Misc 17 - Plane which contains line of intersection of planes

Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

  1. Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
  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 10 years. He provides courses for Maths and Science at Teachoo.