The formula for adding two vectors, as defined in Kells' Analytical Geometry is


This makes sense since we're concerned with both the direction and magnitude of the vectors. When I got to the first exercise though, I was given the question

A man walks east 6 miles then north 6 miles. Draw 2 vectors representing each of the trips, then add the vectors to obtain a vector of the single equivalent trip. Give the magnitude and direction of the vector found.

If $AB=6$ and $BC=6$, then $AB+BC=12$ based on the formula. However, the answer instead utilized the Pythagorean Theorem and arrived at the figurative hypotenuse:

$$ 6\sqrt{2} \text{ miles at } 45^{\circ} \text{ NE} $$

$12 \neq 6\sqrt{2}$ the last time I checked

I understand how they arrived at the answer, but their formula combined with the answer to this problem is causing some serious cognitive dissonance that I cant seem to pinpoint. What key ingredient am I missing here?

  • $\begingroup$ The usual euclidean vector space that we frequently work with, $\Bbb R^2$, has addition defined as $(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$ and has norm defined as $\|(x,y)\|=\sqrt{x^2+y^2}$. For your problem, East6miles + North6miles might be represented as $(6,0)+(0,6)=(6,6)$, noting that $\|(6,0)\|=\sqrt{6^2+0^2}=6$ and that $\|(6,6)\|=\sqrt{6^2+6^2}=6\sqrt{2}$. $\endgroup$
    – JMoravitz
    Jan 14, 2018 at 3:22

3 Answers 3


What is missing and confusing you are the arrows.

$\overrightarrow{AB}$ is a vector while $AB$ is a distance.

Vectors verify the additive relation: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

Vectors and distances are linked by the equality: ${AB}^{\ 2}=\overrightarrow{AB}\cdot\overrightarrow{AB}$ where $\cdot$ is the dot product.

Expressed in the language of vectors, Pythagoras theorem is just the expansion of $(x+y)^2$ formula, indeed one can write:

$\begin{align}{AC}^{\ 2} &=\overrightarrow{AC}\cdot\overrightarrow{AC}\\ &=(\overrightarrow{AB}+\overrightarrow{BC})\cdot(\overrightarrow{AB}+\overrightarrow{BC})\\ &=\overrightarrow{AB}\cdot\overrightarrow{AB}+2\ \overrightarrow{AB}\cdot\overrightarrow{BC}+\overrightarrow{BC}\cdot\overrightarrow{BC}\\ &={AB}^{\ 2}+2\ \underbrace{\overrightarrow{AB}\cdot\overrightarrow{BC}}_{(AB)\perp(BC)\implies 0}+{BC}^{\ 2} \end{align}$

And we get ${AC}^{\ 2}={AB}^{\ 2}+{BC}^{\ 2}$

In the same way you write $[AB]$ for the segment joining $A$ to $B$ and $(AB)$ for the straight line passing through these two points, to distinguish from the distance $AB$, you should note the vector $\overrightarrow{AB}$.

If you make the effort of keeping the correct notations, then confusion should be limited.


Please note that in some books (especially old editions), due to composition constraints, they were unable/unwilling to adopt the complex drawing of superposed arrows, in that case bold font is generally used instead.

  • thus distance would be noted $AB$
  • while vector would be noted $\mathbf{AB}$

Please check your book carefully for such typeface characteristic.

  • $\begingroup$ Thanks that helps see the distinction. Can you explain what you did on the last part of the simplification which cancels AB * BC (assume superposed arrows) ? $\endgroup$ Jan 14, 2018 at 3:54
  • 1
    $\begingroup$ When two vectors are orthogonal their dot product is zero. Here for instance $(6,0)\cdot(0,6)=6\times 0+0\times 6=0+0=0$. The scalar product also has this property $\vec{AB}\cdot\vec{BC}=AB\times BC\times\cos(\theta)$ where $\theta$ is the angle between the vectors when they are originated from $O$, when orthogonal this angle is $\frac\pi2$ and the cosinus vanishes. (Side note for the curious: when vectors are not orthogonal you fall back on the law of cosines :p) $\endgroup$
    – zwim
    Jan 14, 2018 at 4:02

The addition "+" for vectors is defined differently than the "+" defined for scalar numbers.

Here we already have the definition for "+" of two vectors $AB$ and $BC$:

$$AB + BC = AC$$

You have $AB$ with direction EAST and magnitude $6$, $BC$ with direction NORTH and magnitude $6$, so

$$AB + BC=AC$$, and $AC$ is of direction $45^{\circ}$ EAST-NORTH, magnitude $6\sqrt{2}$


We have $AB+BC=AC$ for vectors.

However, we do not have $|AB|+|BC|=|AC|$ in general.

If our goal is to travel from point $A$ to point $C$, we can make a stop at point $B$ and this would cost us more distance.

To minimize the distance, we can travel directly from $A$ to $C$ directly.

  • $\begingroup$ I think so far this is the best answer. The "single equivalent trip" ... aspect of the question is misleading since if the total trip is shorter, and the directions aren't the same, it's not an equivalent trip... only the destination is. $\endgroup$ Jan 14, 2018 at 3:44
  • $\begingroup$ ah, that depends on the definition of equivalence, of what do we really care. Equivalence in the question is defined in the sense of "two journeys are equivalet if the beginning and end point of the two journeys are the same" regardless of the path/ distance of path. $\endgroup$ Jan 14, 2018 at 3:47

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