Question

In Exercises 3–8, let A, B, C, D, ... , Z be points in $R^3$. Simplify the given quantity.

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}$

Short answer

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$

Step-by-step solution

Given

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}$

Simplify

From the law of vector addition:

$$\begin{gathered} \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CA} \\ =\overrightarrow{AC}+\overrightarrow{CA} \\ =\overrightarrow{AC}-\overrightarrow{AC}\left(\overrightarrow{CA}=-\overrightarrow{AC}\right) \\ =0 \end{gathered}$$