Equation of plane - Passing Through Intersection Of Planes
Equation of plane - Passing Through Intersection Of Planes
Last updated at August 8, 2026 by Teachoo
Transcript
Question 10 Find the vector equation of plane passing through the intersection of the planes ๐ โ . (2๐ ฬ + 2๐ ฬ โ 3๐ ฬ) = 7, ๐ โ .(2๐ ฬ + 5๐ ฬ + 3๐ ฬ) = 9 and through the point (2, 1, 3).The vector equation of a plane passing through the intersection of planes ๐ โ. (๐1) โ = d1 and ๐ โ. (๐2) โ = d2 and also through the point (x1, y1, z1) is ๐ โ.((๐๐) โ + ๐(๐๐) โ) = d1 + ๐d2 Given, the plane passes through The vector equation of a plane passing through the intersection of planes ๐ โ. (๐1) โ = d1 and ๐ โ. (๐2) โ = d2 and also through the point (x1, y1, z1) is ๐ โ.((๐๐) โ + ๐(๐๐) โ) = d1 + ๐d2 Given, the plane passes through ๐ โ. (2๐ ฬ + 2๐ ฬ โ 3๐ ฬ) = 7 Comparing with ๐ โ.(๐1) โ = ๐1, (๐1) โ = 2๐ ฬ + 2๐ ฬ โ 3๐ ฬ & d1 = 7 ๐ โ. (2๐ ฬ + 5๐ ฬ + 3๐ ฬ) = 9 Comparing with ๐ โ.(๐2) โ = ๐2, (๐2) โ = 2๐ ฬ + 5๐ ฬ + 3๐ ฬ & d2 = 9 So, equation of the plane is ๐ โ.["(2" ๐ ฬ+"2" ๐ ฬ" " โ"3" ๐ ฬ")" +"๐(2" ๐ ฬ" " + 5๐ ฬ" " + "3" ๐ ฬ")" ] = 7 + ๐.9 ๐ โ. ["2" ๐ ฬ" " +" 2" ๐ ฬ" " โ "3" ๐ ฬ + 2"๐" ๐ ฬ + 5"๐" ๐ ฬ + 3"๐" ๐ ฬ ] = 7 + 9"๐" ๐ โ. ["(2" +"2๐" )๐ ฬ" " +"(2" +"5๐" )๐ ฬ +"(โ" ๐+"3๐" )๐ ฬ ] = 9"๐" + 7 Now, to find ๐ , put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ (x๐ ฬ + y๐ ฬ + z๐ ฬ).["(2 " + "2๐" )๐ ฬ" " + "(2" +"5๐" )๐ ฬ +"(โ" 3" " +" 3๐" )๐ ฬ ] = 9๐ + 7 x"(2 "+" 2๐")" "+ "y (2 "+" 5๐")๐ ฬ + ๐ง"("โ3+"3๐")๐ ฬ = 9๐ + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2๐) + 1(2 + 5๐) + 3(โ3 + 3๐) = 9๐ + 7 4 + 4๐ + 2 + 5๐ + (โ9) + 9๐ = 9๐ + 7 18๐ โ 9๐ = 7 + 3 9๐ = 10 โด ๐ = ๐๐/๐ Putting value of ๐ in (1), ๐ โ. [(2+ 2. 10/9) ๐ ฬ + (2+ 5. 10/9) ๐ ฬ +("โ" 3+ 3. 10/9) ๐ ฬ ] = 9.10/9 + 7 ๐ โ. [(2 + 20/9) ๐ ฬ +(2 + 50/9) ๐ ฬ +("โ" 3 + 30/9) ๐ ฬ ] = 10 + 7 ๐ โ. [38/9 ๐ ฬ+ 68/9 ๐ ฬ + 3/9 ๐ ฬ ] = 17 1/9 ๐ โ. (38๐ ฬ + 68๐ ฬ + 3๐ ฬ ) = 17 ๐ โ.(38๐ ฬ+68๐ ฬ+3๐ ฬ ) = 17 ร 9 ๐ โ.(๐๐๐ ฬ+๐๐๐ ฬ+๐๐ ฬ ) = 153