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

Transcript

Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, โ€“2) and the normal to the plane is ๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [๐‘Ÿ โƒ— โˆ’(๐‘ฅ1๐‘– ฬ‚+๐‘ฆ1๐‘— ฬ‚+๐‘ง1๐‘˜ ฬ‚)]. (A๐‘– ฬ‚ + B๐‘— ฬ‚ + C๐‘˜ ฬ‚) = 0 or (๐‘Ÿ โƒ— โˆ’ ๐‘Ž โƒ—).๐‘› โƒ— = 0 ("A" ๐‘ƒ) โƒ— is perpendicular to "n" โƒ— So, ("A" P) โƒ— . "n" โƒ— = 0 ("r" โƒ— โˆ’ "a" โƒ—)."n" โƒ— = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, โ€“ 2) and the normal to the plane is ๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ 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, 0, โˆ’2) So, ๐’‚ โƒ— = 1๐‘– ฬ‚ + 0๐‘— ฬ‚ โ€“ 2๐‘˜ ฬ‚ Normal to plane = ๐‘– ฬ‚ + ๐‘— ฬ‚ โ€“ ๐‘˜ ฬ‚ ๐’ โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚ Vector equation of plane is (๐’“ โƒ— โˆ’ ๐’‚ โƒ—) . ๐’ โƒ— = 0 [๐‘Ÿ โƒ—โˆ’("1" ๐‘– ฬ‚+"0" ๐‘— ฬ‚โˆ’"2" ๐‘˜ ฬ‚)].(๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) = 0 [๐’“ โƒ—โˆ’(๐’Š ฬ‚โˆ’"2" ๐’Œ ฬ‚)] . (๐’Š ฬ‚+๐’‹ ฬ‚โˆ’๐’Œ ฬ‚) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [๐‘Ÿ โƒ—โˆ’(๐‘– ฬ‚โˆ’"2" ๐‘˜ ฬ‚)] . (๐‘– ฬ‚+๐‘— ฬ‚โˆ’๐‘˜ ฬ‚) = 0 Put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ [("x" ๐‘– ฬ‚+"y" ๐‘— ฬ‚+"z" ๐‘˜ ฬ‚ )โˆ’("1" ๐‘– ฬ‚+"0" ๐‘— ฬ‚โˆ’"2" ๐‘˜ ฬ‚)].(1๐‘– ฬ‚ + 1๐‘— ฬ‚ โ€“ 1๐‘˜ ฬ‚) = 0 [("x" โˆ’1) ๐‘– ฬ‚+("y" โˆ’0) ๐ฝ ฬ‚+(๐‘งโˆ’(โˆ’2))๐‘˜ ฬ‚ ].(1๐‘– ฬ‚ + 1๐‘— ฬ‚ โ€“ 1๐‘˜ ฬ‚) = 0 1(x โˆ’ 1) + 1(y โˆ’ 0) โ€“ 1 (z + 2) = 0 x โˆ’ 1 + y โ€“ z โ€“ 2 = 0 x + y โ€“ z = 3 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, 0, โ€“2) x1 = 1, y1 = 0, z1 = 2 And normal is ๐‘– ฬ‚ + ๐‘— ฬ‚ โ€“ ๐‘˜ ฬ‚ So, Direction ratios of line perpendicular to plane = 1, 1, โ€“1 โˆด A = 1, B = 1, C = โ€“1 Therefore, equation of line in Cartesian form is 1(x โˆ’ 1) + 1(y โˆ’ 0) โˆ’ 1(z โˆ’ (โˆ’2)) = 0 x + y โ€“ z = 3 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 9 years. He provides courses for Maths and Science at Teachoo.