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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 5 (Method 1) Find the vector equation of the line passing through the point (1, 2, ā€“4) and perpendicular to the two lines: (š‘„ āˆ’ 8)/3 = (š‘¦ + 19)/(āˆ’16) = (š‘§ āˆ’ 10)/7 and (š‘„ āˆ’ 15)/3 = (š‘¦ āˆ’ 29)/8 = (š‘§ āˆ’ 5)/(āˆ’5) The vector equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to a vector š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†š’ƒ āƒ— The line passes through (1, 2, āˆ’4) So, š’‚ āƒ— = 1š’Š Ģ‚ + 2š’‹ Ģ‚ āˆ’ 4š’Œ Ģ‚ Given, line is perpendicular to both lines āˆ“ š‘ āƒ— is perpendicular to both lines We know that š‘„ āƒ— Ɨ š‘¦ āƒ— is perpendicular to both š‘„ āƒ— & š‘¦ āƒ— So, š’ƒ āƒ— is cross product of both lines (š‘„ āˆ’ 8)/3 = (š‘¦ + 19)/(āˆ’16) = (š‘§ āˆ’ 10)/7 and (š‘„ āˆ’ 15)/3 = (š‘¦ āˆ’ 29)/8 = (š‘§ āˆ’ 5)/(āˆ’5) Required normal = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@3&āˆ’16&7@3&8&āˆ’5)| = š‘– Ģ‚ (ā€“16(āˆ’5) ā€“ 8(7)) ā€“ š‘— Ģ‚ (3(-5) ā€“ 3(7)) + š‘˜ Ģ‚(3(8) ā€“ 3(ā€“16)) = š‘– Ģ‚ (80 ā€“ 56) ā€“ š‘— Ģ‚ (ā€“15 ā€“ 21) + š‘˜ Ģ‚(24 + 48) = 24š’Š Ģ‚ + 36š’‹ Ģ‚ + 72š’Œ Ģ‚ Thus, š’ƒ āƒ— = 24š’Š Ģ‚ + 36š’‹ Ģ‚ + 72š’Œ Ģ‚ Now, Putting value of š‘Ž āƒ— & š‘ āƒ— in formula š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— āˆ“ š‘Ÿ āƒ— = (1š’Š Ģ‚ + 2š’‹ Ģ‚ ā€“ 4š’Œ Ģ‚) + šœ† (24š’Š Ģ‚ + 36š’‹ Ģ‚ + 72š’Œ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ ā€“ 4š‘˜ Ģ‚) + šœ†12 (2š‘– Ģ‚ + 3š‘— Ģ‚ + 6š‘˜ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ ā€“ 4š‘˜ Ģ‚) + šœ† (2š‘– Ģ‚ + 3š‘— Ģ‚ + 6š‘˜ Ģ‚) Therefore, the equation of the line is (š’Š Ģ‚ + 2š’‹ Ģ‚ ā€“ 4š’Œ Ģ‚) + šœ† (2š’Š Ģ‚ + 3š’‹ Ģ‚ + 6š’Œ Ģ‚). Misc 5 (Method 2) Find the vector equation of the line passing through the point (1, 2, ā€“4) and perpendicular to the two lines: (š‘„ āˆ’ 8)/3 = (š‘¦ + 19)/(āˆ’16) = (š‘§ āˆ’ 10)/7 and (š‘„ āˆ’ 15)/3 = (š‘¦ āˆ’ 29)/8 = (š‘§ āˆ’ 5)/(āˆ’5) The vector equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to a vector š‘ āƒ— is š’“ āƒ— = š’‚ āƒ— + šœ†š’ƒ āƒ— The line passes through (1, 2, āˆ’4) So, š’‚ āƒ— = 1š’Š Ģ‚ + 2š’‹ Ģ‚ āˆ’ 4š’Œ Ģ‚ Let š’ƒ āƒ— = xš’Š Ģ‚ + yš’‹ Ģ‚ + zš’Œ Ģ‚ Two lines with direction ratios š‘Ž1 , š‘1 , š‘1 & š‘Ž2 , š‘2 , š‘2 are perpendicular if š’‚šŸ š’‚šŸ + š’ƒšŸš’ƒšŸ + š’„šŸ š’„šŸ = 0 Given, line š‘ āƒ— is perpendicular to (š‘„ āˆ’ 8)/3 = (š‘¦ + 19)/16 = (š‘§ āˆ’ 10)/7 and (š‘„ āˆ’ 15)/3 = (š‘¦ āˆ’ 29)/8 = (š‘§ āˆ’ 5)/( āˆ’ 5) So, 3x āˆ’ 16y + 7z = 0 and 3x + 8y āˆ’ 5z = 0 š‘„/(80 āˆ’ 56 ) = š‘¦/(21 āˆ’ ( āˆ’15) ) = š‘§/(24 āˆ’ ( āˆ’48) ) š‘„/(24 ) = š‘¦/36 = š‘§/72 š‘„/2 = š‘¦/3 = š‘§/6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, š’ƒ āƒ— = xš’Š Ģ‚ + yš’‹ Ģ‚ + zš’Œ Ģ‚ = 2kš’Š Ģ‚ + 3kš’‹ Ģ‚ + 6kš’Œ Ģ‚ Now, Putting value of š‘Ž āƒ— & š‘ āƒ— in formula š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— āˆ“ š‘Ÿ āƒ— = (š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚) + šœ† (2kš‘– Ģ‚ + 3kš‘— Ģ‚ + 6kš‘˜ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚) + šœ†k (2š‘– Ģ‚ + 3š‘— Ģ‚ + 6š‘˜ Ģ‚) = (š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚) + šœ† (2š‘– Ģ‚ + 3š‘— Ģ‚ + 6š‘˜ Ģ‚) Therefore, the equation of the line is (š’Š Ģ‚ + 2š’‹ Ģ‚ āˆ’ 4š’Œ Ģ‚) + šœ†(2š’Š Ģ‚ + 3š’‹ Ģ‚ + 6š’Œ Ģ‚)

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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.