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Ex 11.3, 10 - Find vector equation of plane passing through - Equation of plane - Passing Through Intersection Of Planes

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.3, 10 Find the vector equation of the plane passing through the intersection of the planes 𝑟﷯ . (2 𝑖﷯ + 2 𝑗﷯ − 3 𝑘﷯) = 7, 𝑟﷯ .(2 𝑖﷯ + 5 𝑗﷯ + 3 𝑘﷯) = 9 and through the point (2, 1, 3). The vector equation of a plane passing through the intersection of planes 𝑟﷯. 𝑛1﷯ = d1 and 𝑟﷯. 𝑛2﷯ = d2 and also through the point (x1, y1, z1) is 𝒓﷯.( 𝒏𝟏﷯ + 𝜆 𝒏𝟐﷯) = d1 + 𝜆d2 Given, the plane passes through So, equation of the plane is 𝑟﷯. (2 𝑖﷯+2 𝑗﷯ −3 𝑘﷯)+𝜆(2 𝑖﷯ + 5 𝑗﷯ + 3 𝑘﷯)﷯ = 7 + 𝜆.9 𝑟﷯. 2 𝑖﷯ + 2 𝑗﷯ − 3 𝑘﷯ + 2𝜆 𝑖﷯ + 5𝜆 𝑗﷯ + 3𝜆 𝑘﷯ ﷯ = 7 + 9𝜆 𝒓﷯. (2+2𝜆) 𝒊﷯ +(2+5𝜆) 𝒋﷯ +(−𝟑+3𝜆) 𝒌﷯ ﷯ = 9𝜆 + 7 Now, to find 𝜆 , put 𝒓﷯ = x 𝒊﷯ + y 𝒋﷯ + z 𝒌﷯ (x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯). (2 + 2𝜆) 𝑖﷯ + (2+5𝜆) 𝑗﷯ +(−3 + 3𝜆) 𝑘﷯ ﷯ = 9𝜆 + 7 x(2 + 2𝜆) + y (2 + 5𝜆) 𝑗﷯ + 𝑧(−3+3𝜆) 𝑘﷯ = 9𝜆 + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2𝜆) + 1(2 + 5𝜆) + 3( − 3 + 3x) = 9𝜆 + 7 4 + 4𝜆 + 2 + 5𝜆 + ( − 9) + 9𝜆 = 9𝜆 + 7 18𝜆 − 9𝜆 = 7 + 3 9𝜆 = 10 ∴ 𝜆 = 𝟏𝟎﷮𝟗﷯ Putting value of 𝜆 in (1), 𝑟﷯. 2+ 2. 10﷮9﷯﷯ 𝑖﷯ + 2+ 5. 10﷮9﷯﷯ 𝑗﷯ + − 3 + 3. 10﷮9﷯﷯ 𝑘﷯﷯ = 9. 10﷮9﷯ + 7 𝑟﷯. 2 + 20﷮9﷯﷯ 𝑖﷯ + 2 + 50﷮9﷯﷯ 𝑗﷯ + − 3 + 30﷮9﷯﷯ 𝑘﷯﷯ = 10 + 7 𝑟﷯. 38﷮9﷯ 𝑖﷯+ 68﷮9﷯ 𝑗﷯ + 3﷮9﷯ 𝑘﷯﷯ = 17 1﷮9﷯ 𝑟﷯. 38 𝑖﷯ + 68 𝑗﷯ + 3 𝑘﷯﷯ = 17 𝑟﷯. 38 𝑖﷯+68 𝑗﷯+3 𝑘﷯﷯ = 17 × 9 𝒓﷯. 𝟑𝟖 𝒊﷯+𝟔𝟖 𝒋﷯+𝟑 𝒌﷯﷯ = 153 Therefore, the vector equation of the required plane is 𝑟﷯. 38 𝑖﷯+68 𝑗﷯+3 𝑘﷯﷯= 153

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