How Do You Find the Sum, Difference, and Magnitude of Vectors?

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Addition combines the movements represented by two vectors. Subtraction finds the vector needed to move from one endpoint to another, so order matters: generally, $\mathbf{u}-\mathbf{v}\neq\mathbf{v}-\mathbf{u}$.

Component form makes vector operations systematic. In the sum, each output component records the combined movement along one coordinate axis. In the difference, each component records the change from the second vector to the first. The magnitude formula follows from Pythagoras’ theorem: first combine two perpendicular components, then include the third. Because magnitude is a length, it is always non-negative and has the same units as the vector’s components.

A useful check is the triangle inequality: $\lvert\mathbf{u}+\mathbf{v}\rvert\leq\lvert\mathbf{u}\rvert+\lvert\mathbf{v}\rvert$. If a calculated magnitude is negative, or if this inequality fails, recheck the component arithmetic. Equality occurs only when the vectors point in the same direction.

Worked Example

Let $\mathbf{u}=(3,-2,4)$ and $\mathbf{v}=(-1,5,2)$.

The sum is

The difference is

The magnitude of $\mathbf{u}$ is

These calculations distinguish a vector from a scalar. A vector has magnitude and direction, whereas magnitude alone is a scalar. Always check signs carefully when subtracting negative components, as in $3-(-1)=4$.

For position vectors, if points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$, then the displacement from $A$ to $B$ is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$. Reversing the order gives the opposite direction.

IB Exam Technique

This is AHL 3.10: Vector definitions, so it is HL-only and may be tested on any GDC-active HL paper. Show the component operation before giving the answer, keep exact magnitudes such as $\sqrt{29}$ unless a decimal is requested, and round only the final value. A common misconception is to add the vector magnitudes; generally, $\lvert\mathbf{u}+\mathbf{v}\rvert\neq\lvert\mathbf{u}\rvert+\lvert\mathbf{v}\rvert$.