Last updated at June 15, 2018 by Teachoo

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Misc 13 (Method 1) Find the equation of the plane passing through the point ( 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through ( 1 , 1 , 1 ) is given by A(x ) + B (y ) + C(z ) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through ( 1, 3, 2) So, equation of plane is A(x + 1) + B (y 3) + C(z 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes. So, their normal to plane would be perpendicular to normals of both planes. We know that is perpendicular to both & So, required is normal is cross product of normals of planes x + 2y + 3z = 5 and 3x + 3y + z = 0 Required normal = 1 2 3 3 3 1 = (2(1) 3(3)) (1(1) 3(3)) + (1(3) 3(2)) = (2 9) (1 9) + (3 6) = 7 + 8 3 Hence, direction ratios = 7, 8, 3 A = 7, B = 8, C = 3 Putting above values in (1) A(x + 1) + B (y 3) + C(z 2) = 0 7k (x + 1) + 8k (y 3) 3k (z 2) = 0 k 7 +1 +8 3 3( 2) = 0 7(x + 1) + 8(y 3) 3(z 2) = 0 7x 7 + 8y 24 3z + 6 = 0 7x + 8y 3z 25 = 0 0 = 7x 8y + 3z + 25 7x 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x 8y + 3z + 25 = 0 Misc 13 (Method 2) Find the equation of the plane passing through the point ( 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through ( 1 , 1 , 1 ) is given by A(x ) + B (y ) + C(z ) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through ( 1, 3, 2) So, equation of plane is A(x + 1) + B (y 3) + C(z 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes. Now, it is given that plane A(x + 1) + B (y 3) + C(z 2) = 0 is perpendicular to plane x + 2y + 3z = 5 Hence, A 1 + B 2 + C 3 = 0 A + 2B + 3C = 0 Similarly, Given that plane A(x + 1) + B (y 3) + C(z 2) = 0 is perpendicular to plane 3x + 3y + z = 0. Hence, A 3 + B 3 + C 1 = 0 3A + 3B + C = 0 So, our equations are A + 2B + 3C = 0 3A + 3B + C = 0 Solving 2 9 = 9 1 = 3 6 = = = k So, A = 7k, B = 8k, C = 3k Putting above values in (1) A(x + 1) + B (y 3) + C(z 2) = 0 7k (x + 1) + 8k (y 3) 3k (z 2) = 0 k 7 +1 +8 3 3( 2) = 0 7(x + 1) + 8(y 3) 3(z 2) = 0 7x 7 + 8y 24 3z + 6 = 0 7x + 8y 3z 25 = 0 0 = 7x 8y + 3z + 25 7x 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x 8y + 3z + 25 = 0

Chapter 11 Class 12 Three Dimensional Geometry

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.