Ex 10.3, 10 - If a, b, c are such, a + b is perpendicular to c

Ex 10.3, 10 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.3, 10 If š‘Ž āƒ— = 2š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚, š‘ āƒ— = āˆ’š‘– Ģ‚ + 2š‘— Ģ‚ + š‘˜ Ģ‚ and š‘ āƒ— = 3š‘– Ģ‚ + š‘— Ģ‚ are such that š‘Ž āƒ— +šœ†š‘ āƒ— is perpendicular to š‘ āƒ— , then find the value of šœ†.š‘Ž āƒ— = 2š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ š‘ āƒ— = āˆ’š‘– Ģ‚ + 2š‘— Ģ‚ + š‘˜ Ģ‚ = āˆ’1š‘– Ģ‚ + 2š‘— Ģ‚ + 1š‘˜ Ģ‚ š‘ āƒ— = 3š‘– Ģ‚ + š‘— Ģ‚ = 3š‘– Ģ‚ + 1š‘— Ģ‚ + 0š‘˜ Ģ‚ Now, (š‘Ž āƒ— + šœ†š‘ āƒ—) = (2š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚) + šœ† (-1š‘– Ģ‚ + 2š‘— Ģ‚ + 1š‘˜ Ģ‚) = 2š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ āˆ’ šœ†š‘– Ģ‚ + 2šœ†š‘— Ģ‚ + šœ†š‘˜ Ģ‚ = (2 āˆ’ šœ†) š‘– Ģ‚ + (2 + 2šœ†) š‘— Ģ‚ + (3 + šœ†) š‘˜ Ģ‚ Since (š‘Ž āƒ— + šœ†š‘ āƒ—) is perpendicular to š‘ āƒ— (š‘Ž āƒ— + šœ†š‘ āƒ—). š‘ āƒ— = 0 [(2āˆ’šœ†) š‘– Ģ‚+(2+2šœ†) š‘— Ģ‚+(3+šœ†)š‘˜ Ģ‚ ] . (3š‘– Ģ‚ + 1š‘— Ģ‚ + 0š‘˜ Ģ‚) = 0 (2 āˆ’ šœ†).3 + (2 + 2šœ†).1 + (3 + šœ† ).0 = 0 3.2 āˆ’ 3šœ† + 2 + 2šœ† + 0 = 0 6 – 3šœ† + 2 + 2šœ† = 0 8 āˆ’ šœ† = 0 šœ† = 8 ∓ šœ† = 8 (Dot product of perpendicular vectors is 0)

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