# arrow_back Prove the orthogonal projection theorem

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Prove the orthogonal projection theorem

Proof Since the vector $\mathbf{w}_{1}$ is to be a scalar multiple of $\mathbf{a}$, it must have the form $$\mathbf{w}_{1}=k \mathbf{a}$$ Our goal is to find a value of the scalar $k$ and a vector $\mathbf{w}_{2}$ that is orthogonal to a such that $$\mathbf{u}=\mathbf{w}_{1}+\mathbf{w}_{2}$$ We can determine $k$ by $$\mathbf{u}=\mathbf{w}_{1}+\mathbf{w}_{2}=k \mathbf{a}+\mathbf{w}_{2}$$ and then $$\mathbf{u} \cdot \mathbf{a}=\left(k \mathbf{a}+\mathbf{w}_{2}\right) \cdot \mathbf{a}=k\|\mathbf{a}\|^{2}+\left(\mathbf{w}_{2} \cdot \mathbf{a}\right)$$ Since $\mathbf{w}_{2}$ is to be orthogonal to a, the last term in (9) must be 0 , and hence $k$ must satisfy the equation $$\mathbf{u} \cdot \mathbf{a}=k\|\mathbf{a}\|^{2}$$ from which we obtain $$k=\frac{\mathbf{u} \cdot \mathbf{a}}{\|\mathbf{a}\|^{2}}$$ as the only possible value for $k$. The proof can be completed $$\mathbf{w}_{2}=\mathbf{u}-\mathbf{w}_{1}=\mathbf{u}-k \mathbf{a}=\mathbf{u}-\frac{\mathbf{u} \cdot \mathbf{a}}{\|\mathbf{a}\|^{2}} \mathbf{a}$$ and then confirming that $\mathbf{w}_{2}$ is orthogonal to a by showing that $\mathbf{w}_{2} \cdot \mathbf{a}=0$

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