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Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important
Ex 11.2, 5 Important
Ex 11.2, 9 (i) Important
Ex 11.2, 10 Important
Ex 11.2, 12 Important You are here
Ex 11.2, 13 Important
Ex 11.2, 15 Important
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Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
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Question 15 Important Deleted for CBSE Board 2024 Exams
Question 4 (a) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 14 (a) Important Deleted for CBSE Board 2024 Exams
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Question 20 Important Deleted for CBSE Board 2024 Exams
Misc 3 Important
Misc 4 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Misc 5 Important
Question 16 Important Deleted for CBSE Board 2024 Exams
Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Aug. 14, 2023 by Teachoo
Ex 11.2, 12 Find the shortest distance between the lines π β = (π Μ + 2π Μ + π Μ) + π (π Μ β π Μ + π Μ) and π β = (2π Μ β π Μ β π Μ) + π (2π Μ + π Μ + 2π Μ) Shortest distance between the lines with vector equations π β = (π1) β + π (π1) βand π β = (π2) β + π(π2) β is |(((ππ) β Γ (ππ) β ).((ππ) β β (ππ) β ))/|(ππ) β Γ (ππ) β | | Given, π β = (π Μ + 2π Μ + π Μ) + π(π Μ β π Μ + π Μ) Comparing with π β = (π1) β + π (π1) β, (π1) β = 1π Μ + 2π Μ + 1π Μ & (π1) β = 1π Μ β 1π Μ + 1π Μ π β = (2π Μ β π Μ β π Μ) + π (2π Μ + π Μ + 2π Μ) Comparing with π β = (π2) β + π(π2) β , (π2) β = 2π Μ β 1π Μ β 1π Μ & (π2) β = 2π Μ + 1π Μ + 2π Μ Now, (ππ) β β (ππ) β = (2π Μ β 1π Μ β 1π Μ) β (1π Μ + 2π Μ + 1π Μ) = (2 β 1) π Μ + (β1β 2)π Μ + (β1 β 1) π Μ = 1π Μ β 3π Μ β 2π Μ (ππ) β Γ (ππ) β = |β 8(π Μ&π Μ&π Μ@1& β1&1@2&1&2)| = π Μ [(β1Γ 2)β(1Γ1)] β π Μ [(1Γ2)β(2Γ1)] + π Μ [(1Γ1)β(2Γβ1)] = π Μ [β2β1] β π Μ [2β2] + π Μ [1+2] = β3π Μ β 0π Μ + 3π Μ Magnitude of ((π1) β Γ (π2) β) = β((β3)2+(0)2+32) |(ππ) β Γ (ππ) β | = β(9+0+9) = β18 = β(9 Γ 2) = 3βπ Also, ((ππ) β Γ (ππ) β) . ((ππ) β β (ππ) β) = (β 3π Μβ0π Μ+3π Μ).(1π Μ β 3π Μ β 2π Μ) = (β3Γ1)".(" 0Γβ"3)" + (3 Γ β2) = β3 β 0 β 6 = β9 So, Shortest distance = |(((π_1 ) β Γ (π_2 ) β ).((π_2 ) β β (π_1 ) β ))/|(π_1 ) β Γ (π_2 ) β | | = |( βπ)/(πβπ)| = 3/β2 = 3/β2 Γ β2/β2 = (πβπ)/π Therefore, shortest distance between the given two lines is (3β2)/2.