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Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7 Deleted for CBSE Board 2023 Exams
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Misc 9 Important
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Misc 13 Important Deleted for CBSE Board 2023 Exams You are here
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Misc 22 (MCQ) Important Deleted for CBSE Board 2023 Exams
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Miscellaneous
Last updated at Feb. 1, 2020 by Teachoo
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