Last updated at Dec. 24, 2019 by Teachoo

Transcript

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

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important You are here

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23 Important

Chapter 11 Class 12 Three Dimensional Geometry

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About the Author

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.