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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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

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

 

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Transcript

Question 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.