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

Transcript

Misc 20 (Method 1) Find the vector equation of the line passing through the point (1, 2, โ€“4) and perpendicular to the two lines: (๐‘ฅ โˆ’ 8)/3 = (๐‘ฆ + 19)/(โˆ’16) = (๐‘ง โˆ’ 10)/7 and (๐‘ฅ โˆ’ 15)/3 = (๐‘ฆ โˆ’ 29)/8 = (๐‘ง โˆ’ 5)/(โˆ’5) The vector equation of a line passing through a point with position vector ๐‘Ž โƒ— and parallel to a vector ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†๐’ƒ โƒ— The line passes through (1, 2, โˆ’4) So, ๐‘Ž โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ Given, line is perpendicular to both lines โˆด ๐‘ โƒ— is perpendicular to both lines We know that ๐‘Ž โƒ— ร— ๐‘ โƒ— is perpendicular to both ๐‘Ž โƒ— & ๐‘ โƒ— So, ๐‘ โƒ— is cross product of both lines (๐‘ฅ โˆ’ 8)/3 = (๐‘ฆ + 19)/(โˆ’16) = (๐‘ง โˆ’ 10)/7 and (๐‘ฅ โˆ’ 15)/3 = (๐‘ฆ โˆ’ 29)/8 = (๐‘ง โˆ’ 5)/(โˆ’5) Required normal = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@3&โˆ’16&7@3&8&โˆ’5)| = ๐‘– ฬ‚ (โ€“16(โˆ’5) โ€“ 8(7)) โ€“ ๐‘— ฬ‚ (3(-5) โ€“ 3(7)) + ๐‘˜ ฬ‚(3(8) โ€“ 3(โ€“16)) = ๐‘– ฬ‚ (80 โ€“ 56) โ€“ ๐‘— ฬ‚ (โ€“15 โ€“ 21) + ๐‘˜ ฬ‚(24 + 48) = 24๐‘– ฬ‚ + 36๐‘— ฬ‚ + 72๐‘˜ ฬ‚ Thus, ๐‘ โƒ— = 24๐‘– ฬ‚ + 36๐‘— ฬ‚ + 72๐‘˜ ฬ‚ Now, Putting value of ๐‘Ž โƒ— & ๐‘ โƒ— in formula ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ†๐‘ โƒ— โˆด ๐‘Ÿ โƒ— = (1๐‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚) + ๐œ† (24๐‘– ฬ‚ + 36๐‘— ฬ‚ + 72๐‘˜ ฬ‚) = (๐‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚) + ๐œ†12 (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) = (๐‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚) + ๐œ† (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) Therefore, the equation of the line is (๐’Š ฬ‚ + 2๐’‹ ฬ‚ โ€“ 4๐’Œ ฬ‚) + ๐œ† (2๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 6๐’Œ ฬ‚). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, โ€“4) and perpendicular to the two lines: (๐‘ฅ โˆ’ 8)/3 = (๐‘ฆ + 19)/(โˆ’16) = (๐‘ง โˆ’ 10)/7 and (๐‘ฅ โˆ’ 15)/3 = (๐‘ฆ โˆ’ 29)/8 = (๐‘ง โˆ’ 5)/(โˆ’5) The vector equation of a line passing through a point with position vector ๐‘Ž โƒ— and parallel to a vector ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†๐’ƒ โƒ— The line passes through (1, 2, โˆ’4) So, ๐‘Ž โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ Let ๐‘ โƒ— = x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ Two lines with direction ratios ๐‘Ž1 , ๐‘1 , ๐‘1 & ๐‘Ž2 , ๐‘2 , ๐‘2 are perpendicular if ๐’‚๐Ÿ ๐’‚๐Ÿ + ๐’ƒ๐Ÿ๐’ƒ๐Ÿ + ๐’„๐Ÿ ๐’„๐Ÿ = 0 Given, line ๐‘ โƒ— is perpendicular to (๐‘ฅ โˆ’ 8)/3 = (๐‘ฆ + 19)/16 = (๐‘ง โˆ’ 10)/7 and (๐‘ฅ โˆ’ 15)/3 = (๐‘ฆ โˆ’ 29)/8 = (๐‘ง โˆ’ 5)/( โˆ’ 5) So, 3x โˆ’ 16y + 7z = 0 and 3x + 8y โˆ’ 5z = 0 ๐‘ฅ/(80 โˆ’ 56 ) = ๐‘ฆ/(21 โˆ’ ( โˆ’15) ) = ๐‘ง/(24 โˆ’ ( โˆ’48) ) ๐‘ฅ/(24 ) = ๐‘ฆ/36 = ๐‘ง/72 ๐‘ฅ/2 = ๐‘ฆ/3 = ๐‘ง/6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, ๐‘ โƒ— = x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ = 2k๐‘– ฬ‚ + 3k๐‘— ฬ‚ + 6k๐‘˜ ฬ‚ Now, Putting value of ๐‘Ž โƒ— & ๐‘ โƒ— in formula ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ†๐‘ โƒ— โˆด ๐‘Ÿ โƒ— = (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) + ๐œ† (2k๐‘– ฬ‚ + 3k๐‘— ฬ‚ + 6k๐‘˜ ฬ‚) = (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) + ๐œ†k (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) = (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) + ๐œ† (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 6๐‘˜ ฬ‚) Therefore, the equation of the line is (๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚) + ๐œ†(2๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 6๐’Œ ฬ‚)

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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.