Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Question 37 (Choice 1) Find the shortest distance between the lines 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) And 𝑟 ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 𝜇(3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) If the lines intersect find their point of intersection Shortest distance between lines with vector equations 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ and 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | 𝒓 ⃗ = (3𝒊 ̂ + 2𝒋 ̂ − 4𝒌 ̂) + 𝜆 (𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆(𝑏1) ⃗ , (𝑎1) ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ & (𝑏1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ 𝒓 ⃗ = (5𝒊 ̂ − 2𝒋 ̂) + 𝝁 (3𝒊 ̂ + 2𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ , (𝑎2) ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 0𝑘 ̂ & (𝑏2) ⃗ = 3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (5𝑖 ̂ − 2𝑗 ̂ + 0𝑘 ̂) − (3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) = (5 − 3) 𝑖 ̂ + (−2 − 2)𝑗 ̂ + (0 − (−4)) 𝑘 ̂ = 2𝒊 ̂ − 4𝒋 ̂ + 4𝒌 ̂ ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1& 2&2@3&2&6)| = 𝑖 ̂ [(2 × 6)−(2 × 2)] − 𝑗 ̂ [(1 × 6)−(3 × 2)] + 𝑘 ̂ [(1 × 2)−(3 × 2)] = 𝑖 ̂ [ 12−4] − 𝑗 ̂ [6−6] + 𝑘 ̂ [2−6] = 𝑖 ̂ (8) − 𝑗 ̂ (0) + 𝑘 ̂(−4) = 8𝒊 ̂ − 4𝒌 ̂ Magnitude of (𝑏1) ⃗ × (𝑏2) ⃗ = √(8^2+0^2+〖(−4)〗^2 ) |(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | = √(64+16) = √80 = 4√5 Also, ((𝒃𝟏) ⃗×(𝒃𝟐) ⃗ ) . ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ) = (8𝑖 ̂ − 4𝑘 ̂). (2𝑖 ̂ − 4𝑗 ̂ + 4𝑘 ̂) = (8 × 2) + (0 × −4) + (−4 × 4) = −16 + 0 + (−16) = 0 Shortest distance = |(((𝑏1) ⃗ × (𝑏2) ⃗ ) . ((𝑎2) ⃗ − (𝑎1) ⃗ ))/|(𝑏1) ⃗ × (𝑏2) ⃗ | | = |( 0)/(4√5)| = 0 Therefore, the shortest distance between the given two lines is 0 Finding Point of Intersection Since our lines are 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) And 𝑟 ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 𝜇(3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) Thus, 3𝒊 ̂ + 2𝒋 ̂ − 4𝒌 ̂ + 𝝀 (𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) = 5𝒊 ̂ − 2𝒋 ̂ + 𝝁(3𝒊 ̂ + 2𝒋 ̂ + 6𝒌 ̂) (3 + 𝜆)𝑖 ̂ + (2 + 2𝜆) 𝑗 ̂ + (−4 + 2𝜆) 𝑘 ̂ = (5 + 3𝜇)𝑖 ̂ + (−2 + 2𝜇)𝑗 ̂ + 6𝜇𝑘 ̂ Thus, 3 + 𝜆 = 5 + 3𝜇 2 + 2𝜆 = −2 + 2𝜇 −4 + 2𝜆 = 6𝜇 Solving (1) and (2) 𝜆 − 3𝜇 = 5 − 3 𝝀 − 3𝝁 = 2 And, 2 + 2𝜆 = −2 + 2𝜇 2𝜆 − 2𝜇 = −4 𝝀 − 𝝁 = −2 Solving both equations 𝝀=−𝟒, 𝝁=−𝟐 Putting 𝝀=−𝟒 in 𝒓 ⃗ 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ − 4(𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ − 4𝑖 ̂ − 8𝑗 ̂ − 8𝑘 ̂ 𝑟 ⃗ = −𝑖 ̂ − 6𝑗 ̂ − 12𝑘 ̂ So, Point of intersection is (−1, −6, −12)