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Misc 4 Find the shortest distance between lines 𝑟 ⃗ = 6𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ + 𝜆 (𝑖 ̂ – 2𝑗 ̂ + 2𝑘 ̂) and 𝑟 ⃗ = –4𝑖 ̂ – 𝑘 ̂ + 𝜇 (3𝑖 ̂ – 2𝑗 ̂ – 2𝑘 ̂) .Shortest distance between lines with vector equations 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ and 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | 𝒓 ⃗ = (6𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) + 𝜆 (𝒊 ̂ − 2𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆(𝑏1) ⃗ , (𝑎1) ⃗ = 6𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ & (𝑏1) ⃗ = 1𝑖 ̂ − 2𝑗 ̂ + 2𝑘 ̂ 𝒓 ⃗ = (−4𝒊 ̂ − 𝒌 ̂) + 𝝁 (3𝒊 ̂ − 2𝒋 ̂ − 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ , (𝑎2) ⃗ = − 4𝑖 ̂ + 0𝑗 ̂ − 1𝑘 ̂ & (𝑏2) ⃗ = 3𝑖 ̂ − 2𝑗 ̂ − 2𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (−4𝑖 ̂ + 0𝑗 ̂ − 1𝑘 ̂) − (6𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) = (−4 − 6) 𝑖 ̂ + (0 − 2)𝑗 ̂ + (−1 − 2) 𝑘 ̂ = − 10𝒊 ̂ − 2𝒋 ̂ − 3𝒌 ̂ ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1& −2&2@3&−2&−2)| = 𝑖 ̂ [(−2×−2)−(−2×2)] − 𝑗 ̂ [(1×−2)−(3×2)] + 𝑘 ̂ [(1×−2)−(3×−2)] = 𝑖 ̂ [ 4+4] − 𝑗 ̂ [−2−6] + 𝑘 ̂ [−2+6] = 𝑖 ̂ (8) − 𝑗 ̂ (−8) + 𝑘 ̂(4) = 8𝒊 ̂ + 8𝒋 ̂ + 4𝒌 ̂ Magnitude of (𝑏1) ⃗ × (𝑏2) ⃗ = √(8^2+8^2+4^2 ) |(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | = √(64+64+16) = √144 = 𝟏𝟐 Also, ((𝒃𝟏) ⃗×(𝒃𝟐) ⃗ ) . ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ) = (8𝑖 ̂ + 8𝑗 ̂ + 4𝑘 ̂).(− 10𝑖 ̂ − 2𝑗 ̂ − 3𝑘 ̂) = (8 × − 10) + (8 × − 2) + (4 × − 3) = − 80 + (−16) + (-12) = − 108 Shortest distance = |(((𝑏1) ⃗ × (𝑏2) ⃗ ) . ((𝑎2) ⃗ − (𝑎1) ⃗ ))/|(𝑏1) ⃗ × (𝑏2) ⃗ | | = |( −𝟏𝟎𝟖)/𝟏𝟐| = |−9| = 9 Therefore, the shortest distance between the given two lines is 9.

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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