂) . (1𝑖 ̂ + 0𝑗 ̂ − 1𝑘 ̂) = (3 × 1) + (−1 × 0) + (−7 × −1) = 3 + 0 + 7 = 10 Therefore, Shortest distance = |(((𝑏1) ⃗ × (𝑏2) ⃗ ).((𝑎2) ⃗ − (𝑎1) ⃗ ))/|(𝑏1) ⃗ × (𝑏2) ⃗ | | = |10/√59| = 𝟏𝟎/√𝟓𝟗

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.