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

Transcript

Misc 17 Find the equation of the plane which contains the line of intersection of the planes ๐‘Ÿ โƒ— . (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) โ€“ 4 = 0 , ๐‘Ÿ โƒ— . (2๐‘– ฬ‚ + ๐‘— ฬ‚ โ€“ ๐‘˜ ฬ‚) + 5 = 0 and which is perpendicular to the plane ๐‘Ÿ โƒ— . (5๐‘– ฬ‚ + 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚) + 8 = 0 .Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z โˆ’ d1) + ๐œ† (A2x + B2y + C2z โ€“ d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 ๐’“ โƒ—. (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 3๐’Œ ฬ‚) โˆ’ 4 = 0 ๐‘Ÿ โƒ—. (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) = 4 Putting ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚).(๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) = 4 (x ร— 1) + (y ร— 2) + (z ร— 3) = 4 1x + 2y + 3z = 4 Comparing with ๐ด_1 "x"+"B1y"+๐ถ_1 "z = d1" ๐ด_1 = 1, ๐ต_1= 2 , ๐ถ_1 = 3 , ๐‘‘_1 = 4 ๐’“ โƒ—. (2๐’Š ฬ‚ + ๐’‹ ฬ‚ โˆ’ ๐’Œ ฬ‚) + 5 = 0 ๐‘Ÿ โƒ—. (2๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) = โˆ’ 5 โˆ’๐‘Ÿ โƒ—. (2๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) = 5 ๐‘Ÿ โƒ—. ( โˆ’2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 Putting ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚, (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚).(-2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 (x ร—โˆ’ 2) + (Y ร— โˆ’ 1) + (z ร— 1) = 5 โˆ’2x โˆ’ 1y + 1z = 5 Comparing with ๐ด_2 "x"+ "B2y"+ ๐ถ_2 "z = d2" ๐ด_2 = โˆ’2, ๐ต_2= โˆ’1 , ๐ถ_2 = 1 , ๐‘‘_2 = 5 Equation of plane is (A1x + B1y + C1z โˆ’ d1) + ๐œ† (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z โˆ’ 4) + ๐œ† ( โˆ’ 2x โˆ’ 1y + 1z โˆ’ 5) = 0 (1 โˆ’ 2๐œ†) x + (2 โˆ’ ๐œ†)y + (3 + ๐œ†) z + ( โˆ’4 โˆ’ 5๐œ†) = 0 Now, the plane is perpendicular to the plane ๐‘Ÿ โƒ—.(5๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) + 8 = 0 So, normal to plane ๐‘ โƒ— will be perpendicular to normal ๐‘› โƒ— of ๐‘Ÿ โƒ—.(5๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) + 8 = 0 Now, ๐‘Ÿ โƒ—.(5๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) + 8 = 0 ๐‘Ÿ โƒ— .(5๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) = โ€“8 โˆ’ ๐‘Ÿ โƒ— .(5๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) = 8 ๐‘Ÿ โƒ— .( โˆ’5๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) = 8 Finding direction cosines of ๐‘ โƒ— & ๐‘› โƒ— Since, ๐‘ โƒ— is perpendicular to ๐‘› โƒ— ๐‘Ž1 ๐‘Ž2 + b1 b2 + c1 c2 = 0 (1 โˆ’ 2๐œ†) ร— โˆ’5 + (2 โˆ’ ๐œ†) ร— โˆ’3 + (3 + ๐œ†) ร— 6 = 0 Theory : Two lines with direction ratios ๐‘Ž1, b1, c1 and ๐‘Ž2, b2, c2 are perpendicular if ๐‘Ž1 ๐‘Ž2 + b1b2 + c1 c2 = 0 ๐‘ต โƒ— = (1 โˆ’ 2๐œ†) ๐’Š ฬ‚ + (2 โˆ’ ๐œ†) ๐’‹ ฬ‚ + (3 + ๐œ†) ๐’Œ ฬ‚ Direction ratios = 1 โˆ’ 2๐œ†, 2 โˆ’ ๐œ†, 3 + ๐œ† โˆด ๐‘Ž1 = 1 โˆ’ 2๐œ†, b1 = 2 โˆ’ ๐œ†, c1 = 3 + ๐œ† ๐’ โƒ— = โˆ’ 5๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 6๐’Œ ฬ‚ Direction ratios = โˆ’5, โˆ’3, 6 โˆด ๐‘Ž2 = โˆ’ 5, b2 = โˆ’3, c2 = 6, โˆ’ 5 + 10๐œ† โˆ’ 6 + 3๐œ† + 18 + 6๐œ† = 0 19๐œ† + 7 = 0 โˆด ๐œ† = (โˆ’๐Ÿ•)/๐Ÿ๐Ÿ— Putting value of ๐œ† in (1), (1 โˆ’ 2๐œ†) x + (2 โˆ’ ๐œ†)y + (3 + ๐œ†) z + ( โˆ’4 โˆ’ 5๐œ†) = 0 (1โˆ’2 ร—(โˆ’7)/19) x + (2โˆ’(( โˆ’7)/19)) y + (3+(( โˆ’ 7)/19)) z + ( โˆ’4โˆ’5ร—(โˆ’7)/19) = 0 (1 + 14/19) x + (2 + 7/19)y + (3 โˆ’ 7/19)z + ( โˆ’ 4 + 35/19) = 0 33/19 x + 45/19 y + 50/19 z โˆ’ 41/19 = 0 1/19 (33x + 45y + 50z โˆ’ 41) = 0 33x + 45y + 50z โˆ’ 41 = 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.