The magnitude is the vector's length — a single positive number, regardless of direction — written with modulus signs: $|\mathbf{a}|$ or $|\overrightarrow{AB}|$. Two vectors pointing in completely different directions can still have the same magnitude, since magnitude only measures size, not orientation.

A column vector $\binom{x}{y}$ forms the two shorter sides of a right-angled triangle, so its magnitude follows directly from Pythagoras' theorem:

For $\mathbf{v}=\binom{24}{-9}$:

Squaring always gives a positive result, even for a negative component — $(-9)^2=+81$, never $-81$.

Combine the vectors into a single column vector first, then take the magnitude of that result — magnitude does not distribute over addition, so $|\mathbf{a}+\mathbf{b}| \ne |\mathbf{a}|+|\mathbf{b}|$ in general.

Example: given $\mathbf{a}=\binom{5}{-3}$ and $\mathbf{b}=\binom{-2}{4}$, find $|\mathbf{a}+\mathbf{b}|$.

If the magnitude is given but one component is unknown, substitute into $\sqrt{x^2+y^2}$ and solve the resulting equation — remembering that squaring loses the sign, so two possible values usually exist.

Example: $\mathbf{w}=\binom{5}{k}$ has magnitude $13$. Find the possible values of $k$.

Both $k=12$ and $k=-12$ satisfy the original equation, since $(-12)^2=144$ just as much as $12^2$ does — unless the context (a diagram showing $k$ clearly positive, for instance) rules one out, both values should be given.

The distance between points $A(x_1,y_1)$ and $B(x_2,y_2)$ is exactly the magnitude of the vector joining them, $\overrightarrow{AB}=\binom{x_2-x_1}{y_2-y_1}$ — so no separate "distance formula" needs to be memorised on top of this one.

Example: find the distance between $A(2,-1)$ and $B(7,11)$.

Recognising $5$–$12$–$13$ as a Pythagorean triple here avoids a calculator step entirely — a habit worth building for common triples like $3$-$4$-$5$, $5$-$12$-$13$ and $8$-$15$-$17$.

Scaling a vector by $k$ scales its magnitude by exactly $|k|$ (the absolute value of $k$, since a magnitude can never be negative): $|k\mathbf{a}| = |k||\mathbf{a}|$. This makes sense geometrically — doubling a displacement doubles its length, and reversing its direction (multiplying by $-1$) leaves the length completely unchanged.

Worked Example

Given $\mathbf{a} = \binom{3}{4}$, find $|3\mathbf{a}|$ and $|-2\mathbf{a}|$.

1. First find $|\mathbf{a}| = \sqrt{3^2+4^2} = \sqrt{25}=5$.
2. $|3\mathbf{a}| = 3 \times 5 = 15$.
3. $|-2\mathbf{a}| = |-2|\times 5 = 2\times5=10$ (the negative sign reverses direction but does not affect length).