Example 10 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Last updated at Aug. 23, 2021 by Teachoo
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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Serial order wise
Transcript
Example 10 Find the angle between the pair of lines (π₯ + 3)/3 = (π¦ β 1)/5 = (π§ + 3)/4 and (π₯ + 1)/1 = (π¦ β 4)/1 = (π§ β 5)/2Angle between the pair of lines (π₯ β π₯1)/π1 = (π¦ β π¦1)/π1 = (π§ β π§1)/π1 and (π₯ β π₯2)/π2 = (π¦ β π¦2)/π2 = (π§ β π§2)/π2 is given by cos ΞΈ = |(π_π π_π + π_π π_π +γ πγ_π π_π)/(β(γπ_πγ^π + γπ_πγ^π+ γπ_πγ^π ) β(γπ_πγ^π +γγ πγ_πγ^π+ γπ_πγ^π ))| (π + π)/π = (π β π)/π = (π + π)/π (π₯ β (β3))/3 = (π¦ β 1)/5 = (π§ β (β3))/4 Comparing with (π₯ β π₯1)/π1 = (π¦ β π¦1)/π1 = (π§ β π§1)/π1 x1 = β3, y1 = 1, z1 = β3 & π1 = 3, b1 = 5, c1 = 4 (π + π)/π = (π β π)/π = (π β π)/π (π₯ β (β1))/1 = (π¦ β 4)/1 = (π§ β 5)/2 Comparing with (π₯ β π₯2)/π2 = (π¦ β π¦2)/π2 = (π§ β π§2)/π2 π₯2 = β1, y2 = 4, z2 = 5 & π2 = 1, π2 = 1, π2 = 2 Now, cos ΞΈ = |(π_1 π_2 + π_1 π_2 +γ πγ_1 π_2)/(β(γπ_1γ^2 + γπ_1γ^2+ γπ_1γ^2 ) β(γπ_2γ^2 +γγ πγ_2γ^2+ γπ_2γ^2 ))| = |((3 Γ 1) + (5 Γ 1) + (4 Γ 2))/(β(3^2 + 5^2 + 4^2 ) Γ β(1^2 + 1^2 + 2^2 ))| = |(3 + 5 + 8)/(β(9 + 25 + 16) β(1 + 1 + 4))| = |16/(β50 β6)| = |16/(5β2 Γ β2 β3)| = |16/(5 Γ 2 Γ β3)| = 8/(5 β3) = 8/(5 β3) Γ β3/β3 = (8β3)/(15 ) So, cos ΞΈ = (8β3)/(15 ) β΄ ΞΈ = cos-1((πβπ)/(ππ )) Therefore, the angle between the given pair of line is cosβ1 ((8β3)/(15 ))
