Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

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Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7

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Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8 Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 9

 

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

Transcript

Ex 11.3, 5 Find the vector and cartesian equations of the planes (b) That passes through the point (1, 4, 6) and the normal vector to the plane is ๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ Vector equation Equation of plane passing through point A whose position vector is ๐’‚ โƒ— & perpendicular to ๐’ โƒ— is (๐’“ โƒ— โˆ’ ๐’‚ โƒ—) . ๐’ โƒ— = 0 Given, Plane passes through (1, 4, 6) So, ๐’‚ โƒ— = 1๐‘– ฬ‚ + 4๐‘— ฬ‚ + 6๐‘˜ ฬ‚ Normal to plane = ๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐’ โƒ— = ๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ Vector equation of plane is (๐‘Ÿ โƒ— โˆ’ ๐‘Ž โƒ—).๐‘› โƒ— = 0 [๐’“ โƒ—โˆ’(๐’Š ฬ‚+๐Ÿ’๐’‹ ฬ‚+๐Ÿ”๐’Œ ฬ‚)] . (๐’Š ฬ‚ โˆ’ 2๐’‹ ฬ‚ + ๐’Œ ฬ‚) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) : Vector equation is [๐‘Ÿ โƒ—โˆ’(๐‘– ฬ‚+4๐‘— ฬ‚+6๐‘˜ ฬ‚)] . (๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 0 Put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ [(๐‘ฅ๐‘– ฬ‚+๐‘ฆ๐‘— ฬ‚+๐‘ง๐‘˜ ฬ‚ )โˆ’(1๐‘– ฬ‚+4๐‘— ฬ‚+6๐‘˜ ฬ‚)]. (1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 0 [(๐‘ฅโˆ’1) ๐‘– ฬ‚ +(๐‘ฆโˆ’4) ๐‘— ฬ‚+(๐‘งโˆ’6)๐‘˜ ฬ‚ ]. (1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 0 1(x โˆ’ 1) + (โˆ’2)(y โˆ’ 4) + 1 (z โˆ’ 6) = 0 x โˆ’ 1 โˆ’ 2(y โˆ’ 4) + z โˆ’ 6 = 0 x โˆ’ 2y + z + 1 = 0 โˆด Equation of plane in Cartesian form is x โˆ’ 2y + z + 1 = 0. Cartesian form (Method 2) Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x โˆ’ x1) + B(y โˆ’ y1) + C (z โˆ’ z1) = 0 Since the plane passes through (1, 4, 6) x1 = 1, y1 = 4, z1 = 6 And normal is ๐‘– ฬ‚ โ€“ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ So, Direction ratios of line perpendicular to plane = 1, โ€“2, 1 โˆด A = 1, B = โ€“2, C = 1 Therefore, equation of line in Cartesian form is 1(x โˆ’ 1) โ€“ 2 (y โˆ’ 4) + 1(x โˆ’ 6) = 0 x โˆ’ 2y + z + 1 = 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 10 years. He provides courses for Maths and Science at Teachoo.