1. **Problem statement:** Find the matrix \(\hat{P}\) representing the projection onto the line \(x + \sqrt{3} y = 0\) by considering the transformations of the unit vectors \(\hat{i} = \begin{pmatrix}1 \\ 0\end{pmatrix}\) and \(\hat{j} = \begin{pmatrix}0 \\ 1\end{pmatrix}\). 2. **Identify the direction vector of the line:** The line equation is \(x + \sqrt{3} y = 0\). Rearranged: \(x = -\sqrt{3} y\). A direction vector \(\mathbf{d}\) along the line satisfies \(\mathbf{d} = \begin{pmatrix} -\sqrt{3} \\ 1 \end{pmatrix}\). 3. **Normalize the direction vector:** Calculate its length: $$\|\mathbf{d}\| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.$$ So the unit vector along the line is: $$\mathbf{u} = \frac{1}{2} \begin{pmatrix} -\sqrt{3} \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}.$$ 4. **Projection formula:** The projection matrix onto the line spanned by unit vector \(\mathbf{u}\) is: $$\hat{P} = \mathbf{u} \mathbf{u}^T = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \begin{pmatrix} u_1 & u_2 \end{pmatrix} = \begin{pmatrix} u_1^2 & u_1 u_2 \\ u_2 u_1 & u_2^2 \end{pmatrix}.$$ 5. **Calculate each element:** $$u_1 = -\frac{\sqrt{3}}{2}, \quad u_2 = \frac{1}{2}.$$ So, $$u_1^2 = \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4},$$ $$u_1 u_2 = -\frac{\sqrt{3}}{2} \times \frac{1}{2} = -\frac{\sqrt{3}}{4},$$ $$u_2^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.$$ 6. **Write the projection matrix:** $$\hat{P} = \begin{pmatrix} \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{4} & \frac{1}{4} \end{pmatrix}.$$ 7. **Check by projecting \(\hat{i}\) and \(\hat{j}\):** - \(\hat{P} \hat{i} = \begin{pmatrix} \frac{3}{4} \\ -\frac{\sqrt{3}}{4} \end{pmatrix}\) is the projection of \(\hat{i}\). - \(\hat{P} \hat{j} = \begin{pmatrix} -\frac{\sqrt{3}}{4} \\ \frac{1}{4} \end{pmatrix}\) is the projection of \(\hat{j}\). This confirms the matrix correctly projects vectors onto the line \(x + \sqrt{3} y = 0\). **Final answer:** $$\hat{P} = \begin{pmatrix} \frac{3}{4} & -\frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{4} & \frac{1}{4} \end{pmatrix}.$$