2.4 Products of Vectors

The Scalar Product of Two Vectors (the Dot Product)

Scalar multiplication of two vectors yields a scalar product.

In the definition of the dot product, the direction of angle $\phi$ does not matter, and $\phi$ can be measured from either of the two vectors to the other because $\text{cos}\phi =\text{cos}(\text{−}\phi )=\text{cos}(2\pi -\phi )$. The dot product is a negative number when $90\text{°}<\phi \le 180\text{°}$ and is a positive number when $0\text{°}\le \phi <90\text{°}$. Moreover, the dot product of two parallel vectors is $\overrightarrow{A}·\overrightarrow{B}=AB\text{cos}0\text{°}=AB$, and the dot product of two antiparallel vectors is $\overrightarrow{A}·\overrightarrow{B}=AB\text{cos}180\text{°}=\text{−}AB$. The scalar product of two orthogonal vectors vanishes: $\overrightarrow{A}·\overrightarrow{B}=AB\text{cos}90\text{°}=0$. The scalar product of a vector with itself is the square of its magnitude:

Figure 2.27 The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection $A_{││}$ of vector $\overrightarrow{A}$ onto the direction of vector $\overrightarrow{B}$. (c) The orthogonal projection $B_{││}$ of vector $\overrightarrow{B}$ onto the direction of vector $\overrightarrow{A}$.

In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanish because these unit vectors are orthogonal:

In these equations, we use the fact that the magnitudes of all unit vectors are one: $\left|\widehat{i}\right|=\left|\widehat{j}\right|=\left|\widehat{k}\right|=1$. For unit vectors of the axes, Equation 2.28 gives the following identities:

The scalar product $\overrightarrow{A}·\overrightarrow{B}$ can also be interpreted as either the product of B with the projection $A_{ǁ}$ of vector $\overrightarrow{A}$ onto the direction of vector $\overrightarrow{B}$ (Figure 2.27(b)) or the product of A with the projection $B_{ǁ}$ of vector $\overrightarrow{B}$ onto the direction of vector $\overrightarrow{A}$ (Figure 2.27(c)):

For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the unit vector $\widehat{i}$, and the scalar y-component of a vector is its dot product with the unit vector $\widehat{j}$:

Scalar multiplication of vectors is commutative,

and obeys the distributive law:

We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot product of two vectors in terms of their scalar components.

When the vectors in Equation 2.27 are given in their vector component forms,

we can compute their scalar product as follows:

Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselves give one (see Equation 2.29 and Equation 2.30), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to

We can use Equation 2.33 for the scalar product in terms of scalar components of vectors to find the angle between two vectors. When we divide Equation 2.27 by AB, we obtain the equation for $\text{cos}\phi$, into which we substitute Equation 2.33:

Angle $\phi$ between vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is obtained by taking the inverse cosine of the expression in Equation 2.34.

Example 2.16

Angle between Two Forces

Three dogs are pulling on a stick in different directions, as shown in Figure 2.28. The first dog pulls with force ${\overrightarrow{F}}_1=(10.0\widehat{i}-20.4\widehat{j}+2.0\widehat{k})\text{N}$, the second dog pulls with force ${\overrightarrow{F}}_2=(\mathrm{-15.0}\widehat{i}-6.2\widehat{k})\text{N}$, and the third dog pulls with force ${\overrightarrow{F}}_3=(5.0\widehat{i}+12.5\widehat{j})\text{N}$. What is the angle between forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$?

Figure 2.28 Three dogs are playing with a stick.

Strategy

The components of force vector ${\overrightarrow{F}}_1$ are $F_{1x}=10.0\text{N}$, $F_{1y}=\mathrm{-20.4}\text{N}$, and $F_{1z}=2.0\text{N}$, whereas those of force vector ${\overrightarrow{F}}_2$ are $F_{2x}=\mathrm{-15.0}\text{N}$, $F_{2y}=0.0\text{N}$, and $F_{2z}=\mathrm{-6.2}\text{N}$. Computing the scalar product of these vectors and their magnitudes, and substituting into Equation 2.34 gives the angle of interest.

