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

Transcript

Ex 11.3, 6 (Introduction) Find the equations of the planes that passes through three points. (a) (1, 1, โ€“ 1), (6, 4, โ€“ 5), (โ€“ 4, โ€“ 2, 3) Vector equation of a plane passing through three points with position vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— is ("r" โƒ— โˆ’ ๐‘Ž โƒ—) . [(๐‘ โƒ—โˆ’๐‘Ž โƒ—)ร—(๐‘ โƒ—โˆ’๐‘Ž โƒ—)] = 0 Ex 11.3, 6 Find the equations of the planes that passes through three points. (a) (1, 1, โ€“1), (6, 4, โ€“5), (โ€“4, โ€“2, 3) Vector equation of a plane passing through three points with position vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— is ("r" โƒ— โˆ’ ๐’‚ โƒ—) . [(๐’ƒ โƒ—โˆ’๐’‚ โƒ—)ร—(๐’„ โƒ—โˆ’๐’‚ โƒ—)] = 0 Now, the plane passes through the points (๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) = (6๐‘– ฬ‚ + 4๐‘— ฬ‚ โ€“ 5๐‘˜ ฬ‚) โˆ’ (1๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚) = (6 โˆ’1)๐‘– ฬ‚ + (4 โˆ’ 1)๐‘— ฬ‚ + (โˆ’5 โˆ’ (โˆ’1)) ๐‘˜ ฬ‚ = 5๐’Š ฬ‚ + 3๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚ A (1, 1, โˆ’1) ๐‘Ž โƒ— = 1๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚ B (6, 4, โˆ’5) ๐‘ โƒ— = 6๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ C ( โˆ’4, โˆ’2, 3) ๐‘ โƒ— = โˆ’4๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ (๐’„ โƒ— โˆ’ ๐’‚ โƒ—) = (โˆ’4๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) โˆ’ (1๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚) = (โˆ’4 โˆ’ 1)๐‘– ฬ‚ +(โˆ’2 โˆ’ 1)๐‘— ฬ‚ + (3 โˆ’ (โˆ’1)) ๐‘˜ ฬ‚ = โˆ’5๐’Š ฬ‚ โˆ’ 3๐’‹ ฬ‚ + 4๐’Œ ฬ‚ (๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) ร— (๐’„ โƒ— โˆ’ ๐’‚ โƒ—) = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@5&3&โˆ’4@โˆ’5&โˆ’3&4)| = โ€“ |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@5&3&โˆ’4@ 5& 3&โˆ’4)| = ๐ŸŽ โƒ— This implies, the three points are collinear. Using property: Since the two rows of the determinant are same, the value of determinant is zero. โˆด Vector equation of plane is [๐‘Ÿ โƒ—โˆ’(๐‘– ฬ‚+๐‘— ฬ‚ โˆ’๐‘˜ ฬ‚ )] . 0 โƒ— = 0 Since, the above equation is satisfied for all values of ๐‘Ÿ โƒ—, Therefore, there will be infinite planes passing through the given 3 collinear points. Ex 11.3, 6 Find the equations of the planes that passes through three points. (b) (1, 1, 0), (1, 2, 1), (โ€“2, 2, โ€“1) Vector equation of a plane passing through three points with position vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— is ("r" โƒ— โˆ’ ๐’‚ โƒ—) . [(๐’ƒ โƒ—โˆ’๐’‚ โƒ—)ร—(๐’„ โƒ— โˆ’๐’‚ โƒ—)] = 0 Now, the plane passes through the points A (1, 1, 0) ๐‘Ž โƒ— = 1๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚ B (1, 2, 1) ๐‘ โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚ C (โˆ’2, 2, โˆ’1) ๐‘ โƒ— = โˆ’2๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚ (๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) = (1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚) โˆ’ (1๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = (1 โˆ’ 1) ๐‘– ฬ‚ + (2 โˆ’ 1)๐‘— ฬ‚ + (1 โˆ’ 0) ๐‘˜ ฬ‚ = 0๐’Š ฬ‚ + 1๐’‹ ฬ‚ + 1๐’Œ ฬ‚ (๐’„ โƒ— โˆ’ ๐’‚ โƒ—) = (โ€“2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚) โˆ’ (1๐‘– ฬ‚ + 1๐‘— ฬ‚ +0๐‘˜ ฬ‚) = (โˆ’2 โˆ’ 1)๐‘– ฬ‚ + (2 โˆ’ 1)๐‘— ฬ‚ + (โˆ’1 โˆ’ 0) ๐‘˜ ฬ‚ = โˆ’3๐’Š ฬ‚ + 1๐’‹ ฬ‚ โˆ’ 1๐’Œ ฬ‚ (๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) ร— (๐’„ โƒ— โˆ’ ๐’‚ โƒ—) = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@0&1&1@ โˆ’ 3&1& โˆ’ 1)| = ๐‘– ฬ‚ [(1ร—โˆ’1)โˆ’(1ร—1)] โ€“ ๐‘— ฬ‚ [(0ร—โˆ’1)โˆ’(โˆ’3 ร—1)] + (๐‘˜ ) ฬ‚[(0ร—1)โˆ’(โˆ’3 ร—1)] = ๐‘– ฬ‚(โ€“1 โ€“ 1) โ€“ ๐‘— ฬ‚ (0 + 3) + ๐‘˜ ฬ‚ ( 0 + 3) = โ€“2๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 3๐’Œ ฬ‚ โˆด Vector equation of plane is [๐‘Ÿ โƒ—โˆ’(1๐‘– ฬ‚+1๐‘— ฬ‚+0๐‘˜ ฬ‚ ) ].(โˆ’2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+3๐‘˜ ฬ‚ ) = 0 [๐’“ โƒ—โˆ’(๐’Š ฬ‚+๐’‹ ฬ‚ ) ].(โˆ’๐Ÿ๐’Š ฬ‚โˆ’๐Ÿ‘๐’‹ ฬ‚+๐Ÿ‘๐’Œ ฬ‚ ) = ๐ŸŽ Finding Cartesian equation Put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ [๐‘Ÿ โƒ—โˆ’(๐‘– ฬ‚+๐‘— ฬ‚ ) ].(โˆ’2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+3๐‘˜ ฬ‚ ) = 0 [(๐‘ฅ๐‘– ฬ‚+๐‘ฆ๐‘— ฬ‚+๐‘ง๐‘˜ ฬ‚ )โˆ’(๐‘– ฬ‚+๐‘— ฬ‚)]. (โˆ’2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+3๐‘˜ ฬ‚ ) = 0 [(๐‘ฅโˆ’1) ๐‘– ฬ‚ +(๐‘ฆโˆ’1) ๐‘— ฬ‚+๐‘ง๐‘˜ ฬ‚ ]. (โˆ’2๐‘– ฬ‚โˆ’3๐‘— ฬ‚+3๐‘˜ ฬ‚ ) = 0 โ€“2(x โˆ’ 1) + (โˆ’3)(y โˆ’ 1) + 3(z) = 0 โ€“2x + 2 โˆ’ 3y + 3 + 3z = 0 2x + 3y โ€“ 3z = 5 โˆด Equation of plane in Cartesian form is 2x + 3y โ€“ 3z = 5

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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.