Magnitude of a Vector

A vector has both a size and a direction. The magnitude of a vector is the number that describes that size, in other words, how long the vector is. If you picture a vector as an arrow drawn on a coordinate grid, the magnitude is just the length of that arrow, regardless of which way it points.

Because magnitude measures a length, it is always zero or positive, it is never negative. The magnitude of a vector \(\vec{v}\) is written \(|\vec{v}|\) or sometimes \(\|\vec{v}\|\).

If a vector is written in component form as \(\vec{v} = (x, y)\), its horizontal and vertical components form the two legs of a right triangle, and the vector itself is the hypotenuse. That means the magnitude comes straight from the Pythagorean theorem:

\( |\vec{v}| = \sqrt{x^2 + y^2} \)

This is the same idea used when you measure horizontal and vertical distances between two points on a grid, since a vector's components are just those horizontal and vertical distances.

Example 1

Find the magnitude of \(\vec{v} = (3, 4)\).

\( |\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Example 2

Find the magnitude of \(\vec{u} = (-5, 12)\).

\( |\vec{u}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

Notice that a negative component becomes positive once it is squared, so the sign of a component never makes the magnitude negative.

The same pattern extends naturally into three dimensions. For a vector \(\vec{v} = (x, y, z)\), simply add a third squared term under the square root:

\( |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \)

Example 3

Find the magnitude of \(\vec{v} = (2, -3, 6)\).

\( |\vec{v}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)

Sometimes a vector is described by its starting point and ending point rather than by components right away. If a vector goes from point \(A(x_1, y_1)\) to point \(B(x_2, y_2)\), first find the component form by subtracting coordinates:

\( \vec{AB} = (x_2 - x_1, \; y_2 - y_1) \)

Then apply the usual magnitude formula to that result.

Example 4

Find the magnitude of the vector from \(A(1, 2)\) to \(B(4, 6)\).

\( \vec{AB} = (4 - 1, \; 6 - 2) = (3, 4) \)

\( |\vec{AB}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \)

This is exactly the distance formula in disguise: the magnitude of a vector between two points is just the distance between those two points.

Once you know a vector's magnitude, you can rescale the vector to have a length of exactly \(1\) by dividing every component by the magnitude. This produces a unit vector pointing in the same direction as the original. Magnitude also plays a key role when describing a vector by its length and its direction angle instead of by \(x\) and \(y\) components.

In physics-style problems, several forces acting on an object combine into a single resultant vector, and the magnitude of that resultant force is found using the same square-root formula applied to the combined components. The same square-root idea also underlies the magnitude of a cross product between two vectors, though that calculation involves an extra step beyond simple addition of components.

• 2D vector \((x, y)\): magnitude \(= \sqrt{x^2 + y^2}\)
• 3D vector \((x, y, z)\): magnitude \(= \sqrt{x^2 + y^2 + z^2}\)
• Vector between two points: subtract coordinates first, then apply the formula
• Magnitude is always \(\ge 0\), and a magnitude of \(1\) means the vector is a unit vector

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