Solution

The magnitudes of forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ are

$$F_1=\sqrt{F_{1x}^2+F_{1y}^2+F_{1z}^2}=\sqrt{{10.0}^2+{20.4}^2+{2.0}^2}\text{N}=22.8\text{N}$$

and

$$F_2=\sqrt{F_{2x}^2+F_{2y}^2+F_{2z}^2}=\sqrt{{15.0}^2+{6.2}^2}\text{N}=16.2\text{N}.$$

Substituting the scalar components into Equation 2.33 yields the scalar product

$$\begin{array}{cc}\hfill {\overrightarrow{F}}_1·{\overrightarrow{F}}_2& =F_{1x}{F}_{2x}+F_{1y}{F}_{2y}+F_{1z}{F}_{2z}\hfill \\ & =(10.0\text{N})(\mathrm{-15.0}\text{N})+(\mathrm{-20.4}\text{N})(0.0\text{N})+(2.0\text{N})(\mathrm{-6.2}\text{N})\hfill \\ & =\mathrm{-162.4}{\text{N}}^2.\hfill \end{array}$$

Finally, substituting everything into Equation 2.34 gives the angle

$$\text{cos}\phi =\frac{{\overrightarrow{F}}_1·{\overrightarrow{F}}_2}{F_1{F}_2}=\frac{\mathrm{-162.4}{\text{N}}^2}{(22.8\text{N})(16.2\text{N})}=\mathrm{-0.439}\Rightarrow \phi ={\text{cos}}^{\mathrm{-1}}(\mathrm{-0.439})=116.0\text{°}.$$

Significance

Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the +x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces in the problem is the same if the +x-direction is to the west and the +y-direction is to the south.

The Vector Product of Two Vectors (the Cross Product)

Vector multiplication of two vectors yields a vector product.

According to Equation 2.35, the vector product vanishes for pairs of vectors that are either parallel $\left(\phi =0\text{°}\right)$ or antiparallel $\left(\phi =180\text{°}\right)$ because $\text{sin}0\text{°}=\text{sin}180\text{°}=0$.

Figure 2.29 The vector product of two vectors is drawn in three-dimensional space. (a) The vector product $\overrightarrow{A}\times \overrightarrow{B}$ is a vector perpendicular to the plane that contains vectors $\overrightarrow{A}$ and $\overrightarrow{B}$. Small squares drawn in perspective mark right angles between $\overrightarrow{A}$ and $\overrightarrow{C}$, and between $\overrightarrow{B}$ and $\overrightarrow{C}$ so that if $\overrightarrow{A}$ and $\overrightarrow{B}$ lie on the floor, vector $\overrightarrow{C}$ points vertically upward to the ceiling. (b) The vector product $\overrightarrow{B}\times \overrightarrow{A}$ is a vector antiparallel to vector $\overrightarrow{A}\times \overrightarrow{B}$.

On the line perpendicular to the plane that contains vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ there are two alternative directions—either up or down, as shown in Figure 2.29—and the direction of the vector product may be either one of them. In the standard right-handed orientation, where the angle between vectors is measured counterclockwise from the first vector, vector $\overrightarrow{A}\times \overrightarrow{B}$ points upward, as seen in Figure 2.29(a). If we reverse the order of multiplication, so that now $\overrightarrow{B}$ comes first in the product, then vector $\overrightarrow{B}\times \overrightarrow{A}$ must point downward, as seen in Figure 2.29(b). This means that vectors $\overrightarrow{A}\times \overrightarrow{B}$ and $\overrightarrow{B}\times \overrightarrow{A}$ are antiparallel to each other and that vector multiplication is not commutative but anticommutative. The anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:

The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown in Figure 2.30, a corkscrew is placed in a direction perpendicular to the plane that contains vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, and its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew.

Figure 2.30 The corkscrew right-hand rule can be used to determine the direction of the cross product $\overrightarrow{A}\times \overrightarrow{B}$. Place a corkscrew in the direction perpendicular to the plane that contains vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, and turn it in the direction from the first to the second vector in the product. The direction of the cross product is given by the progression of the corkscrew. (a) Upward movement means the cross-product vector points up. (b) Downward movement means the cross-product vector points downward.

Example 2.18

The Torque of a Force

The mechanical advantage that a familiar tool called a wrench provides (Figure 2.31) depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is applied. The distance R from the nut to the point where force vector $\overrightarrow{F}$ is attached is represented by the radial vector $\overrightarrow{R}$. The physical vector quantity that makes the nut turn is called torque (denoted by $\overrightarrow{\tau })$, and it is the vector product of the distance between the pivot to force with the force: $\overrightarrow{\tau }=\overrightarrow{R}\times \overrightarrow{F}$.

