In Section Vectors, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. We begin with the following definition.

For example, if \(\overrightarrow {v} = \left <3,4\right >\) and \(\overrightarrow {w} = \left <1,-2\right >\),then \(\overrightarrow {v} \cdot \overrightarrow {w} = \left <3,4\right > \cdot \left <1,-2\right > = (3)(1) + (4)(-2) = -5\).

Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity \(\overrightarrow {v} \cdot \overrightarrow {w}\) is often called the scalar product of \(\overrightarrow {v}\) and \(\overrightarrow {w}\). The dot product enjoys the following properties.

Like most of the theorems involving vectors, the proof of Theorem dotprodprops amounts to using the definition of the dot product and properties of real number arithmetic.

For example, to show the commutative property, let \(\overrightarrow {v} = \left <v_1,v_2\right >\) and \(\overrightarrow {w} = \left <w_1,w_2\right >\). Then

The distributive property is proved similarly and is left as an exercise.

For the scalar property, assume that \(\overrightarrow {v} = \left <v_1,v_2\right >\) and \(\overrightarrow {w} = \left <w_1,w_2\right >\) and \(k\) is a scalar. Then

We leave the proof of \(k(\overrightarrow {v} \cdot \overrightarrow {w}) = \overrightarrow {v} \cdot (k \overrightarrow {w})\) as an exercise.

For the last property, we note that if \(\overrightarrow {v} = \left <v_1,v_2\right >\), then \(\overrightarrow {v} \cdot \overrightarrow {v} = \left <v_1,v_2\right > \cdot \left <v_1,v_2\right > = v_1^2 + v_2^2 = \|\overrightarrow {v}\|^2\), where the last equality comes courtesy of Definition polarformvector.

The following example puts Theorem dotprodprops to good use. As in Example vectoreqnex, we work out the problem in great detail and encourage the reader to supply the justification for each step.

If we take a step back from the pedantry in Example dotprodpropex, we see that the bulk of the work is needed to show that \((\overrightarrow {v} - \overrightarrow {w}) \cdot (\overrightarrow {v} - \overrightarrow {w}) = \overrightarrow {v} \cdot \overrightarrow {v} -2(\overrightarrow {v} \cdot \overrightarrow {w}) + \overrightarrow {w}\cdot \overrightarrow {w}\). If this looks familiar, it should.

Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand \((\overrightarrow {v} - \overrightarrow {w}) \cdot (\overrightarrow {v} - \overrightarrow {w})\) for vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\) match those required to expand \((v-w)(v-w)\) for real numbers \(v\) and \(w\), and hence we get similar looking results.

The identity verified in Example dotprodpropex plays a large role in the development of the geometric properties of the dot product, which we now explore.

Suppose \(\overrightarrow {v}\) and \(\overrightarrow {w}\) are two nonzero vectors. If we draw \(\overrightarrow {v}\) and \(\overrightarrow {w}\) with the same initial point, we define the angle between \(\overrightarrow {v}\) and \(\overrightarrow {w}\) to be the angle \(\theta \) determined by the rays containing the vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\), as illustrated below. We require \(0 \leq \theta \leq \pi \). (Think about why this is needed in the definition.)

The following theorem gives us some insight into the geometric role the dot product plays.

We prove Theorem dotproductgeo in cases. If \(\theta = 0\), then \(\overrightarrow {v}\) and \(\overrightarrow {w}\) have the same direction. It follows ^[1] (Since \(\overrightarrow {v} = \| \overrightarrow {v} \| \hat {v}\) and \(\overrightarrow {w} = \| \overrightarrow {w} \| \hat {w}\), if \(\hat {v} = \hat {w}\) then \(\overrightarrow {w} = \|\overrightarrow {w}\| \hat {v} = \frac {\| \overrightarrow {w} \|}{\| \overrightarrow {v} \|} (\| \overrightarrow {v} \| \hat {v}) = \frac {\| \overrightarrow {w} \|}{\| \overrightarrow {v} \|} \overrightarrow {v}\). In this case, \(k = \frac {\| \overrightarrow {w} \|}{\| \overrightarrow {v} \|} > 0\).) that there is a real number \(k > 0\) so that \(\overrightarrow {w} = k \overrightarrow {v}\). Hence, \(\overrightarrow {v} \cdot \overrightarrow {w} = \overrightarrow {v} \cdot (k \overrightarrow {v}) = k (\overrightarrow {v} \cdot \overrightarrow {v}) = k \| \overrightarrow {v} \|^2\).

