Last updated at Dec. 8, 2016 by Teachoo

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Misc 17 Find the equation of the plane which contains the line of intersection of the planes đďˇŻ . ( đďˇŻ + 2 đďˇŻ + 3 đďˇŻ) â 4 = 0 , đďˇŻ . (2 đďˇŻ + đďˇŻ â đďˇŻ) + 5 = 0 and which is perpendicular to the plane đďˇŻ . (5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) + 8 = 0 . Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z â d1) + đ (A2x + B2y + C2z â d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 Equation of plane is (A1x + B1y + C1z â d1) + đ (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z â 4) + đ ( â 2x â 1y + 1z â 5) = 0 (1 â 2đ) x + (2 â đ)y + (3 + đ) z + ( â4 â 5đ) = 0 Now, the plane is perpendicular to the plane đďˇŻ.(5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) + 8 = 0 So, normal to plane đďˇŻ will be perpendicular to normal đďˇŻ of đďˇŻ.(5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) + 8 = 0 Now, đďˇŻ.(5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) + 8 = 0 đďˇŻ .(5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) = â8 â đďˇŻ .(5 đďˇŻ + 3 đďˇŻ â 6 đďˇŻ) = 8 đďˇŻ .( â5 đďˇŻ â 3 đďˇŻ + 6 đďˇŻ) = 8 Finding direction cosines of đďˇŻ & đďˇŻ Since, đďˇŻ is perpendicular to đďˇŻ đ1 đ2 + b1 b2 + c1 c2 = 0 (1 â 2đ) Ă â5 + (2 â đ) Ă â3 + (3 + đ) Ă 6 = 0 â 5 + 10đ â 6 + 3đ + 18 + 6đ = 0 19đ + 7 = 0 â´ đ = âđďˇŽđđďˇŻ Putting value of đ in (1), (1 â 2đ) x + (2 â đ)y + (3 + đ) z + ( â4 â 5đ) = 0 1â2 Ă â7ďˇŽ19ďˇŻďˇŻ x + 2â â7ďˇŽ19ďˇŻďˇŻďˇŻy + 3+ â 7ďˇŽ19ďˇŻďˇŻďˇŻz + â4â5Ă â7ďˇŽ19ďˇŻďˇŻ = 0 1 + 14ďˇŽ19ďˇŻďˇŻ x + 2 + 7ďˇŽ19ďˇŻďˇŻy + 3 â 7ďˇŽ19ďˇŻďˇŻz + â 4 + 35ďˇŽ19ďˇŻďˇŻ = 0 33ďˇŽ19ďˇŻ x + 45ďˇŽ19ďˇŻ y + 50ďˇŽ19ďˇŻ z â 41ďˇŽ19ďˇŻ = 0 1ďˇŽ19ďˇŻ (33x + 45y + 50z â 41) = 0 33x + 45y + 50z â 41 = 0 Therefore, the equation of the plane is 33x + 45y + 50z = 41.

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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