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Ex 11.3

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at March 22, 2023 by Teachoo

Ex 11.3, 10 Find the vector equation of plane passing through the intersection of the planes π β . (2π Μ + 2π Μ β 3π Μ) = 7, π β .(2π Μ + 5π Μ + 3π Μ) = 9 and through the point (2, 1, 3).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 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. (2π Μ + 2π Μ β 3π Μ) = 7 Comparing with π β.(π1) β = π1, (π1) β = 2π Μ + 2π Μ β 3π Μ & d1 = 7 π β. (2π Μ + 5π Μ + 3π Μ) = 9 Comparing with π β.(π2) β = π2, (π2) β = 2π Μ + 5π Μ + 3π Μ & d2 = 9 So, equation of the plane is π β.["(2" π Μ+"2" π Μ" " β"3" π Μ")" +"π(2" π Μ" " + 5π Μ" " + "3" π Μ")" ] = 7 + π.9 π β. ["2" π Μ" " +" 2" π Μ" " β "3" π Μ + 2"π" π Μ + 5"π" π Μ + 3"π" π Μ ] = 7 + 9"π" π β. ["(2" +"2π" )π Μ" " +"(2" +"5π" )π Μ +"(β" π+"3π" )π Μ ] = 9"π" + 7 Now, to find π , put π β = xπ Μ + yπ Μ + zπ Μ (xπ Μ + yπ Μ + zπ Μ).["(2 " + "2π" )π Μ" " + "(2" +"5π" )π Μ +"(β" 3" " +" 3π" )π Μ ] = 9π + 7 x"(2 "+" 2π")" "+ "y (2 "+" 5π")π Μ + π§"("β3+"3π")π Μ = 9π + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2π) + 1(2 + 5π) + 3(β3 + 3π) = 9π + 7 4 + 4π + 2 + 5π + (β9) + 9π = 9π + 7 18π β 9π = 7 + 3 9π = 10 β΄ π = ππ/π Putting value of π in (1), π β. [(2+ 2. 10/9) π Μ + (2+ 5. 10/9) π Μ +("β" 3+ 3. 10/9) π Μ ] = 9.10/9 + 7 π β. [(2 + 20/9) π Μ +(2 + 50/9) π Μ +("β" 3 + 30/9) π Μ ] = 10 + 7 π β. [38/9 π Μ+ 68/9 π Μ + 3/9 π Μ ] = 17 1/9 π β. (38π Μ + 68π Μ + 3π Μ ) = 17 π β.(38π Μ+68π Μ+3π Μ ) = 17 Γ 9 π β.(πππ Μ+πππ Μ+ππ Μ ) = 153