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Ex 11.3
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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