Working from the other end of the equation, \(\| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos (\theta ) = \| \overrightarrow {v} \| \|k \overrightarrow {v} \| \cos (0) = \| \overrightarrow {v} \| (|k| \| \overrightarrow {v} \|) (1) = k \| \overrightarrow {v} \|^2\), where \(\|k \overrightarrow {v} \| = |k| \| \overrightarrow {v} \|\) courtesy of Theorem magdirprops, and \(|k| = k\) since \(k > 0\).

Hence, in the case \(\theta = 0\), we have shown \(\overrightarrow {v} \cdot \overrightarrow {w} = k \| \overrightarrow {v} \|^2\) and \(\| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos (\theta )= k \| \overrightarrow {v} \|^2\). Putting these two equations together shows that \(\overrightarrow {v} \cdot \overrightarrow {w} = \|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta )\) holds in this case.

If \(\theta = \pi \), \(\overrightarrow {v}\) and \(\overrightarrow {w}\) have the exact opposite directions, so there is a real number \(k< 0\) with \(\overrightarrow {w} = k \overrightarrow {v}\).

As before, we compute \(\overrightarrow {v} \cdot \overrightarrow {w} = \overrightarrow {v} \cdot (k \overrightarrow {v}) = k (\overrightarrow {v} \cdot \overrightarrow {v}) = k \| \overrightarrow {v} \|^2\). Since \(k< 0\) here, we have \(|k| = -k\). Hence, we find \(\| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos (\theta ) = \| \overrightarrow {v} \| \| k \overrightarrow {v} \| \cos (\pi ) = \| \overrightarrow {v} \| (|k| \| \overrightarrow {v} \|) (-1) = \| \overrightarrow {v} \| (-k) \| \overrightarrow {v} \| (-1) = k \| \overrightarrow {v} \|^2\).

Once again, both \(\overrightarrow {v} \cdot \overrightarrow {w} = k \| \overrightarrow {v} \|^2\) and \(\| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos (\theta )= k \| \overrightarrow {v} \|^2\), so \(\overrightarrow {v} \cdot \overrightarrow {w} = \|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta )\) in this case.

Next, if \(0 < \theta < \pi \), the vectors \(\overrightarrow {v}\), \(\overrightarrow {w}\) and \(\overrightarrow {v} - \overrightarrow {w}\) determine a triangle with side lengths \(\| \overrightarrow {v} \|\), \(\| \overrightarrow {w} \|\) and \(\| \overrightarrow {v} - \overrightarrow {w} \|\), respectively, as seen in the diagram below.

The Law of Cosines yields \(\| \overrightarrow {v} - \overrightarrow {w} \|^2 = \|\overrightarrow {v}\|^2 + \|\overrightarrow {w}\|^2 - 2\|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta )\). From Example dotprodpropex, we also have that \(\|\overrightarrow {v} - \overrightarrow {w}\|^2 = \|\overrightarrow {v}\|^2 -2 (\overrightarrow {v} \cdot \overrightarrow {w}) + \|\overrightarrow {w}\|^2\).

Equating these two expressions for \(\| \overrightarrow {v} - \overrightarrow {w} \|^2\) gives \(\|\overrightarrow {v}\|^2 + \|\overrightarrow {w}\|^2 - 2\|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta ) = \|\overrightarrow {v}\|^2 -2 (\overrightarrow {v} \cdot \overrightarrow {w}) + \|\overrightarrow {w}\|^2\) which reduces to \(- 2\|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta ) = -2 (\overrightarrow {v} \cdot \overrightarrow {w})\). Hence, \(\overrightarrow {v} \cdot \overrightarrow {w} = \|\overrightarrow {v}\| \|\overrightarrow {w}\| \cos (\theta )\), as required.

An immediate consequence of Theorem dotproductgeo is the following.

We obtain the formula in Theorem anglebetweenvectorthm by solving the equation given in Theorem dotproductgeo for \(\theta \).

Since \(\overrightarrow {v}\) and \(\overrightarrow {w}\) are nonzero, so are \(\| \overrightarrow {v} \|\) and \(\|\overrightarrow {w}\|\). Hence, we may divide both sides of \(\overrightarrow {v} \cdot \overrightarrow {w} = \| \overrightarrow {v} \| \|\overrightarrow {w} \| \cos (\theta )\) by \(\| \overrightarrow {v} \| \|\overrightarrow {w} \|\). Since \(0 \leq \theta \leq \pi \) by definition, the values of \(\theta \) exactly match the range of the arccosine function. Hence,

Using Theorem dotprodprops, we can rewrite

giving us the alternative formula listed in Theorem anglebetweenvectorthm: \(\theta = \arccos (\hat {v} \cdot \hat {w})\). We are overdue for an example.

A few remarks about Example anglebetweenvectorex are in order. Note that for nonzero vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\), the lengths \(\| \overrightarrow {v} \|\) and \(\| \overrightarrow {w} \|\) are always positive. Since Theorem dotproductgeo tells us that \(\overrightarrow {v} \cdot \overrightarrow {w} = \| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos (\theta )\), we know the sign of \(\overrightarrow {v} \cdot \overrightarrow {w}\) is the same as the sign of \(\cos (\theta )\).

Geometrically, if \(\overrightarrow {v} \cdot \overrightarrow {w} < 0\), then \(\cos (\theta ) < 0\) so \(\theta \) is an obtuse angle, demonstrated number anglebetweenvectorexone above.

If \(\overrightarrow {v} \cdot \overrightarrow {w} = 0\), then \(\cos (\theta ) = 0\) so \(\theta = \frac {\pi }{2}\) as in number anglebetweenvectorextwo. In this case, the vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\) are called orthogonal. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. Hence, if \(\overrightarrow {v}\) and \(\overrightarrow {w}\) are orthogonal, we write \(\overrightarrow {v} \perp \overrightarrow {w}\).

Note there is no ‘zero product property’ for the dot product. As with the vectors in number anglebetweenvectorextwo above, it is quite possible to have \(\overrightarrow {v} \cdot \overrightarrow {w} = 0\) but neither \(\overrightarrow {v}\) nor \(\overrightarrow {w}\) be \(\overrightarrow {0}\).

Finally, if \(\overrightarrow {v} \cdot \overrightarrow {w} > 0\), then \(\cos (\theta ) > 0\) so \(\theta \) is an acute angle, as in the case of number anglebetweenvectorexthree above.

We summarize all of our observations in the schematic below.

Of the three cases diagrammed above, the one which has the most mathematical significance moving forward is the orthogonal case. Hence, we state the corresponding theorem below.

Basically, Theorem dotprodorththm tells us that ‘the dot product detects orthogonality.’ This is a helpful interpretation to keep in mind as you continue your study of vectors in later courses.

We have already argued one direction of Theorem dotprodorththm, namely if \(\overrightarrow {v} \cdot \overrightarrow {w} = 0\) then \(\overrightarrow {v} \perp \overrightarrow {w}\) in the comments following Example anglebetweenvectorex.

To show the converse, we note if \(\overrightarrow {v} \perp \overrightarrow {w}\), then the angle between \(\overrightarrow {v}\) and \(\overrightarrow {w}\), \(\theta = \frac {\pi }{2}\). From Theorem dotproductgeo, we have that \(\overrightarrow {v} \cdot \overrightarrow {w} = \| \overrightarrow {v} \| \| \overrightarrow {w} \| \cos \left ( \frac {\pi }{2} \right ) = \| \overrightarrow {v} \| \| \overrightarrow {w} \| \cdot (0) = 0\), as required.

We can use Theorem dotprodorththm in the following example to provide a different proof about the relationship between the slopes of perpendicular lines. ^[2] (See Exercise perpendicularlineproof in Section AppLines.)

1 Vector Projections

While Theorem 4 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors \(\overrightarrow {v}\) and \(\overrightarrow {w}\) drawn with a common initial point \(O\) below. For the moment, assume that the angle between \(\overrightarrow {v}\) and \(\overrightarrow {w}\), \(\theta \), is acute.

We wish to develop a formula for the vector \(\overrightarrow {p}\), indicated below, which is called the orthogonal projection of \(\overrightarrow {v}\) onto \(\overrightarrow {w}\). The vector \(\overrightarrow {p}\) is obtained geometrically as follows: drop a perpendicular from the terminal point \(T\) of \(\overrightarrow {v}\) to the vector \(\overrightarrow {w}\) and call the point of intersection \(R\). The vector \(\overrightarrow {p}\) is then defined as \(\overrightarrow {p} = \overrightarrow {OR}\).

Like any vector, \(\overrightarrow {p}\) is determined by its magnitude \(\| \overrightarrow {p} \|\) and its direction \(\hat {p}\) according to the formula \(\overrightarrow {p} = \| \overrightarrow {p} \| \hat {p}\). Since we want \(\hat {p}\) to have the same direction as \(\overrightarrow {w}\), we have \(\hat {p} = \hat {w}\).

To determine \(\| \overrightarrow {p} \|\), we apply Definition ?? to the right triangle \(\triangle ORT\). We find \(\cos (\theta ) = \frac {\| \overrightarrow {p} \|}{\| \overrightarrow {v} \|}\), or, equivalently, \(\| \overrightarrow {p} \| = \| \overrightarrow {v} \| \cos (\theta )\). Using Theorems 2 and 1, we get:

Hence, \(\| \overrightarrow {p} \| = \overrightarrow {v} \cdot \hat {w}\), and since \(\hat {p} = \hat {w}\), we have \(\overrightarrow {p} = \| \overrightarrow {p} \| \hat {p} = (\overrightarrow {v} \cdot \hat {w}) \hat {w}\).

Now suppose that the angle \(\theta \) between \(\overrightarrow {v}\) and \(\overrightarrow {w}\) is obtuse, and consider the diagram below.

In this case, we see that \(\hat {p} = - \hat {w}\) and using the triangle \(\triangle ORT\), we find \(\| \overrightarrow {p} \| = \| \overrightarrow {v} \| \cos (\theta ')\). Since \(\theta + \theta ' = \pi \), it follows that \(\cos (\theta ') = -\cos (\theta )\), which means \(\| \overrightarrow {p} \| = \| \overrightarrow {v} \| \cos (\theta ') = - \| \overrightarrow {v} \| \cos (\theta )\).

Rewriting this last equation in terms of \(\overrightarrow {v}\) and \(\overrightarrow {w}\) as before, we get \(\|\overrightarrow {p} \| = -(\overrightarrow {v} \cdot \hat {w})\). Putting this together with \(\hat {p} = - \hat {w}\), we get \(\overrightarrow {p} = \| \overrightarrow {p} \| \hat {p} = -(\overrightarrow {v} \cdot \hat {w}) (-\hat {w}) = (\overrightarrow {v} \cdot \hat {w}) \hat {w}\) in this case as well.

