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

Examples

Example 1

Example, 2 Important

Example, 3

Example, 4 Important

Example, 5 Important

Example, 6 Important

Example, 7

Example 8 Important You are here

Example 9

Example 10 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Important Deleted for CBSE Board 2024 Exams

Question 14 Deleted for CBSE Board 2024 Exams

Question 15 Important Deleted for CBSE Board 2024 Exams

Question 16 Deleted for CBSE Board 2024 Exams

Question 17 Important Deleted for CBSE Board 2024 Exams

Question 18 Important Deleted for CBSE Board 2024 Exams

Question 19 Important Deleted for CBSE Board 2024 Exams

Question 20 Important Deleted for CBSE Board 2024 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at May 29, 2023 by Teachoo

Example 8 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 ))