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

Ex 11.2, 9 Important

Ex 11.2, 10 (i) Important

Ex 11.2, 10 (ii)

Ex 11.2, 11 (i) Important

Ex 11.2, 11 (ii)

Ex 11.2, 12 Important

Ex 11.2, 13

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 16 You are here

Ex 11.2, 17 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 23, 2021 by Teachoo

Ex 11.2, 16 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&[email protected]&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.