Ex 10.2, 9 - For a = 2i - j + 2k, and b = -i + j - k, find unit vector

Ex 10.2, 9 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.2, 9 For given vectors, š‘Ž āƒ— = 2š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 2š‘˜ Ģ‚ and š‘ āƒ— = āˆ’š‘– Ģ‚ + š‘— Ģ‚ āˆ’ š‘˜ Ģ‚ , find the unit vector in the direction of the vector š‘Ž āƒ— + š‘ āƒ—š‘Ž āƒ— = 2š‘– Ģ‚ āˆ’ j Ģ‚ + 2š‘˜ Ģ‚ = 2š‘– Ģ‚ – 1š‘— Ģ‚ + 2š‘˜ Ģ‚ š‘ āƒ— = āˆ’š‘– Ģ‚ + š‘— Ģ‚ – š‘˜ Ģ‚ = āˆ’1š‘– Ģ‚ + 1š‘— Ģ‚ – 1š‘˜ Ģ‚ Now, (š‘Ž āƒ— + š‘ āƒ—) = (2 – 1) š‘– Ģ‚ + (-1 + 1) š‘— Ģ‚ + (2 – 1) š‘˜ Ģ‚ = 1š‘– Ģ‚ + 0š‘— Ģ‚ + 1š‘˜ Ģ‚ Let š‘ āƒ— = š‘Ž āƒ— + š‘ āƒ— ∓ c āƒ— = 1š‘– Ģ‚ + 0š‘— Ģ‚ + 1š‘˜ Ģ‚ Magnitude of š‘ āƒ— = √(12+02+12) |š‘ āƒ— | = √(1+0+1) = √2 Unit vector in direction of š‘ āƒ— = 1/|š‘ āƒ— | . š‘ āƒ— š‘ Ģ‚ = 1/√2 [1š‘– Ģ‚+0š‘— Ģ‚+1š‘˜ Ģ‚ ] š‘ Ģ‚ = 1/√2 š‘– Ģ‚ + 0š‘— Ģ‚ + 1/√2 š‘˜ Ģ‚ š‘ Ģ‚ = šŸ/āˆššŸ š’Š Ģ‚ + šŸ/āˆššŸ š’Œ Ģ‚ Thus, unit vector in direction of š‘ āƒ— = 1/√2 š‘– Ģ‚ + 1/√2 š‘˜ Ģ‚

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