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arrow_back Prove the orthogonal projection theorem

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Prove the orthogonal projection theorem

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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 $\mathbf{w}_{2} \cdot \mathbf{a}=0$

by Platinum
(106,844 points)

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