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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Transcript

Misc 15 Prove that (๐‘Ž โƒ— + ๐‘ โƒ—) โ‹… (๐‘Ž โƒ— + ๐‘ โƒ—) =|๐‘Ž โƒ—|2 + |๐‘ โƒ—|2 , if and only if ๐‘Ž โƒ—, ๐‘ โƒ— are perpendicular, given ๐‘Ž โƒ— โ‰  0 โƒ—, ๐‘ โƒ— โ‰  0 โƒ— (๐‘Ž โƒ— + ๐‘ โƒ—) โ‹… (๐‘Ž โƒ— + ๐‘ โƒ—) = ๐‘Ž โƒ— . ๐‘Ž โƒ— + ๐‘Ž โƒ— . ๐‘ โƒ— + ๐’ƒ โƒ— . ๐’‚ โƒ— + ๐‘ โƒ— . ๐‘ โƒ— = ๐‘Ž โƒ— . ๐‘Ž โƒ— + ๐‘Ž โƒ— . ๐‘ โƒ— + ๐’‚ โƒ— . ๐’ƒ โƒ— + ๐‘ โƒ— . ๐‘ โƒ— = ๐’‚ โƒ— . ๐’‚ โƒ— + 2๐‘Ž โƒ— . ๐‘ โƒ— + ๐’ƒ โƒ— . ๐’ƒ โƒ— =|๐’‚ โƒ—|2 + 2๐‘Ž โƒ— . ๐‘ โƒ— + |๐’ƒ โƒ—|2 Since ๐‘Ž โƒ— and ๐‘ โƒ— are perpendicular, ๐’‚ โƒ— . ๐’ƒ โƒ— = 0 (Using prop: ๐‘Ž โƒ—.๐‘ โƒ— = ๐‘ โƒ—.๐‘Ž โƒ—) (Using prop: ๐‘Ž โƒ—.๐‘Ž โƒ— =|๐‘Ž โƒ— |^2) Putting ๐‘Ž โƒ— . ๐‘ โƒ— = 0 in (1) (๐’‚ โƒ— + ๐’ƒ โƒ—) . (๐’‚ โƒ— + ๐’ƒ โƒ—) = |๐‘Ž โƒ—|2 + 2.(0) + |๐‘ โƒ—|2 = |๐’‚ โƒ—|2 + |๐’ƒ โƒ—|2 Hence proved

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