Ex 11.2, 17 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

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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& −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 .