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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Misc 13 (Method 1) 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 (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) is given by A(x โˆ’ ๐’™_๐Ÿ) + B (y โˆ’ ๐’š_๐Ÿ) + C(z โ€“ ๐’›_๐Ÿ) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ€“1, 3, 2) So, equation of plane is A(x + 1) + B (y โ€“ 3) + C(z โˆ’ 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C The equation of a plane passing through (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) is given by A(x โˆ’ ๐’™_๐Ÿ) + B (y โˆ’ ๐’š_๐Ÿ) + C(z โ€“ ๐’›_๐Ÿ) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ€“1, 3, 2) So, equation of plane is A(x + 1) + B (y โ€“ 3) + C(z โˆ’ 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‘Ž โƒ— ร— ๐‘ โƒ— is perpendicular to both ๐‘Ž โƒ— & ๐‘ โƒ— 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) = โ€“7๐‘– ฬ‚ + 8๐‘— ฬ‚ โ€“ 3๐‘˜ ฬ‚ Hence, direction ratios = โ€“7, 8, โ€“3 โˆด A = โ€“7, B = 8, C = โ€“3 Putting above values in (1) A(x + 1) + B (y โ€“ 3) + C(z โˆ’ 2) = 0 โˆ’7k (x + 1) + 8k (y โˆ’ 3) โˆ’ 3k (z โˆ’2) = 0 k [โˆ’7(๐‘ฅ+1)+8(๐‘ฆโˆ’3)โˆ’3(๐‘งโˆ’2)] = 0 โˆ’7(x + 1) + 8(y โˆ’ 3) โˆ’ 3(z โˆ’ 2) = 0 โˆ’7x โˆ’ 7 + 8y โˆ’ 24 โˆ’ 3z + 6 = 0 โˆ’7x + 8y โˆ’3z โˆ’ 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 (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) is given by A(x โˆ’ ๐’™_๐Ÿ) + B (y โˆ’ ๐’š_๐Ÿ) + C(z โ€“ ๐’›_๐Ÿ) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (โ€“1, 3, 2) So, equation of plane is A(x + 1) + B (y โ€“ 3) + C(z โˆ’ 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 โˆ’ 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 โˆ’ 2) = 0 is perpendicular to plane 3x + 3y + z = 0. Two lines with direction ratios ๐‘Ž_1, ๐‘_1, ๐‘_1 and ๐‘Ž_2, ๐‘_2, ๐‘_2 are perpendicular if ๐‘Ž_1 ๐‘Ž_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‘Ž_1, ๐‘_1, ๐‘_1 and ๐‘Ž_2, ๐‘_2, ๐‘_2 are perpendicular if ๐‘Ž_1 ๐‘Ž_2 + ๐‘_1 ๐‘_2 + ๐‘_1 ๐‘_2 = 0 ๐ด/(2 โˆ’ 9) = ๐ต/(9 โˆ’ 1) = ๐ถ/(3 โˆ’ 6) ๐‘จ/(โˆ’๐Ÿ•) = ๐‘ฉ/๐Ÿ– = ๐‘ช/(โˆ’๐Ÿ‘) = k So, A = โˆ’7k, B = 8k, C = โˆ’3k Putting above values in (1) A(x + 1) + B (y โ€“ 3) + C(z โˆ’ 2) = 0 โˆ’7k (x + 1) + 8k (y โˆ’ 3) โˆ’ 3k (z โˆ’2) = 0 k [โˆ’7(๐‘ฅ+1)+8(๐‘ฆโˆ’3)โˆ’3(๐‘งโˆ’2)] = 0 โˆ’7(x + 1) + 8(y โˆ’ 3) โˆ’ 3(z โˆ’ 2) = 0 โˆ’7x โˆ’ 7 + 8y โˆ’ 24 โˆ’ 3z + 6 = 0 โˆ’7x + 8y โˆ’3z โˆ’ 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

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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.