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Misc 13 - Find equation of plane passing through (-1, 3, 2) - Teacho

Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7
Misc 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.