Sum of 2 mutually orthogonal vectors

Question

I have a question:

Resolve the vector $(3,-4,2)$ into the sum of two mutually orthogonal vectors, one of which is parallel to the vector $(1,2,3)$

^{Fitz-Gerald, G. F., Grundy, I. H., Clarke, G. T., & Peckham, I. A. (2005). Mathematical methods for engineers and scientists / with contributions from G.T. Clarke and I.H. Grundy. Pearson Education Australia. (p. 35)}

Which I have broken down into the following assumptions of the 2 orthogonal vectors, $\textbf{a}$ and $\textbf{b}$.

1. $\textbf{a} \cdot \textbf{b} = 0$ (As they are perpendicular)
2. $\textbf{a}+\textbf{b} = 3\textbf{i}-4\textbf{j}+2\textbf{k}$ (Given in question)
3. $c \cdot \textbf{a} = \textbf{i} + 2\textbf{j} + 3\textbf{k}$ The parallel vector (Given in question)

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I'm not sure where to continue from here. How do I obtain the 2 orthogonal vectors?

Answer 1

Think geometrically.

Let $u=(3,-4,2)$ and $v=(1,2,3)$. If $w$ denotes the orthogonal projection of $u$ on the line directed by $v$, then the decomposition $u=w+(u-w)$ is exactly what you need.

So you just have to compute $w$, but $w$ is given by the formula $$w = \dfrac{\langle u,v \rangle}{||v||^2} v = \dfrac{1}{14}(1,2,3)$$

(where $\langle .,. \rangle$ denotes the canonical dot product on $\mathbb{R}^3$).

Then $u-w = \dfrac{1}{14}(41, -58, 25)$, and you can check that $$\boxed{u = \dfrac{1}{14}(1,2,3) + \dfrac{1}{14}(41, -58, 25)}$$

answers your question.