Coordinated Multivariable Calculus

Subsection1.3.1The Dot Product

The definition of the dot product for vectors in \(\R^2\) given in Preview Activity1.3.1 can be extended to vectors in \(\R^n\text{.}\)

For instance, we find that

Notice that the resulting quantity is a scalar. Our work in Preview Activity1.3.1 examined dot products of two-dimensional vectors.

The dot product is a natural way to define a product of two vectors. In addition, it behaves in ways that are similar to the product of, say, real numbers.

Moreover, the dot product gives us valuable geometric information about the vectors and their relative orientation. For instance, let's consider what happens when we dot a vector with itself:

In other words, the dot product of a vector with itself gives the square of the length of the vector: \(\vu\cdot\vu=|\vu|^2\text{.}\)

Subsection1.3.2The angle between vectors

The dot product can help us understand the angle between two vectors. For instance, if we are given two non-zero vectors \(\vu\) and \(\vv\text{,}\) there are two angles that these vectors create, as depicted at left in Figure1.3.5. We will call \(\theta\text{,}\) the smaller of these angles, the angle between these vectors. Notice that \(\theta\) lies between 0 and \(\pi\text{.}\)

It is sometimes useful to think of Equation(1.3.1) as giving us an expression for the angle between two vectors:

The real beauty of this expression is this: the dot product is a very simple algebraic operation to perform yet it provides us with important geometric information namely the angle between the vectors that would be difficult to determine otherwise.

Subsection1.3.3The Dot Product and Orthogonality

When the angle between two vectors is a right angle, it is frequently the case that something important is happening. In this case, we say the vectors are orthogonal. For instance, orthogonality often plays a role in optimization problems; to determine the shortest path from a point in \(\R^3\) to a given plane, we move along a line orthogonal to the plane.

As Activity1.3.4 indicates, the dot product provides a simple means to determine whether two vectors are orthogonal to one another. In this case, \(\vu\cdot\vv=|\vu||\vv|\cos(\pi/2) = 0\text{,}\) so we make the following important observation.

More generally, the sign of the dot product gives us useful information about the relative orientation of the vectors. If we remember that

we see that for nonzero vectors \(\vu\) and \(\vv\text{,}\)

This is illustrated in Figure1.3.6.

Subsection1.3.4Work, Force, and Displacement

In physics, work is a measure of the energy required to apply a force to an object through a displacement. For instance, Figure1.3.7 shows a force \(\vF\) displacing an object from point \(A\) to point \(B\text{.}\) The displacement is then represented by the vector \(\overrightarrow{AB}\text{.}\)

It turns out that the work required to displace the object is

This means that the work is determined only by the magnitude of the force applied parallel to the displacement. Consequently, if we are given two vectors \(\vu\) and \(\vv\text{,}\) we would like to write \(\vu\) as a sum of two vectors, one of which is parallel to \(\vv\) and one of which is orthogonal to \(\vv\text{.}\) We take up this task after the next activity.

Subsection1.3.5Projections

Suppose we are given two non-zero vectors \(\vu\) and \(\vv\) as shown at left in Figure1.3.8. Motivated by our discussion of work, we would like to write \(\vu\) as a sum of two vectors, one of which is parallel to \(\vv\) and one of which is orthogonal. That is, we would like to write

where \(\proj_{\vv}\vu\) is parallel to \(\vv\) and \(\proj_{\perp\vv}\vu\) is orthogonal to \(\vv\text{.}\) We call the vector \(\proj_{\vv}\vu\) the projection of \(\vu\) onto \(\vv\). Note that, as the diagram at right in Figure1.3.8 illustrates, it is also possible to create a projection even if the angle between the vectors \(\vu\) and \(\vv\) exceeds \(\frac\pi2\text{.}\)

To find the vector \(\proj_{\vv} \vu\text{,}\) we will dot both sides of Equation(1.3.2) with the vector \(\vv\text{,}\) to find that

Notice that \((\proj_{\perp\vv}\vu)\cdot\vv = 0\) since \(\proj_{\perp\vv}\vu\) is orthogonal to \(\vv\text{.}\) Also, \(\proj_{\vv}\vu\) must be a scalar multiple of \(\vv\) since it is parallel to \(\vv\text{,}\) so we will write \(\proj_{\vv}\vu = s\vv\text{.}\) It follows that

which means that

and hence

It is sometimes useful to write \(\proj_{\vv}\vu\) as a scalar times a unit vector in the direction of \(\vv\text{.}\) We call this scalar the component of \(\vu\) along \(\vv\) and denote it as \(\comp_{\vv}\vu\text{.}\) We therefore have

so that

Moreover, since

it follows that

This shows that once we have computed \(\proj_{\vv} \vu\text{,}\) we can find \(\proj_{\perp \vv} \vu\) simply by calculating the difference of two known vectors.

Subsection1.3.6Summary

• The dot product of two vectors in \(\R^n\text{,}\) \(\vu = \langle u_1, u_2, \ldots, u_n \rangle\) and \(\vv = \langle v_1, v_2, \ldots, v_n \rangle\text{,}\) is the scalar

  \begin{equation*} \vu \cdot \vv = u_1v_1 + u_2v_2 + \cdots + u_nv_n. \end{equation*}

• The dot product is related to the length of a vector since \(\vu\cdot\vu = |\vu|^2\text{.}\)

• The dot product provides us with information about the angle between the vectors since

  \begin{equation*} \vu\cdot\vv = |\vu| \ |\vv|\cos(\theta), \end{equation*}

  where \(\theta\) is the angle between \(\vu\) and \(\vv\text{.}\)

• Two vectors are orthogonal if the angle between them is \(\pi/2\text{.}\) In terms of the dot product, the non-zero vectors \(\vu\) and \(\vv\) are orthogonal if and only if \(\vu \cdot \vv = 0\text{.}\)

• The projection of a vector \(\vu\) in \(\R^n\) onto a non-zero vector \(\vv\) in \(\R^n\) is the vector

  \begin{equation*} \proj_{\vv} \vu = \frac{\vu \cdot \vv}{\vv\cdot\vv} \vv. \end{equation*}