Answer on Question #44082 – Math – Vector Calculus
Prove that (a+3b)×(a+b)+(3a−5b)×(a−b)=0
Solution.
(a+3b)×(a+b)=(a×a)+3(b×a)+3(b×b)+(a×b);(3a−5b)×(a−b)=3(a×a)−3(a×b)−5(b×a)+5(b×b).
Since, (a×a)=0 and (b×b)=0, we obtain
(a+3b)×(a+b)+(3a−5b)×(a−b)=3(b×a)+(a×b)−3(a×b)−5(b×a).
As
(a×b)=−(b×a),
we have
(a+3b)×(a+b)+(3a−5b)×(a−b)=−3(a×b)+(a×b)−3(a×b)+5(a×b)=0.
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