The formula for adding two vectors, as defined in Kells' Analytical Geometry is
$AB+BC=AC$
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?
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$\begingroup$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.
$\color{red}{\text{warning:}}$
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.
$\endgroup$ 2 $\begingroup$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 $6\sqrt{2}$
$\endgroup$ $\begingroup$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.
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