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

Transcript

Example 27 (Method 1) Find the equation of the plane that contains the point (1, โ€“1, 2) and is perpendicular to each of the planes 2x + 3y โ€“ 2z = 5 and x + 2y โ€“ 3z = 8. 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 (1, โˆ’1, 2) So, equation of plane is A(x โˆ’1) + B (y + 1) + 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 normal of planes 2x + 3y โ€“ 2z = 5 and x + 2y โ€“ 3z = 8. Required normal = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@2&3&โˆ’2@1&2&โˆ’3)| = ๐‘– ฬ‚ (3(โ€“3) โ€“ 2(โ€“2)) โ€“ ๐‘— ฬ‚ (2(โ€“3) โ€“ 1(โ€“2)) + ๐‘˜ ฬ‚(2(2) โ€“ 1(3)) = ๐‘– ฬ‚ (โ€“9 + 4) โ€“ ๐‘— ฬ‚ (โ€“6 + 2) + ๐‘˜ ฬ‚(4 โ€“ 3) = โ€“5๐‘– ฬ‚ + 4๐‘— ฬ‚ + ๐‘˜ ฬ‚ Hence, direction ratios = โ€“5, 4, 1 โˆด A = โ€“5, B = 4, C = 1 Putting above values in (1), A(x โˆ’1) + B (y + 1) + C(z โˆ’ 2) = 0 โˆ’5(x โˆ’ 1) + 4 (y + 1) + 1 (z โˆ’ 2) = 0 โˆ’5x + 5 + 4y + 4 + z โˆ’ 2 = 0 โˆ’5x + 4y + z + 7 = 0 โˆ’5x + 4y + z = โˆ’7 โˆ’(5x โˆ’4y โˆ’ z) = โˆ’7 5x โˆ’ 4y โˆ’ z = 7 Therefore, the equation of the required plane is 5x โˆ’ 4y โˆ’ z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, โ€“ 1, 2) and is perpendicular to each of the planes 2x + 3y โ€“ 2z = 5 and x + 2y โ€“ 3z = 8. 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 (1, โˆ’1, 2) So, equation of plane is A(x โˆ’1) + B (y + 1) + 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 โˆ’1) + B (y + 1) + C(z โˆ’ 2) = 0 is perpendicular to plane 2x + 3y โ€“ 2z = 5 Hence, A ร— 2 + B ร— 3 + C ร— (โ€“2) = 0 2A + 3B โˆ’ 2C = 0 Similarly, Given that plane A(x โˆ’1) + B (y + 1) + C(z โˆ’ 2) = 0 is perpendicular to plane x + 2y โ€“ 3z = 8 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— 1 + B ร— 2 + C ร— (โ€“3) = 0 A + 2B โˆ’ 3C = 0 So, our equations are 2A + 3B โˆ’2C = 0 A + 2B โˆ’ 3C = 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/(โˆ’9 โˆ’ (โˆ’4)) = ๐ต/(โˆ’2 โˆ’ (โˆ’6)) = ๐ถ/(4 โˆ’ 3) ๐ด/(โˆ’9 + 4) = ๐ต/(โˆ’2 + 6) = ๐ถ/1 ๐ด/(โˆ’5) = ๐ต/4 = ๐ถ/1 = k So, A = โ€“5k , B = 4k , C = k Putting above values in (1), A(x โˆ’1) + B (y + 1) + C(z โˆ’ 2) = 0 โˆ’5k(x โˆ’ 1) + 4k (y + 1) + k (z โˆ’ 2) = 0 k[โˆ’5(x โˆ’ 1) + 4(y + 1) + (z โˆ’ 2)] = 0 โˆ’5x + 5 + 4y + 4 + z โˆ’ 2 = 0 โˆ’5x + 4y + z + 7 = 0 โˆ’5x + 4y + z = โˆ’7 โˆ’(5x โˆ’4y โˆ’ z) = โˆ’7 5x โˆ’ 4y โˆ’ z = 7 Therefore, the equation of the required plane is 5x โˆ’ 4y โˆ’ z = 7.

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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.