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


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Ex 11.2, 15 Find the shortest distance between the lines whose vector equations are 𝑟 ⃗ = (1 − t) 𝑖 ̂ + (t − 2) 𝑗 ̂ + (3 − 2t) 𝑘 ̂ and 𝑟 ⃗ = (s + 1) 𝑖 ̂ + (2s – 1) 𝑗 ̂ – (2s + 1) 𝑘 ̂ Shortest distance between lines with vector equations 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ and 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ is |("(" (𝒃𝟏) ⃗× (𝒃𝟐) ⃗")" ."(" (𝒂𝟐) ⃗ − (𝒂𝟏) ⃗")" )/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | 𝒓 ⃗ = (𝟏 − t) 𝒊 ̂ + (𝒕−𝟐)𝒋 ̂ + (3 − 2t) 𝒌 ̂ = 1𝑖 ̂ − t𝑖 ̂ + t𝑗 ̂ − 2𝑗 ̂ + 3𝑘 ̂ − 2t𝑘 ̂ = (1𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂) + t(−1𝑖 ̂ + 1𝑗 ̂ − 2𝑘 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + t (𝑏1) ⃗, (𝑎1) ⃗ = 1𝑖 ̂ – 2𝑗 ̂ + 3𝑘 ̂ & (𝑏1) ⃗ = − 1𝑖 ̂ + 1𝑗 ̂ − 2𝑘 ̂ 𝒓 ⃗ = (𝒔 + 1) 𝒊 ̂ + (𝟐𝒔" − " 𝟏)𝒋 ̂ − (2s + 1) 𝒌 ̂ = s𝑖 ̂ + 1𝑖 ̂ + 2s𝑗 ̂ − 1𝑗 ̂ − 2s𝑘 ̂ − 1𝑘 ̂ = (1𝑖 ̂ − 1𝑗 ̂ − 1𝑘 ̂) + s(1𝑖 ̂ + 2𝑗 ̂ − 2𝑘 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + s (𝑏2) ⃗, (𝑎2) ⃗ = 1𝑖 ̂ − 1𝑗 ̂ − 1𝑘 ̂ & (𝑏2) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ − 2𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂_𝟏 ) ⃗) = (1𝑖 ̂ − 1𝑗 ̂ − 1𝑘 ̂) − (1𝑖 ̂ − 2𝑗 + 3𝑘 ̂) = (1 − 1) 𝑖 ̂ + ( − 1 + 2)𝑗 ̂ + ( − 1 − 3)𝑘 ̂ = 0𝒊 ̂ + 1𝒋 ̂ − 4𝒌 ̂ ( (𝒃_𝟏 ) ⃗× (𝒃_𝟐 ) ⃗ ) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@ −1&1& −2@1&2& −2)| = 𝑖 ̂[(1×− 2)−(2×− 2)] − 𝑗 ̂[(−1×−2)−(1×− 2)] + 𝑘 ̂[(− 1×2)−(1×1)] = 𝑖 ̂[−2+4] − 𝑗 ̂[2+2] A + 𝑘 ̂[−2−1] = 2𝒊 ̂ − 4𝒋 ̂ − 3𝒌 ̂ Magnitude of ((𝑏1) ⃗×(𝑏2) ⃗) = √(22+(− 4)2+(− 3)2) |(𝒃𝟏) ⃗×(𝒃𝟐) ⃗ | = √(4+16+9) = √𝟐𝟗 Also, ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) . ((𝒂𝟐) ⃗ – (𝒂𝟏) ⃗) = (2𝑖 ̂ − 4𝑗 ̂ − 3𝑘 ̂) . (0𝑖 ̂ + 1𝑗 ̂ − 4𝑘 ̂) = (2 × 0) + (−4 × 1) + (−3 × −4) = −0 + (−4) + 12 = 8 So, shortest distance = |(((𝑏_1 ) ⃗ × (𝑏_2 ) ⃗ ) . ((𝑎_2 ) ⃗ × (𝑎_1 ) ⃗ ).)/((𝑏_1 ) ⃗ × (𝑏_2 ) ⃗ )| = |8/√29| = 𝟖/√𝟐𝟗 Therefore, the shortest distance between the given two lines is 8/√29 .

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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.