Cartesian Vectors in 2-Space and 3-Space

Place a vector's tail at the origin and its head lands on a point $P(x,y)$. That vector, $\overrightarrow{OP}=(x,y)$, is the POSITION VECTOR of $P$, and $x$ and $y$ are its components. Using the unit vectors $\vec{i}=(1,0)$ and $\vec{j}=(0,1)$, the same vector is $x\vec{i}+y\vec{j}$. The vector between two points is head minus tail: from $A(x_1,y_1)$ to $B(x_2,y_2)$, $\overrightarrow{AB}=(x_2-x_1,y_2-y_1)$. Two vectors are equal exactly when all their components are equal, and by Pythagoras the magnitude is $|(x,y)|=\sqrt{x^2+y^2}$.

Going from direction form to components is direct: a vector of magnitude $8$ at $30°$ is $(8\cos 30°,8\sin 30°)=(4\sqrt{3},4)\approx (6.93,4)\text{.}$ The signs take care of themselves in every quadrant, as long as $\theta$ is the standard angle (convert a bearing first). Going back needs care. The vector $(8,6)$ has $r=\sqrt{64+36}=10$ and $\theta ={\tan }^{-1}\left(\frac{6}{8}\right)\approx 36.9°$, fine because it lies in quadrant 1. A calculator's ${\tan }^{-1}$ only returns angles between $-90°$ and $90°$, so for vectors pointing left it gives the wrong quadrant.

Vector components from magnitude and direction

Add a third axis, $z$, perpendicular to both $x$ and $y$. The axes form a RIGHT-HANDED system: curl the fingers of your right hand from the positive $x$-axis toward the positive $y$-axis and your thumb points along the positive $z$-axis. A point is $P(x,y,z)$ and its position vector is $\overrightarrow{OP}=(x,y,z)=x\vec{i}+y\vec{j}+z\vec{k}$, with $\vec{k}=(0,0,1)$. The SAME triple of numbers describes the point and the vector; the context says which you mean. The three coordinate planes are the $xy$-plane ($z=0$), the $xz$-plane ($y=0$) and the $yz$-plane ($x=0$).