Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important
Ex 11.2, 5 Important
Ex 11.2, 9 (i) Important
Ex 11.2, 10 Important
Ex 11.2, 12 Important
Ex 11.2, 13 Important
Ex 11.2, 15 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 4 (a) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 14 (a) Important Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Question 19 Important Deleted for CBSE Board 2024 Exams
Question 20 Important Deleted for CBSE Board 2024 Exams
Misc 3 Important
Misc 4 Important You are here
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Misc 5 Important
Question 16 Important Deleted for CBSE Board 2024 Exams
Chapter 11 Class 12 Three Dimensional Geometry
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.