Misc 15 - Prove that vectors (a + b) . (a + b) = |a|^2 + |b|^2

Misc 15 - Chapter 10 Class 12 Vector Algebra - Part 2

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Misc 15 Prove that (š‘Ž āƒ— + š‘ āƒ—) ā‹… (š‘Ž āƒ— + š‘ āƒ—) =|š‘Ž āƒ—|2 + |š‘ āƒ—|2 , if and only if š‘Ž āƒ—, š‘ āƒ— are perpendicular, given š‘Ž āƒ— ≠ 0 āƒ—, š‘ āƒ— ≠ 0 āƒ— (š‘Ž āƒ— + š‘ āƒ—) ā‹… (š‘Ž āƒ— + š‘ āƒ—) = š‘Ž āƒ— . š‘Ž āƒ— + š‘Ž āƒ— . š‘ āƒ— + š’ƒ āƒ— . š’‚ āƒ— + š‘ āƒ— . š‘ āƒ— = š‘Ž āƒ— . š‘Ž āƒ— + š‘Ž āƒ— . š‘ āƒ— + š’‚ āƒ— . š’ƒ āƒ— + š‘ āƒ— . š‘ āƒ— = š’‚ āƒ— . š’‚ āƒ— + 2š‘Ž āƒ— . š‘ āƒ— + š’ƒ āƒ— . š’ƒ āƒ— =|š’‚ āƒ—|2 + 2š‘Ž āƒ— . š‘ āƒ— + |š’ƒ āƒ—|2 Since š‘Ž āƒ— and š‘ āƒ— are perpendicular, š’‚ āƒ— . š’ƒ āƒ— = 0 (Using prop: š‘Ž āƒ—.š‘ āƒ— = š‘ āƒ—.š‘Ž āƒ—) (Using prop: š‘Ž āƒ—.š‘Ž āƒ— =|š‘Ž āƒ— |^2) Putting š‘Ž āƒ— . š‘ āƒ— = 0 in (1) (š’‚ āƒ— + š’ƒ āƒ—) . (š’‚ āƒ— + š’ƒ āƒ—) = |š‘Ž āƒ—|2 + 2.(0) + |š‘ āƒ—|2 = |š’‚ āƒ—|2 + |š’ƒ āƒ—|2 Hence proved

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