Given two non-perpendicular vectors prove there exist vectors satisfying given relations

Let $A, B \in \mathbb{R}^n$ be two vectors which are not perpendicular. Prove that there exist vector $C,D \in \mathbb{R}^n$ such that $C$ is parallel to $A$ , $D$ is orthogonal to $A$ , and $B = C+D$ .

Proof. Since $C$ is parallel to $A$ we have $C = tA$ for some $t \neq 0$ .
Since $D$ is perpendicular to $A$ we have

$D \cdot A = 0 \quad \implies \quad \sum_{i=1}^n a_i d_i = 0.$

Since $B = C+D$ we have

$c_i + d_i = b_i \qquad \text{for all } 1 \leq i \leq n.$

Substituting our expression for $c_i$ in terms of $a_i$ and $t$ in the last equation we have

$d_i = b_i - ta_i \qquad \text{for all } 1 \leq i \leq n.$

Then, substituting this expression for $d_i$ into the equation from $D$ perpendicular to $A$ we have

$\begin{align*} && \sum_{i=1}^n a_i (b_i - ta_i) &= 0 \\[9pt] \implies && \sum_{i=1}^n a_i b_i - t \sum_{i=1}^n a_i^2 &= 0 \\[9pt] \implies && t &= \frac{A \cdot B}{\lVert A \rVert^2} \end{align*}$

which is nonzero since $A \cdot B \neq 0$ by hypothesis. Thus,

$C = tA, \quad D = B-tA \qquad \text{where} \qquad t = \frac{A \cdot B}{\lVert A \rVert^2}$

is always a solution $. \qquad \blacksquare$

Stumbling Robot

Given two non-perpendicular vectors prove there exist vectors satisfying given relations

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