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Ex 11.2
Ex 11.2, 2
Ex 11.2, 3 Important
Ex 11.2, 4
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Ex 11.2, 6
Ex 11.2, 7 Important
Ex 11.2, 8
Ex 11.2, 9 Important
Ex 11.2, 10 (i) Important
Ex 11.2, 10 (ii)
Ex 11.2, 11 (i) Important
Ex 11.2, 11 (ii)
Ex 11.2, 12 Important
Ex 11.2, 13
Ex 11.2, 14 Important
Ex 11.2, 15 Important
Ex 11.2, 16
Ex 11.2, 17 Important You are here
Last updated at March 16, 2023 by Teachoo
Ex 11.2, 17 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& −[email protected]&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 .