Last updated at Dec. 16, 2024 by Teachoo
Question 9 Find the equation of the plane through the intersection of the planes 3x y + 2z 4 = 0 and x + y + z 2 = 0 and the point (2, 2, 1). Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 and through the point (x1, y1, z1) is (A1x + B1y + C1z d1) + (A2x + B2y + C2z d2) = 0 Given, plane passes through Equation of plane is (3x 1y + 2z 4) + (1x + 1y + 1z 2) = 0 3x y + 2z 4 + x + y + z 2 = 0 (3 + ) x + ( 1 + )y + (2 + )z + ( 4 2 ) = 0 We now find the value of The plane passes through (2, 2, 1) Putting (2, 2, 1) in (1), (3 + ) x + ( 1 + )y + (2 + )z + ( 4 2 ) = 0 (3 + ) 2 + ( 1 + ) 2 + (2 + ) 1 + ( 4 2 ) = 0 6 + 2 2 + 2 + 2 + 4 2 = 0 3 + 2 = 0 3 = 2 = Putting value of in (1), (3 + ) x + ( 1 + )y + (2 + )z + ( 4 2 ) = 0 3+ 2 3 x + 1+ 2 3 y + 2+ 2 3 z + 4 2 2 3 = 0 3 2 3 + 1 2 3 y + 2 2 3 z + 4+ 4 3 = 0 7 3 5 3 + 4 3 8 3 = 0 1 3 (7x 5y + 4z 8) = 0 7x 5y + 4z 8 = 0 The equation of plane is 7x 5y + 4z 8 = 0
Plane
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo