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Misc 20 - Find vector equation of line passing through (1, 2, -4) and

Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6
Misc 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7

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Misc 20 (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‘Ž āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚ Given, line is perpendicular to both lines āˆ“ š‘ āƒ— is perpendicular to both lines We know that š‘Ž āƒ— Ɨ š‘ āƒ— is perpendicular to both š‘Ž āƒ— & š‘ āƒ— So, š‘ 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‘– Ģ‚ + 36š‘— Ģ‚ + 72š‘˜ Ģ‚ Thus, š‘ āƒ— = 24š‘– Ģ‚ + 36š‘— Ģ‚ + 72š‘˜ 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‘– Ģ‚ + 2š‘— Ģ‚ ā€“ 4š‘˜ Ģ‚) + šœ† (24š‘– Ģ‚ + 36š‘— Ģ‚ + 72š‘˜ 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 20 (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‘Ž āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚ Let š‘ āƒ— = xš‘– Ģ‚ + yš‘— Ģ‚ + zš‘˜ 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‘ āƒ— = xš‘– Ģ‚ + yš‘— Ģ‚ + zš‘˜ Ģ‚ = 2kš‘– Ģ‚ + 3kš‘— Ģ‚ + 6kš‘˜ 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.