Find the shortest distance between the lines r   = 3i  + 2j  - 4k  + λ (i  + 2j  -2k )

And r = 5i  + 2j  - u (3i  + 2j  +6k). If the lines intersect find their point of intersection

Find shortest distance between lines r = 3i + 2j - 4k + (i + 2j + 2k)
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5 Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

 

 

Note : This is similar to Ex 11.2, 14 ; Ex 11.2, 16 and Misc 9 of NCERT – Chapter 11 Class 12 Three Dimensional Geometry

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Transcript

Question 37 (Choice 1) Find the shortest distance between the lines 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) And 𝑟 ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 𝜇(3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) If the lines intersect find their point of intersection Shortest distance between lines with vector equations 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆 (𝑏1) ⃗ and 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ is |(((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ ).((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ))/|(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | | 𝒓 ⃗ = (3𝒊 ̂ + 2𝒋 ̂ − 4𝒌 ̂) + 𝜆 (𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎1) ⃗ + 𝜆(𝑏1) ⃗ , (𝑎1) ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ & (𝑏1) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ 𝒓 ⃗ = (5𝒊 ̂ − 2𝒋 ̂) + 𝝁 (3𝒊 ̂ + 2𝒋 ̂ + 6𝒌 ̂) Comparing with 𝑟 ⃗ = (𝑎2) ⃗ + 𝜇(𝑏2) ⃗ , (𝑎2) ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 0𝑘 ̂ & (𝑏2) ⃗ = 3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂ Now, ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗) = (5𝑖 ̂ − 2𝑗 ̂ + 0𝑘 ̂) − (3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂) = (5 − 3) 𝑖 ̂ + (−2 − 2)𝑗 ̂ + (0 − (−4)) 𝑘 ̂ = 2𝒊 ̂ − 4𝒋 ̂ + 4𝒌 ̂ ((𝒃𝟏) ⃗ × (𝒃𝟐) ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1& 2&2@3&2&6)| = 𝑖 ̂ [(2 × 6)−(2 × 2)] − 𝑗 ̂ [(1 × 6)−(3 × 2)] + 𝑘 ̂ [(1 × 2)−(3 × 2)] = 𝑖 ̂ [ 12−4] − 𝑗 ̂ [6−6] + 𝑘 ̂ [2−6] = 𝑖 ̂ (8) − 𝑗 ̂ (0) + 𝑘 ̂(−4) = 8𝒊 ̂ − 4𝒌 ̂ Magnitude of (𝑏1) ⃗ × (𝑏2) ⃗ = √(8^2+0^2+〖(−4)〗^2 ) |(𝒃𝟏) ⃗ × (𝒃𝟐) ⃗ | = √(64+16) = √80 = 4√5 Also, ((𝒃𝟏) ⃗×(𝒃𝟐) ⃗ ) . ((𝒂𝟐) ⃗ − (𝒂𝟏) ⃗ ) = (8𝑖 ̂ − 4𝑘 ̂). (2𝑖 ̂ − 4𝑗 ̂ + 4𝑘 ̂) = (8 × 2) + (0 × −4) + (−4 × 4) = −16 + 0 + (−16) = 0 Shortest distance = |(((𝑏1) ⃗ × (𝑏2) ⃗ ) . ((𝑎2) ⃗ − (𝑎1) ⃗ ))/|(𝑏1) ⃗ × (𝑏2) ⃗ | | = |( 0)/(4√5)| = 0 Therefore, the shortest distance between the given two lines is 0 Finding Point of Intersection Since our lines are 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) And 𝑟 ⃗ = 5𝑖 ̂ − 2𝑗 ̂ + 𝜇(3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) Thus, 3𝒊 ̂ + 2𝒋 ̂ − 4𝒌 ̂ + 𝝀 (𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) = 5𝒊 ̂ − 2𝒋 ̂ + 𝝁(3𝒊 ̂ + 2𝒋 ̂ + 6𝒌 ̂) (3 + 𝜆)𝑖 ̂ + (2 + 2𝜆) 𝑗 ̂ + (−4 + 2𝜆) 𝑘 ̂ = (5 + 3𝜇)𝑖 ̂ + (−2 + 2𝜇)𝑗 ̂ + 6𝜇𝑘 ̂ Thus, 3 + 𝜆 = 5 + 3𝜇 2 + 2𝜆 = −2 + 2𝜇 −4 + 2𝜆 = 6𝜇 Solving (1) and (2) 𝜆 − 3𝜇 = 5 − 3 𝝀 − 3𝝁 = 2 And, 2 + 2𝜆 = −2 + 2𝜇 2𝜆 − 2𝜇 = −4 𝝀 − 𝝁 = −2 Solving both equations 𝝀=−𝟒, 𝝁=−𝟐 Putting 𝝀=−𝟒 in 𝒓 ⃗ 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ + 𝜆 (𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ − 4(𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) 𝑟 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ − 4𝑘 ̂ − 4𝑖 ̂ − 8𝑗 ̂ − 8𝑘 ̂ 𝑟 ⃗ = −𝑖 ̂ − 6𝑗 ̂ − 12𝑘 ̂ So, Point of intersection is (−1, −6, −12)

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.