Slide13.JPG Slide14.JPG Slide15.JPG

Slide16.JPG Slide17.JPG Slide18.JPG Slide19.JPG

You saved atleast 2 minutes of distracting ads by going ad-free. Thank you :)

You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.


Transcript

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‚)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo