Let & be the angle between B und A. We have to show that B - |B|cos & A/|A| is normal to A. Using the dot product we must have (B - |B|cos & A/|A| )·A =0. Since the dot product is distributive with respect to addition, we can write it as: B·A - |B|cos & A/|A|·A = |B| |A|cos & - (|B|cos & A/|A|)|A|^2 = 0, since (|B|cos & A/|A|)|A|^2 = |B| |A|cos &.
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