Misc 13 - Scalar product of vector i + j + k with unit vector

Misc 13 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 13 - Chapter 10 Class 12 Vector Algebra - Part 3

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Misc 13 The scalar product of the vector š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚ with a unit vector along the sum of vectors 2š‘– Ģ‚ + 4š‘— Ģ‚ āˆ’ 5š‘˜ Ģ‚ and Ī»š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ is equal to one. Find the value of Ī». Let š’‚ āƒ— = š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚ š’ƒ āƒ— = 2š‘– Ģ‚ + 4š‘— Ģ‚ – 5š‘˜ Ģ‚ š’„ āƒ— = šœ† š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ (š’ƒ āƒ— + š’„ āƒ—) = (2 + šœ†) š‘– Ģ‚ + (4 + 2) š‘— Ģ‚ + (āˆ’5 + 3) š‘˜ Ģ‚ = (2 + šœ†) š’Š Ģ‚ + 6š’‹ Ģ‚ āˆ’ 2š’Œ Ģ‚ Let š’“ Ģ‚ be unit vector along (š‘ āƒ— + š‘ āƒ—) š‘Ÿ Ģ‚ = 1/(š‘€š‘Žš‘”š‘›š‘–š‘”š‘¢š‘‘š‘’ š‘œš‘“ (š‘ āƒ—" + " š‘ āƒ—)) Ɨ (š‘ āƒ— + š‘ āƒ—) š‘Ÿ Ģ‚ = 1/√((2 + šœ†)^2 + 6^2 + (āˆ’2)^2 ) Ɨ ((2 + šœ†) š‘– Ģ‚ + 6š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) š‘Ÿ Ģ‚ = 1/√(2^2 + šœ†^2 + 4šœ† + 36 + 4) Ɨ ((2 + šœ†) š‘– Ģ‚ + 6š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) š’“ Ģ‚ = šŸ/√(š€^šŸ + šŸ’š€ +šŸ’šŸ’) Ɨ ((2 + šœ†) š’Š Ģ‚ + 6š’‹ Ģ‚ āˆ’ 2š’Œ Ģ‚) Given, š’‚ āƒ—. (š’“ Ģ‚) = 1 (1š‘– Ģ‚ + 1š‘— Ģ‚ + 1š‘˜ Ģ‚). (1/√(šœ†^2 + 4šœ† +44) " Ɨ ((2 + šœ†) " š‘– Ģ‚" + 6" š‘— Ģ‚" āˆ’ 2" š‘˜ Ģ‚")" ) = 1 1/√(šœ†^2 + 4šœ† +44) (1š‘– Ģ‚ + 1š‘— Ģ‚ + 1š‘˜ Ģ‚).((šœ† +2) š‘– Ģ‚ + 6š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) = 1 (1š‘– Ģ‚ + 1š‘— Ģ‚ + 1š‘˜ Ģ‚).((šœ† +2) š‘– Ģ‚ + 6š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) = √(šœ†^2 + 4šœ† +44) 1.(šœ† + 2) + 1.6 + 1.(āˆ’2) = √(šœ†^2 + 4šœ† +44) šœ† + 2 + 6 āˆ’ 2 = √(šœ†^2 + 4šœ† +44) šœ† + 6 = √(š€^šŸ + šŸ’š€ +šŸ’šŸ’) Squaring both sides (šœ† + 6)2 = (√(šœ†^2 + 4šœ† +44))^2 šœ†2 + 36 + 12šœ† = šœ†^2 + 4šœ† +44 8šœ† = 8 šœ† = 8/8 šœ† = 1 So, šœ† = 1

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