Misc 6 - Find a vector of magnitude 5 units, parallel to

Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 3

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Misc 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors š‘Ž āƒ— = 2š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ š‘˜ Ģ‚ and š‘ āƒ— = š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + š‘˜ Ģ‚.Given š‘Ž āƒ— = 2š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ š‘˜ Ģ‚(, š‘) āƒ— = š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + š‘˜ Ģ‚ Resultant of š’‚ āƒ— & š’ƒ āƒ— = š’‚ āƒ— + š’ƒ āƒ— (š’‚ āƒ— + š’ƒ āƒ—) = (2 + 1)š‘– Ģ‚ + (3 āˆ’ 2)š‘— Ģ‚ + (āˆ’1 + 1)š‘˜ Ģ‚ = 3š’Š Ģ‚ + 1š’‹ Ģ‚ + 0š’Œ Ģ‚ Let š’„ āƒ— = (š’‚ āƒ— + š’ƒ āƒ—) ∓ š‘ āƒ— = 3š‘– Ģ‚ + 1š‘— Ģ‚ + 0š‘˜ Ģ‚ Magnitude of š‘ āƒ— = √(32+12+02) |š‘ āƒ— | = √(9+1) = √10 Unit vector in direction of š‘ āƒ— = 1/|š‘ āƒ— | Ɨ š‘ āƒ— š‘ Ģ‚ = 1/√10 Ɨ [3š‘– Ģ‚+1š‘— Ģ‚+0š‘˜ Ģ‚ ] š’„ Ģ‚ = šŸ‘/āˆššŸšŸŽ š’Š Ģ‚ + šŸ/āˆššŸšŸŽ š’‹ Ģ‚ + 0š’Œ Ģ‚ Vector with magnitude 1 = 3/√10 š‘– Ģ‚ + 1/√10 š‘— Ģ‚ + 0š‘˜ Ģ‚ Vector with magnitude 5 = 5 Ɨ [3/√10 " " š‘– Ģ‚" + " 1/√10 š‘— Ģ‚" + 0" š‘˜ Ģ‚ ] = 15/√10 š‘– Ģ‚ + 5/√10 š‘— Ģ‚ + 0š‘˜ Ģ‚ = 15/√10 š‘– Ģ‚ + 5/√10 š‘— Ģ‚ Rationalizing = 15/√10 Ɨ √10/√10 š‘– Ģ‚ + 5/√10 "Ɨ " √10/√10 š‘— Ģ‚ = (15√10)/10 š‘– Ģ‚ + (5√10)/10 š‘— Ģ‚ = (šŸ‘āˆššŸšŸŽ)/šŸ š’Š Ģ‚ + āˆššŸšŸŽ/šŸ š’‹ Ģ‚ Hence the required vector is (šŸ‘āˆššŸšŸŽ)/šŸ š’Š Ģ‚ + āˆššŸšŸŽ/šŸ š’‹ Ģ‚

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