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

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 You are here

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

Question 10 Find the vector equation of the plane passing through the intersection of the planes π β . (π Μ + π Μ + π Μ) = 6 and π β . (2π Μ + 3π Μ + 4π Μ) = β 5, and the point (1, 1, 1).The vector equation of a plane passing through the intersection of planes π β. (π1) β = d1 and π β. (π2) β = d2 and also through the point (x1, y1, z1) is π β.((ππ) β + π(ππ) β) = d1 + πd2 Given, the plane passes through π β.(π Μ + π Μ + π Μ) = 6 Comparing with π β.(π1) β = d1, (ππ) β = π Μ + π Μ + π Μ & d1 = 6 π β.(2π Μ + 3π Μ + 4π Μ) = β5 βπ β.(2π Μ + 3π Μ + 4π Μ) = 5 π β .(β 2π Μ β 3π Μ β 4π Μ) = 5 Comparing with π β.(π2) β = d2 (ππ) β = β 2π Μ β 3π Μ β 4π Μ & d2 = 5 Equation of plane is π β. [(π Μ+π Μ+π Μ )+"π" (β2π Μβ3π Μβ4π Μ)] = 6 + π5 π β. [(π Μ" " +π Μ" " +π Μ )β"π" (ππ Μ+ππ Μ+ππ Μ)] = 6 + 5π Now to find π , put π β = xπ Μ + yπ Μ + zπ Μ (xπ Μ + yπ Μ + zπ Μ). [(π Μ+π Μ+π Μ )β"π" (2π Μ+3π Μ+4π Μ)] = 5π + 6 (xπ Μ + yπ Μ + zπ Μ).(π Μ+π Μ+π Μ ) β π (xπ Μ + yπ Μ + zπ Μ).(2π Μ+3π Μ+4π Μ) = 5π + 6 (x Γ 1) + (y Γ 1) + (z Γ 1) β π[(π₯Γ2)+(π¦Γ3)+(π§Γ4)] = 5π + 6 x + y + z β π[2π₯+3π¦+4π§] = 5π + 6 x + y + z β 2ππ₯ β 3πy β 4πz = 5π + 6 (1 β 2π)x + (1 β 3π)y + (1 β 4π) z = 5π + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 β 2π)x + (1 β 3π)y + (1 β 4π) z = 5π + 6 (1 β2π) Γ 1 + (1 β 3π) Γ 1 + (1 β 4π) Γ 1 = 5π + 6 1 β2π + 1 β 3π + 1 β 4π= 5π + 6 3 β 9π = 5π + 6 β14π = 3 β΄ π = (βπ)/ππ Putting value of π in (1), π β. [(π Μ" " +" " π Μ" " +" " π Μ )β(( β3)/14)(2π Μ+3π Μ+"4" π Μ)]= 6 + 5 Γ ( β3)/14 π β. [(π Μ+π Μ+" " π Μ )+3/14(2π Μ+3π Μ+"4" π Μ)]= 6 β 15/14 π β. [π Μ+π Μ" " +π Μ+ 6/14 π Μ+9/14 π Μ+12/14 π Μ ]= 69/14 π β. [(1+6/14) π Μ +(1+9/14) π Μ+(1+12/14) π Μ ]= 69/14 π β. [20/14 π Μ + 23/14 π Μ + 26/14 π Μ ]= 69/14 π β. [1/14(20π Μ+23π Μ+26π Μ)]= 69/14 1/14 π β. (20π Μ + 23π Μ + 26π Μ) = 69/14 π β. (20π Μ + 23π Μ + 26π Μ) = 69 Therefore, the vector equation of the required plane is π β.(πππ Μ + πππ Μ + πππ Μ) = ππ