Unit Vector Component (Normalization)

\hat{u}_x = \frac{v_x}{|\vec{v}|}

ûₓ =

vₓ =

|v| =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Normalizing strips a vector of its length and keeps only its heading. Divide each component by the magnitude and you get a unit vector: (3, 4) has length 5, so its normalized form is (0.6, 0.8), and 0.6² + 0.8² = 1 exactly. Hamilton called such a direction-only object a versor, and the idea is everywhere in computer graphics — surface normals, camera directions and lighting calculations are all stored as unit vectors so that a dot product returns a clean cosine rather than a cosine tangled up with two lengths.

Reversed, the relation rebuilds a vector from a heading and a distance: a unit component of 0.8 on a vector of length 25 means v_x = 20. The one forbidden input is a magnitude of zero. The zero vector points nowhere, so normalizing it is not a rounding problem to be nudged past — it is genuinely undefined, and code that ignores that fact produces NaNs that surface much later as invisible geometry.

Unit Vector Component (Normalization)

\hat{u}_x = \frac{v_x}{|\vec{v}|}

Where

$\hat{u}_x$ = Unit vector component
$v_x$ = Vector component
$|\vec{v}|$ = Vector magnitude

Missing one of these? Work it out first, then come back

Unit vector component — Magnitude of a 2D Vector, Magnitude of a 3D Vector
Vector component — Magnitude of a 2D Vector, Magnitude of a 3D Vector
Vector magnitude — Magnitude of a 2D Vector, Magnitude of a 3D Vector

Part of this formula set

Vector operations

12 formulasMagnitude, direction, components and unit vectors, plus the dot and cross products and the projections and angles they give you.