Magnitude of a 2D Vector

|\vec{v}| = \sqrt{v_x^2 + v_y^2}

|v| =

vₓ =

v_y =

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

Learning zone

A vector is an arrow: a direction plus a length. Drop it onto axes and it splits into two ordinary numbers, the components v_x and v_y, which form the legs of a right triangle whose hypotenuse is the arrow itself — so the magnitude is √(v_x² + v_y²). A hiker who walks 3 km east and 4 km north ends up 5 km from camp, not 7 km, because displacement is a vector and its parts add head-to-tail rather than arithmetically. That gap between 5 and 7 is the single most common student error in the whole subject.

Run it backwards and a magnitude plus one component recovers the other: a vector of length 13 with v_y = 5 must have v_x = √(169 − 25) = 12. The solver returns the positive root, since a component's sign is a matter of which way the axis points and the arithmetic cannot know it. Note also that no component may exceed the magnitude — a leg longer than the hypotenuse describes a triangle that does not exist.

Magnitude of a 2D Vector

|\vec{v}| = \sqrt{v_x^2 + v_y^2}

Where

$|\vec{v}|$ = Vector magnitude
$v_x$ = x-component
$v_y$ = y-component

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

Vector magnitude — Magnitude of a 3D Vector, Unit Vector Component (Normalization)
x-component — Magnitude of a 3D Vector, Dot Product of Two 2D Vectors (Components)
y-component — Magnitude of a 3D Vector, Dot Product of Two 2D Vectors (Components)

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.