$\mathbf{a}=(\cos \theta, \sin \theta)$ is a unit vector along the line $L$, so our first problem is to find the orthogonal projection of $\mathbf{e}_{1}$ along a. Since $$ \|\mathbf{a}\|=\sqrt{\sin ^{2} \theta+\cos ^{2} \theta}=1 \text { and } \mathbf{e}_{1} \cdot \mathbf{a}=(1,0) \cdot(\cos \theta, \sin \theta)=\cos \theta $$ it follows that this projection is $$ \operatorname{proj}_{\mathbf{a}} \mathbf{e}_{1}=\frac{\mathbf{e}_{1} \cdot \mathbf{a}}{\|\mathbf{a}\|^{2}} \mathbf{a}=(\cos \theta)(\cos \theta, \sin \theta)=\left(\cos ^{2} \theta, \sin \theta \cos \theta\right) $$ Similarly, since $\mathbf{e}_{2} \cdot \mathbf{a}=(0,1) \cdot(\cos \theta, \sin \theta)=\sin \theta$, it follows that $$ \operatorname{proj}_{\mathbf{a}} \mathbf{e}_{2}=\frac{\mathbf{e}_{2} \cdot \mathbf{a}}{\|\mathbf{a}\|^{2}} \mathbf{a}=(\sin \theta)(\cos \theta, \sin \theta)=\left(\sin \theta \cos \theta, \sin ^{2} \theta\right) $$