Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important You are here
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
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
Example 10 Find the distance between the lines 𝑙_1 and 𝑙_2 given by 𝑟 ⃗ = 𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂ + 𝜆 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂ ) and 𝑟 ⃗ = 3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ + μ (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂)Distance between two parallel lines with vector equations 𝑟 ⃗ = (𝑎_1 ) ⃗ + 𝜆𝒃 ⃗ and 𝑟 ⃗ = (𝑎_2 ) ⃗ + 𝜇𝒃 ⃗ is |(𝒃 ⃗ × ((𝒂_𝟐 ) ⃗ − (𝒂_𝟏 ) ⃗))/|𝒃 ⃗ | | 𝑟 ⃗ = (𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) + 𝜆 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 𝑏 ⃗, (𝑎1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ – 4𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂ 𝑟 ⃗ = (3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂) + 𝜇 (2𝒊 ̂ + 3𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇𝑏 ⃗, (𝑎2) ⃗ = 3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 6𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (3𝑖 ̂ + 3𝑗 ̂ − 5𝑘 ̂) − (1𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) = (3 − 1) 𝑖 ̂ + (3 − 2)𝑗 ̂ + ( − 5 + 4)𝑘 ̂ = 2𝒊 ̂ + 1𝒋 ̂ − 1𝒌 ̂ Magnitude of 𝑏 ⃗ = √(22 + 32 + 62) |𝒃 ⃗ | = √(4+9+36) = √49 = 7 Also, 𝒃 ⃗ × ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&3&6@2&1&−1)| = 𝑖 ̂ [(3×−1)−(1×6)] − 𝑗 ̂ [(2×−1)−(2×6)] + 𝑘 ̂ [(2×1)−(2×3)] = 𝑖 ̂ [−3−6] − 𝑗 ̂ [−2−12] + 𝑘 ̂ [2−6] = 𝑖 ̂ (–9) − 𝑗 ̂ (–14) + 𝑘 ̂(−4) = −𝟗𝒊 ̂ + 14𝒋 ̂ − 4𝒌 ̂ Now, |𝒃 ⃗" × (" (𝒂𝟐) ⃗" − " (𝒂𝟏) ⃗")" | = √((−9)^2+(14)^2+(−4)^2 ) = √(81+196+16) = √𝟐𝟗𝟑 So, Distance = |(𝑏 ⃗ × ((𝑎_2 ) ⃗ − (𝑎_1 ) ⃗))/|𝑏 ⃗ | | = |√293/7| = √𝟐𝟗𝟑/𝟕 Therefore, the distance between the given two parallel lines is √293/7.