To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle $\phi =40\text{°}$ and at a distance of 0.25 m from the nut, as shown in Figure 2.31(a). Find the magnitude and direction of the torque applied to the nut. What would the magnitude and direction of the torque be if the force were applied at angle $\phi =45\text{°}$, as shown in Figure 2.31(b)? For what value of angle $\phi$ does the torque have the largest magnitude?

Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turn counterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.

Strategy

We adopt the frame of reference shown in Figure 2.31, where vectors $\overrightarrow{R}$ and $\overrightarrow{F}$ lie in the xy-plane and the origin is at the position of the nut. The radial direction along vector $\overrightarrow{R}$ (pointing away from the origin) is the reference direction for measuring the angle $\phi$ because $\overrightarrow{R}$ is the first vector in the vector product $\overrightarrow{\tau }=\overrightarrow{R}\times \overrightarrow{F}$. Vector $\overrightarrow{\tau }$ must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both $\overrightarrow{R}$ and $\overrightarrow{F}$ lie. To compute the magnitude $\tau$, we use Equation 2.35. To find the direction of $\overrightarrow{\tau }$, we use the corkscrew right-hand rule (Figure 2.30).

Solution

For the situation in (a), the corkscrew rule gives the direction of $\overrightarrow{R}\times \overrightarrow{F}$ in the positive direction of the z-axis. Physically, it means the torque vector $\overrightarrow{\tau }$ points out of the page, perpendicular to the wrench handle. We identify F = 20.00 N and R = 0.25 m, and compute the magnitude using Equation 2.35:

$$\tau =|\overrightarrow{R}\times \overrightarrow{F}|=RF\text{sin}\phi =(0.25\text{m})(20.00\text{N})\text{sin}40\text{°}=3.21\text{N}·\text{m}.$$

For the situation in (b), the corkscrew rule gives the direction of $\overrightarrow{R}\times \overrightarrow{F}$ in the negative direction of the z-axis. Physically, it means the vector $\overrightarrow{\tau }$ points into the page, perpendicular to the wrench handle. The magnitude of this torque is

$$\tau =|\overrightarrow{R}\times \overrightarrow{F}|=RF\text{sin}\phi =(0.25\text{m})(20.00\text{N})\text{sin}45\text{°}=3.53\text{N}·\text{m}.$$

The torque has the largest value when $\text{sin}\phi =1$, which happens when $\phi =90\text{°}$. Physically, it means the wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicular to the wrench handle. For the situation in this example, this best-torque value is ${\tau }_{\text{best}}=RF=(0.25\text{m})(20.00\text{N})=5.00\text{N}·\text{m}$.

Significance

When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in the following equivalent solution. Notice that once we have identified that vector $\overrightarrow{R}\times \overrightarrow{F}$ lies along the z-axis, we can write this vector in terms of the unit vector $\widehat{k}$ of the z-axis:

$$\overrightarrow{R}\times \overrightarrow{F}=RF\text{sin}\phi \widehat{k}.$$

In this equation, the number that multiplies $\widehat{k}$ is the scalar z-component of the vector $\overrightarrow{R}\times \overrightarrow{F}$. In the computation of this component, care must be taken that the angle $\phi$ is measured counterclockwise from $\overrightarrow{R}$ (first vector) to $\overrightarrow{F}$ (second vector). Following this principle for the angles, we obtain $RF\text{sin}(+40\text{°})=+3.2\text{N}·\text{m}$ for the situation in (a), and we obtain $RF\text{sin}(\mathrm{-45}\text{°})=\mathrm{-3.5}\text{N}·\text{m}$ for the situation in (b). In the latter case, the angle is negative because the graph in Figure 2.31 indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because $+(360\text{°}-45\text{°})=+315\text{°}$ and $\text{sin}(+315\text{°})=\text{sin}(\mathrm{-45}\text{°})$. In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is $\overrightarrow{R}\times \overrightarrow{F}=+3.2\text{N}·\text{m}\widehat{k}$; for the situation in (b), the solution is $\overrightarrow{R}\times \overrightarrow{F}=\mathrm{-3.5}\text{N}·\text{m}\widehat{k}$.

Similar to the dot product (Equation 2.32), the cross product has the following distributive property: