The dot product between two vectors is based on the projection of one vector onto another. Let”s imagine we have two vectors $vc{a}$ and $vc{b}$, and we want to calculate how much of $vc{a}$ is pointing in the same direction as the vector $vc{b}$. We want a quantity that would be positive if the two vectors are pointingin similar directions, zero if they are perpendicular, and negativeif the two vectors are pointing in nearly opposite directions.We will define the dot product between the vectors to capture these quantities.

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But first, notice that the question “how much of $vc{a}$ is pointing in the same direction as the vector $vc{b}$” does not have anything to do with the magnitude (or length) of $vc{b}$; it is based only on its direction.(Recall that a vector has a magnitude and a direction.) The answer to this question should not depend on the magnitude of $vc{b}$, only its direction. To sidestep any confusion caused by the magnitude of $vc{b}$, let”s scale the vector so that it has length one. In other words, let”s replace $vc{b}$ with the unit vector that points in the same direction as $vc{b}$.We”ll call this vector $vc{u}$, which is defined by$$vc{u} = frac{vc{b}}{|vc{b}|}.$$

The dot product of $vc{a}$ with unit vector $vc{u}$, denoted$vc{a} cdot vc{u}$, is defined to be the projection of $vc{a}$ inthe direction of $vc{u}$, or the amount that $vc{a}$ is pointing inthe same direction as unit vector $vc{u}$.Let”s assume for a moment that $vc{a}$ and $vc{u}$ are pointing insimilar directions.Then, you can imagine $vc{a} cdot vc{u}$ as the length of the shadow of $vc{a}$ onto $vc{u}$ if their tails were together and thesun was shining from a direction perpendicular to $vc{u}$.By forming a right triangle with $vc{a}$ and this shadow, youcan use geometry to calculate that egin{gather}vc{a} cdot vc{u} = |vc{a}| cos heta, label{dot_product_unit}end{gather}where $ heta$ is the angle between $vc{a}$ and $vc{u}$.

If $vc{a}$ and $vc{u}$ were perpendicular, there would be no shadow. That corresponds to the case when $cos heta = cos pi/2 =0$and $vc{a} cdot vc{u}=0$. If the angle $ heta$ between $vc{a}$ and $vc{u}$ were larger than $pi/2$,then the shadow wouldn”t hit $vc{u}$. Since in this case $cos heta

In the following interactive applet, you can explore thisgeometric intrepretation of the dot product, and observe how it depends on the vectors and the angle between them. (The reported number shouldn”tdepend on the length of $|vc{b}|$ since we divided by that magnitude.)

The dot product as projection. The dot product of the vectors $vc{a}$ (in blue) and $vc{b}$ (in green), when divided by the magnitude of $vc{b}$, is the projection of $vc{a}$ onto $vc{b}$. This projection is illustrated by the red line segment from the tail of $vc{b}$ to the projection of the head of $vc{a}$ on $vc{b}$. You can change the vectors $vc{a}$ and $vc{b}$ by dragging the points at their ends or dragging the vectors themselves. Notice how the dot product is positive for acute angles and negative for obtuse angles. The reported number does not depend on $|vc{b}|$ only because we”ve divided through by that magnitude.

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More information about applet.

The geometric definition of equation eqref{dot_product_definition} makes the properties of the dot product clear. One can see immediately from the formula that the dot product $vc{a}cdotvc{b}$ is positive for acute angles and negative for obtuse angles.The formula demonstrates that the dot product grows linearly with the length of both vectors and is commutative, i.e., $vc{a} cdot vc{b} = vc{b} cdot vc{a}$.

However, the geometric formula eqref{dot_product_definition} is not convenient for calculating the dot product when we are given the vectors $vc{a}$ and $vc{b}$ in terms of their components.To facilitate such calculations, we derive a formula for the dot product in terms of vector components.With such formula in hand, we can run through examples of calculating the dot product.

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