Miscellaneous
Misc 2 Important
Misc 3 Important
Misc 4 Important You are here
Misc 5 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Question 15 Deleted for CBSE Board 2024 Exams
Question 16 Important Deleted for CBSE Board 2024 Exams
Question 17 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 18 (MCQ) Important Deleted for CBSE Board 2024 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
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.