Resolution of a Vector in Two Dimensions

Setting up. Let $\hat i,\hat j$ be the unit vectors along the positive $x$-axis and positive $y$-axis respectively, both with initial point at the origin $O$. Let $P(x,y)$ be any point in the plane, so $\vec{OP}$ is its position vector.

Theorem. $\vec{OP}$ can be written uniquely as $\vec{OP}=x\hat i+y\hat j,\qquad\text{and}\qquad |\vec{OP}|=\sqrt{x^2+y^2}.$

Proof. Drop perpendiculars from $P$ to the $x$-axis (foot $L$) and to the $y$-axis (foot $M$). Then $\vec{OP}=\vec{OL}+\vec{LP}=\vec{OL}+\vec{OM}$. Since $\hat i,\hat j$ are unit vectors along the axes and $OL=x, OM=y$, we get $\vec{OL}=x\hat i$ and $\vec{OM}=y\hat j$, so $\vec{OP}=x\hat i+y\hat j$.

Uniqueness. Suppose $\vec{OP}=x_1\hat i+y_1\hat j=x_2\hat i+y_2\hat j$ were two representations. Then $(x_1-x_2)\hat i+(y_1-y_2)\hat j=\vec 0$, which (since $\hat i,\hat j$ point along two genuinely different directions) forces $x_1=x_2$ and $y_1=y_2$.

Magnitude. In right triangle $OLP$, $OP^2=OL^2+LP^2=x^2+y^2$, so $|\vec{OP}|=\sqrt{x^2+y^2}$. …