If the angle between \(\overrightarrow {v}\) and \(\overrightarrow {w}\) is \(\frac {\pi }{2}\) then it is easy to show ^[3] (In this case, the point \(R\) coincides with the point \(O\), so \(\overrightarrow {p} = \overrightarrow {OR} = \overrightarrow {OO} = \overrightarrow {0}\).) that \(\overrightarrow {p} = \overrightarrow {0}\). Since \(\overrightarrow {v} \perp \overrightarrow {w}\) in this case, \(\overrightarrow {v} \cdot \overrightarrow {w} = 0\). It follows that \(\overrightarrow {v} \cdot \hat {w} = 0\) and \(\overrightarrow {p} = \overrightarrow {0} = 0 \hat {w} = (\overrightarrow {v} \cdot \hat {w}) \hat {w}\) in this case, too. We have motivated the following.

Definition 2 gives us a good idea what the dot product does. The scalar \(\overrightarrow {v} \cdot \hat {w}\) is a measure of how much of the vector \(\overrightarrow {v}\) is in the direction of the vector \(\overrightarrow {w}\) and is thus called the scalar projection of \(\overrightarrow {v}\) onto \(\overrightarrow {w}\).

While the formula given in Definition 2 is theoretically appealing, because of the presence of the normalized unit vector \(\hat {w}\), computing the projection using the formula \(\text {proj}_{\overrightarrow {w}}(\overrightarrow {v}) = (\overrightarrow {v} \cdot \hat {w}) \hat {w}\) can be messy. We present two other formulas that are often used in practice.

The proof of Theorem 5, which we leave to the reader as an exercise, amounts to using the formula \(\hat {w} = \left (\frac {1}{\| \overrightarrow {w} \|}\right ) \overrightarrow {w}\) and properties of the dot product. It is time for an example.

In Example 4 above, writing \(\overrightarrow {v} = \overrightarrow {p} + \overrightarrow {q}\) is an example of what is called a vector decomposition of \(\overrightarrow {v}\). We generalize this result in the following theorem.

If the vectors \(\overrightarrow {p}\) and \(\overrightarrow {q}\) in Theorem 6 are nonzero, then we can say \(\overrightarrow {p}\) is ‘parallel’ ^[5] (See Exercise ?? in Section ??.) to \(\overrightarrow {w}\) and \(\overrightarrow {q}\) is ‘orthogonal’ to \(\overrightarrow {w}\). In this case, the vector \(\overrightarrow {p}\) is sometimes called the ‘vector component of \(\overrightarrow {v}\) parallel to \(\overrightarrow {w}\)’ and \(\overrightarrow {q}\) is called the ‘vector component of \(\overrightarrow {v}\) orthogonal to \(\overrightarrow {w}\).’

To prove Theorem 6, we take \(\overrightarrow {p} = \text {proj}_{\overrightarrow {w}}(\overrightarrow {v})\) and \(\overrightarrow {q} = \overrightarrow {v} - \overrightarrow {p}\). Then \(\overrightarrow {p}\) is, by definition, a scalar multiple of \(\overrightarrow {w}\). Next, we compute \(\overrightarrow {q} \cdot \overrightarrow {w}\).

Hence, \(\overrightarrow {q} \cdot \overrightarrow {w} = 0\), as required. At this point, we have shown that the vectors \(\overrightarrow {p}\) and \(\overrightarrow {q}\) guaranteed by Theorem 6 exist. Now we need to show that they are unique - that is, there is only one such way to decompose \(\overrightarrow {v}\) in the manner described in Theorem 6.

Suppose \(\overrightarrow {v} = \overrightarrow {p} + \overrightarrow {q} = \overrightarrow {p} \,' + \overrightarrow {q} \,'\) where the vectors \(\overrightarrow {p} \,'\) and \(\overrightarrow {q} \,'\) satisfy the same properties described in Theorem 6 as \(\overrightarrow {p}\) and \(\overrightarrow {q}\). Then \(\overrightarrow {p} - \overrightarrow {p} \,' = \overrightarrow {q} \,' - \overrightarrow {q}\), so \(\overrightarrow {w} \cdot (\overrightarrow {p} - \overrightarrow {p} \,') = \overrightarrow {w} \cdot (\overrightarrow {q} \,' - \overrightarrow {q}) = \overrightarrow {w} \cdot \overrightarrow {q} \,' - \overrightarrow {w} \cdot \overrightarrow {q} = 0 - 0 = 0\). The long and short of this computation is that \(\overrightarrow {w} \cdot (\overrightarrow {p} - \overrightarrow {p} \,') = 0\).

