

Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Feb. 1, 2020 by Teachoo
Transcript
Ex 11.2, 14 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, ((π1) β Γ (π2) β) . ((π2) β β (π1) 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 | | = |( β9)/(3β2)| = 3/β2 = 3/β2 Γ β2/β2 = (πβπ)/π Therefore, shortest distance between the given two lines is (3β2)/2.
Ex 11.2
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 Important Not in Syllabus - CBSE Exams 2021
Ex 11.2, 11 Important Not in Syllabus - CBSE Exams 2021
Ex 11.2, 12 Important
Ex 11.2, 13
Ex 11.2, 14 Important You are here
Ex 11.2, 15 Important
Ex 11.2, 16 Important
Ex 11.2, 17 Important
About the Author