\mathbf{a} \cdot \mathbf{a} = 1^2 + 1^2 + 1^2 = 3

February 22, 2026

$\mathbf{a} \cdot \mathbf{a} = 1^2 + 1^2 + 1^2 = 3$

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Solution:** The component of \(\mathbf{r}\) parallel to \(\mathbf{a}\) is given by the projection of \(\mathbf{r}\) onto \(\mathbf{a}\):
\text{proj}_{\mathbf{a}} \mathbf{r} = \frac{\mathbf{r} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a}
\mathbf{r} \cdot \mathbf{a} = 2 \cdot 1 + 3 \cdot 1 + 6 \cdot 1 = 11
Thus, the projection is:
\text{proj}_{\mathbf{a}} \mathbf{r} = \frac{11}{3} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{11}{3} \\ \frac{11}{3} \\ \frac{11}{3} \end{bmatrix}
\boxed{\begin{bmatrix} \frac{11}{3} \\ \frac{11}{3} \\ \frac{11}{3} \end{bmatrix}}

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