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A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.