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Last updated at Feb. 1, 2020 by Teachoo

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Misc 13 (Method 1) Find the equation of the plane passing through the point (โ1, 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 (๐ฅ_1, ๐ฆ_1, ๐ง_1) is given by A(x โ ๐_๐) + B (y โ ๐_๐) + C(z โ ๐_๐) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + C(z โ 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C The equation of a plane passing through (๐ฅ_1, ๐ฆ_1, ๐ง_1) is given by A(x โ ๐_๐) + B (y โ ๐_๐) + C(z โ ๐_๐) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + C(z โ 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 normal of both planes. We know that ๐ โ ร ๐ โ is perpendicular to both ๐ โ & ๐ โ So, required is normal is cross product of normals of planes x + 2y + 3z = 5 and 3x + 3y + z = 0 Required normal = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@1&2&3@3&3&1)| = ๐ ฬ (2(1) โ 3(3)) โ ๐ ฬ (1(1) โ 3(3)) + ๐ ฬ(1(3) โ 3(2)) = ๐ ฬ (2 โ 9) โ ๐ ฬ (1 โ 9) + ๐ ฬ(3 โ 6) = โ7๐ ฬ + 8๐ ฬ โ 3๐ ฬ Hence, direction ratios = โ7, 8, โ3 โด A = โ7, B = 8, C = โ3 Putting above values in (1) A(x + 1) + B (y โ 3) + C(z โ 2) = 0 โ7k (x + 1) + 8k (y โ 3) โ 3k (z โ2) = 0 k [โ7(๐ฅ+1)+8(๐ฆโ3)โ3(๐งโ2)] = 0 โ7(x + 1) + 8(y โ 3) โ 3(z โ 2) = 0 โ7x โ 7 + 8y โ 24 โ 3z + 6 = 0 โ7x + 8y โ3z โ 25 = 0 0 = 7x โ 8y + 3z + 25 7x โ 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x โ 8y + 3z + 25 = 0 Misc 13 (Method 2) Find the equation of the plane passing through the point (โ 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 (๐ฅ_1, ๐ฆ_1, ๐ง_1) is given by A(x โ ๐_๐) + B (y โ ๐_๐) + C(z โ ๐_๐) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ1, 3, 2) So, equation of plane is A(x + 1) + B (y โ 3) + C(z โ 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 โ 3) + C(z โ 2) = 0 is perpendicular to plane x + 2y + 3z = 5 Hence, A ร 1 + B ร 2 + C ร 3 = 0 A + 2B + 3C = 0 Similarly, Given that plane A(x + 1) + B (y โ 3) + C(z โ 2) = 0 is perpendicular to plane 3x + 3y + z = 0. Two lines with direction ratios ๐_1, ๐_1, ๐_1 and ๐_2, ๐_2, ๐_2 are perpendicular if ๐_1 ๐_2 + ๐_1 ๐_2 + ๐_1 ๐_2 = 0 Hence, A ร 3 + B ร 3 + C ร 1 = 0 3A + 3B + C = 0 So, our equations are A + 2B + 3C = 0 3A + 3B + C = 0 Solving Two lines with direction ratios ๐_1, ๐_1, ๐_1 and ๐_2, ๐_2, ๐_2 are perpendicular if ๐_1 ๐_2 + ๐_1 ๐_2 + ๐_1 ๐_2 = 0 ๐ด/(2 โ 9) = ๐ต/(9 โ 1) = ๐ถ/(3 โ 6) ๐จ/(โ๐) = ๐ฉ/๐ = ๐ช/(โ๐) = k So, A = โ7k, B = 8k, C = โ3k Putting above values in (1) A(x + 1) + B (y โ 3) + C(z โ 2) = 0 โ7k (x + 1) + 8k (y โ 3) โ 3k (z โ2) = 0 k [โ7(๐ฅ+1)+8(๐ฆโ3)โ3(๐งโ2)] = 0 โ7(x + 1) + 8(y โ 3) โ 3(z โ 2) = 0 โ7x โ 7 + 8y โ 24 โ 3z + 6 = 0 โ7x + 8y โ3z โ 25 = 0 0 = 7x โ 8y + 3z + 25 7x โ 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x โ 8y + 3z + 25 = 0

Miscellaneous

Misc 1
Important

Misc 2

Misc 3 Not in Syllabus - CBSE Exams 2021

Misc 4 Important

Misc 5 Important Not in Syllabus - CBSE Exams 2021

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

Serial order wise

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.