Misc 7 - Find unit vector parallel to vector 2a - b + 3c

Misc 7 - Chapter 10 Class 12 Vector Algebra - Part 2

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Misc 7 If š‘Ž āƒ— = š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚, š‘ āƒ— = 2š‘– Ģ‚ āˆ’š‘— Ģ‚ + 3š‘˜ Ģ‚ and š‘ āƒ— = š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + š‘˜ Ģ‚ , find a unit vector parallel to the vector 2š‘Ž āƒ— – š‘ āƒ— + 3š‘ āƒ— . Given š‘Ž āƒ— = š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚ š‘ āƒ— = 2š‘– Ģ‚ + š‘— Ģ‚ + 3š‘˜ Ģ‚ š‘ āƒ— = š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + š‘˜ Ģ‚ Let š’“ āƒ— = 2š’‚ āƒ— āˆ’ š’ƒ āƒ— + 3š’„ āƒ— = 2(š‘– Ģ‚ + š‘— Ģ‚ + š‘˜ Ģ‚) āˆ’ (2š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 3š‘˜ Ģ‚) + 3(š‘– Ģ‚ āˆ’2š‘— Ģ‚ + š‘˜ Ģ‚) = 2š‘– Ģ‚ + 2š‘— Ģ‚ + 2š‘˜ Ģ‚ āˆ’ 2š‘– Ģ‚ + 1š‘— Ģ‚ āˆ’ 3š‘˜ Ģ‚ + 3š‘– Ģ‚ āˆ’ 6š‘— Ģ‚ + 3š‘˜ Ģ‚ = (2 āˆ’ 2 + 3) š‘– Ģ‚ + (2 + 1 āˆ’ 6) š‘— Ģ‚ + (2 āˆ’ 3 + 3) š‘˜ Ģ‚ = 3š’Š Ģ‚ – 3š’‹ Ģ‚ + 2š’Œ Ģ‚ ∓ š’“ āƒ— = 3š’Š Ģ‚ – 3š’‹ Ģ‚ + 2š’Œ Ģ‚ Magnitude of š‘Ÿ āƒ— = √(32+(āˆ’3)2+22) |š’“ āƒ— | = √(9+9+4) = āˆššŸšŸ Unit vector in the direction of š‘Ÿ āƒ— = šŸ/|š’“ āƒ— | x š’“ āƒ— = 1/√22 Ɨ [3š‘– Ģ‚ āˆ’3š‘— Ģ‚+2š‘˜ Ģ‚ ] = 3/√22 š‘– Ģ‚ – 3/√22 š‘— Ģ‚ + 2/√22 š‘˜ Ģ‚ Hence the required vector is šŸ‘/āˆššŸšŸ š’Š Ģ‚ – šŸ‘/āˆššŸšŸ š’‹ Ģ‚ + šŸ/āˆššŸšŸ š’Œ Ģ‚

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