Now there are scalars \(k\) and \(k \,'\) so that \(\overrightarrow {p} = k \overrightarrow {w}\) and \(\overrightarrow {p} \,' = k\,'\overrightarrow {w}\). This means \(\overrightarrow {w} \cdot (\overrightarrow {p} - \overrightarrow {p} \,') = \overrightarrow {w} \cdot ( k \overrightarrow {w} - k \,' \overrightarrow {w}) = \overrightarrow {w} \cdot ([k - k \,'] \overrightarrow {w}) = (k - k \,') (\overrightarrow {w} \cdot \overrightarrow {w}) = (k - k \,') \| \overrightarrow {w} \|^2\).

Since \(\overrightarrow {w} \neq \overrightarrow {0}\), \(\| \overrightarrow {w} \|^2 \neq 0\), which means the only way \(\overrightarrow {w} \cdot (\overrightarrow {p} - \overrightarrow {p} \,') = (k - k \,') \| \overrightarrow {w} \|^2 = 0\) is for \(k - k \,' = 0\), or \(k = k \,'\). This means \(\overrightarrow {p} = k \overrightarrow {w} = k \,' \overrightarrow {w} = \overrightarrow {p} \,'\). Since \(\overrightarrow {q} \,' - \overrightarrow {q} = \overrightarrow {p} - \overrightarrow {p} \,' = \overrightarrow {p} - \overrightarrow {p} = \overrightarrow {0}\), it must be that \(\overrightarrow {q} \,' = \overrightarrow {q}\) as well.

Hence, we have shown there is only one way to write \(\overrightarrow {v}\) as a sum of vectors as described in Theorem 6, so the decomposition listed there is unique.

We close this section with an application of the dot product. In Physics, if a constant force \(F\) is exerted over a distance \(d\), the work \(W\) done by the force is given by \(W = Fd\). Here, the assumption is that the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done.

Consider the scenario sketched below in which the constant force \(\overrightarrow {F}\) is applied to move an object from the point \(P\) to the point \(Q\). Here the force is being applied at an angle \(\theta \) as opposed to being applied directly in the direction of the motion.

To find the work \(W\) done in this scenario, we need to find how much of the force \(\overrightarrow {F}\) is in the direction of the motion \(\overrightarrow {PQ}\). This is precisely what the dot product \(\overrightarrow {F} \cdot \widehat {PQ}\) represents.

Since the distance the object travels is \(\| \overrightarrow {PQ} \|\), we get \(W = (\overrightarrow {F} \cdot \widehat {PQ}) \| \overrightarrow {PQ} \|\). Since \(\overrightarrow {PQ} = \|\overrightarrow {PQ}\| \widehat {PQ}\), we can simplify this formula as follows: \(W = (\overrightarrow {F} \cdot \widehat {PQ}) \| \overrightarrow {PQ} \| = \overrightarrow {F} \cdot ( \| \overrightarrow {PQ} \|\widehat {PQ} ) = \overrightarrow {F} \cdot \overrightarrow {PQ}\).

Using Theorem 2, we can rewrite \(W = \overrightarrow {F} \cdot \overrightarrow {PQ} = \| \overrightarrow {F} \| \| \overrightarrow {PQ} \| \cos (\theta )\), where \(\theta \) is the angle between the applied force \(\overrightarrow {F}\) and the trajectory of the motion \(\overrightarrow {PQ}\). We have proved the following.

We test out our formula for work in the following example.