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Worked solution

Prove orthogonality and explain the zero-vector case

Given n=(b⋅b)a−(a⋅b)b\mathbf n=(\mathbf b\cdot\mathbf b)\mathbf a-(\mathbf a\cdot\mathbf b)\mathbf b, show that n is perpendicular to b, if a,b are two non-zero vectors.

CJC · Review problems · Question 10

Answer

b⋅n=0\mathbf b\cdot\mathbf n=0; n is orthogonal to b, with a right-angle interpretation when n≠0.

Distribute the scalar product

b⋅n=b⋅[(b⋅b)a]−b⋅[(a⋅b)b]=(b⋅b)(b⋅a)−(a⋅b)(b⋅b)=0.\begin{aligned}&\mathbf b\cdot\mathbf n\\ & =\mathbf b\cdot[(\mathbf b\cdot\mathbf b)\mathbf a]-\mathbf b\cdot[(\mathbf a\cdot\mathbf b)\mathbf b]\\ & =(\mathbf b\cdot\mathbf b)(\mathbf b\cdot\mathbf a)-(\mathbf a\cdot\mathbf b)(\mathbf b\cdot\mathbf b)\\ & =0.\end{aligned}

Symmetry b⋅a=a⋅b\mathbf b\cdot\mathbf a=\mathbf a\cdot\mathbf b makes the terms cancel.

How did this part go?

Interpret the conclusion carefully

The identity proves n and b are orthogonal. If a is parallel to b, n=0 even though both input vectors are nonzero; the zero vector has no direction and no defined angle. For a right-angle interpretation, require a,b nonparallel so n is nonzero. The dot-product-zero claim itself holds in every stated case.

How did this part go?

Common mistakes

Nonzero inputs do not ensure that the constructed vector n is nonzero.

Solution notes

All printed proof lines retained and the degenerate parallel-input case explicitly qualified. Q11 begins on the same PDF page; its separate native transcript preserves that original page too. No marks or figure are printed for Q10.

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