Magnitude and direction of a vector

The magnitude of a vector as Pythagoras on its components (left as an exact surd), unit vectors as a vector scaled to length one, direction as an angle or bearing, and finding the angle in a vector triangle with the cosine rule.

Pure · Vectors

Magnitude and direction of a vector

Why it works

The magnitude (or modulus) of a vector, written

|\mathbf{a}|

, is the length of its arrow. Lay the components out as the legs of a right-angled triangle and the arrow is the hypotenuse, so Pythagoras gives

|x\mathbf{i} + y\mathbf{j}| = \sqrt{x^2 + y^2}, \qquad |x\mathbf{i} + y\mathbf{j} + z\mathbf{k}| = \sqrt{x^2 + y^2 + z^2}.

Square each component, add, then square-root — and as with any length, leave it as an exact surd (

\sqrt{50} = 5\sqrt{2}

, not

7.07

) unless a decimal is asked for. Because the components are squared, their signs make no difference to the size.

A unit vector has magnitude

1

. To shrink any vector to a unit vector pointing the same way, divide it by its own length:

\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\,\mathbf{a}.

Running this in reverse gives a tidy trick: a vector of magnitude

m

in the direction of

\mathbf{a}

m\hat{\mathbf{a}} = \dfrac{m}{|\mathbf{a}|}\,\mathbf{a}

.

Direction in two dimensions is an angle. Measured from the positive

x

-axis,

\tan\theta = \dfrac{y}{x}

— but always sketch the components first, because a calculator's

\tan^{-1}

only knows the first and fourth quadrants and will mislead you when

x

y

is negative. (For bearings, see Modelling with vectors — those are measured clockwise from north.)

The angle in a vector triangle. To find an angle such as

\angle ABC

when you know the vectors, find the lengths of the three sides of the triangle and use the cosine rule. For the angle at

B

(the side opposite it is

AC

\cos(\angle ABC) = \frac{|AB|^2 + |BC|^2 - |AC|^2}{2\,|AB|\,|BC|}.

The two sides touching the angle go on the bottom; the side facing it is the one that gets subtracted on top.