   1. Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
2. Serial order wise
3. Ex 11.2

Transcript

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&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. 