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Ex 11.2

Ex 11.2, 1

Ex 11.2, 2

Ex 11.2, 3 Important

Ex 11.2, 4

Ex 11.2, 5 Important

Ex 11.2, 6

Ex 11.2, 7 Important

Ex 11.2, 8 (i) Important

Ex 11.2, 8 (ii)

Ex 11.2, 9 (i) Important

Ex 11.2, 9 (ii)

Ex 11.2, 10 Important

Ex 11.2, 11

Ex 11.2, 12 Important

Ex 11.2, 13 Important

Ex 11.2, 14 You are here

Ex 11.2, 15 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 14, 2023 by Teachoo

Ex 11.2, 14 Find the shortest distance between the lines whose vector equations are 𝑟 ⃗ = (𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) + 𝜆 (𝑖 ̂ – 3𝑗 ̂ + 2𝑘 ̂) and 𝑟 ⃗ = (4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂) + 𝜇 (2𝑖 ̂ + 3𝑗 ̂ + 𝑘 ̂)Shortest distance between the lines with vector equations 𝑟 ⃗ = (𝑎_1 ) ⃗ + 𝜆 (𝑏"1" ) ⃗ and 𝑟 ⃗ = (𝑎"2" ) ⃗ + 𝜇(𝑏"2" ) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | Given, 𝒓 ⃗ = (𝒊 ̂ + 2𝒋 ̂ + 3𝒌 ̂) + 𝜆 (𝒊 ̂ − 3𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗, (𝑎1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ & (𝑏1) ⃗ = 1𝑖 ̂ − 3𝑗 ̂ + 2𝑘 ̂ 𝒓 ⃗ = (4𝒊 ̂ + 5𝒋 ̂ + 6𝒌 ̂) + 𝝁(2𝒊 ̂ + 3𝒋 ̂ + 𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗, (𝑎2) ⃗ = 4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂ & (𝑏2) ⃗ = 2𝑖 ̂ + 3𝑗 ̂ + 1𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (4𝑖 ̂ + 5𝑗 ̂ + 6𝑘 ̂) − (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) = (4 − 1)𝑖 ̂ + (5 − 2) 𝑗 ̂ + (6 − 3) 𝑘 = 3𝒊 ̂ + 3𝒋 ̂ + 3𝒌 ̂ ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1& −3&2@2&3&1)| = 𝑖 ̂ [(−3×1)−(3×2)] − 𝑗 ̂ [(1×1)−(2×2)] + 𝑘 ̂ [(1×3)−(2×−3)] = 𝑖 ̂ [−3−6] − 𝑗 ̂ [1−4] + 𝑘 ̂ [3+6] = 𝑖 ̂ (−9) − 𝑗 ̂ (−3) + 𝑘 ̂(9) = − 9𝒊 ̂ + 3𝒋 ̂ + 9𝒌 ̂ Magnitude of ((𝑏1) ⃗ × (𝑏2) ⃗) = √((− 9)2+32+92) |(𝒃𝟏) ⃗" × " (𝒃𝟐) ⃗ | = √(81+9+81) = √171 = √(9×19 ) = 3√𝟏𝟗 Also, ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) . ((𝒂𝟐) ⃗ – (𝒂𝟏) ⃗) = (−9𝑖 ̂ + 3𝑗 ̂ + 9𝑘 ̂).(3𝑖 ̂ + 3𝑗 ̂ + 3𝑘 ̂) = (−9 × 3) + (3 × 3) + (9 × 3) = −27 + 9 + 27 = 9 So, shortest distance = |("(" (𝑏1) ⃗×" " (𝑏2) ⃗")" ."(" (𝑎2) ⃗ −" " (𝑎1) ⃗")" )/|(𝑏1) ⃗× (𝑏2) ⃗ | | = |9/(3√19)| = 𝟑/√𝟏𝟗 Therefore, shortest distance between the given two lines is 3/√19.