Check sibling questions

Find the shortest distance between the following lines

r ⃗ = (i ̂+j ̂-k ̂ )+s(2i ̂+j ̂+k ̂ )

r âƒ— = (i Ì‚+j Ì‚+k Ì‚ )+t(4i Ì‚+(2j) Ì‚+2k Ì‚ )

This question is similar to Example 12 - Chapter 11 Class 12 - Three Dimensional Geometry

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 10 (Choice 1) Find the shortest distance between the following lines: 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂−𝑘 ̂ )+𝑠(2𝑖 ̂+𝑗 ̂+𝑘 ̂ ) 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂+2𝑘 ̂ )+𝑡(4𝑖 ̂+2𝑗 ̂+2𝑘 ̂ ) Given lines 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂−𝑘 ̂ )+𝑠(2𝑖 ̂+𝑗 ̂+𝑘 ̂ ) 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂+2𝑘 ̂ )+𝑡(4𝑖 ̂+2𝑗 ̂+2𝑘 ̂ ) We can write them as 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂−𝑘 ̂ )+𝑠(𝟐𝒊 ̂+𝒋 ̂+𝒌 ̂ ) 𝑟 ⃗ = (𝑖 ̂+𝑗 ̂+2𝑘 ̂ )+2𝑡(𝟐𝒊 ̂+𝒋 ̂+𝒌 ̂ ) Since parallel vector is same, the lines are parallel Distance between two parallel lines with vector equations 𝑟 ⃗ = (𝑎_1 ) ⃗ + 𝜆𝒃 ⃗ and 𝑟 ⃗ = (𝑎_2 ) ⃗ + 𝜇𝒃 ⃗ is |(𝒃 ⃗ × ((𝒂_𝟐 ) ⃗ − (𝒂_𝟏 ) ⃗))/|𝒃 ⃗ | | Finding (𝒂_𝟏 ) ⃗ , (𝒂_𝟐 ) ⃗ and 𝒃 ⃗ 𝑟 ⃗ = (𝑖 ̂ + 𝑗 ̂ − 𝑘 ̂) + s (2𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 𝑏 ⃗, (𝑎1) ⃗ = 𝑖 ̂ + 𝑗 ̂ – 𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂ 𝑟 ⃗ = (𝑖 ̂ + 𝑗 ̂ + 2𝑘 ̂) + 𝜇 (2𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇𝑏 ⃗, (𝑎2) ⃗ = 𝑖 ̂ + 𝑗 ̂ + 2𝑘 ̂ & 𝑏 ⃗ = 2𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (𝑖 ̂ + 𝑗 ̂ + 2𝑘 ̂) − (𝑖 ̂ + 𝑗 ̂ − 𝑘 ̂) = 𝑖 ̂ − 𝑖 ̂ + 𝑗 ̂ − 𝑗 ̂ + 2𝑘 ̂ + 𝑘 ̂ = 3𝒌 ̂ Magnitude of 𝑏 ⃗ = √(22 +12 +12) |𝒃 ⃗ | = √(4+1+1) = √𝟔 Also, 𝒃 ⃗ × ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&1&1@0&0&3)| = 𝑖 ̂ [(3×1)−(0×1)] − 𝑗 ̂ [(2×3)−(0×1)] + 𝑘 ̂ [(2×0)−(0×1)] = 𝑖 ̂ [3−0] − 𝑗 ̂ [6−0] + 𝑘 ̂ [0−0] = 𝟑𝒊 ̂ − 6𝒋 ̂Now, |𝒃 ⃗" × (" (𝒂𝟐) ⃗" − " (𝒂𝟏) ⃗")" | = √(3^2+6^2 ) = √(9+36) = √45 = √(9 × 5) = 𝟑√𝟓 So, Distance = |(𝑏 ⃗ × ((𝑎_2 ) ⃗ − (𝑎_1 ) ⃗))/|𝑏 ⃗ | | = |(3√5)/√6| = (𝟑√𝟓)/√𝟔 Therefore, the distance between the given two parallel lines is (3√5)/√6.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.