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Question

Let a,b and c be three vectors such that |a|=3b=4|c|=5 and each one of them being perpendicular to the sum of the other two find a+b+c.

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Solution

|a|=3,b=4,|c|=5a(b+c)=0b(a+c)=0c(a+b)=0ab+ac=0ba+bc=0ca+cb=0ab+bc+ca=0a+b+c2=|a|2+|b|2+|c|2+2(ab+bc+ca)=9+16+25+2×0=50a+b+c=52

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