Products of Vectors

The Scalar Product
For the vectors shown in (Figure), find the scalar product [latex]\stackrel{\to }{A}·\stackrel{\to }{F}[/latex].

Strategy
From (Figure), the magnitudes of vectors [latex]\stackrel{\to }{A}[/latex] and [latex]\stackrel{\to }{F}[/latex] are A = 10.0 and F = 20.0. Angle [latex]\theta[/latex], between them, is the difference: [latex]\theta =\phi -\alpha =110\text{°}-35\text{°}=75\text{°}[/latex]. Substituting these values into (Figure) gives the scalar product.

Solution
A straightforward calculation gives us

Angle between Two Forces
Three dogs are pulling on a stick in different directions, as shown in (Figure). The first dog pulls with force [latex]{\stackrel{\to }{F}}_{1}=\left(10.0\stackrel{^}{i}-20.4\stackrel{^}{j}+2.0\stackrel{^}{k}\right)\text{N}[/latex], the second dog pulls with force [latex]{\stackrel{\to }{F}}_{2}=\left(-15.0\stackrel{^}{i}-6.2\stackrel{^}{k}\right)\text{N}[/latex], and the third dog pulls with force [latex]{\stackrel{\to }{F}}_{3}=\left(5.0\stackrel{^}{i}+12.5\stackrel{^}{j}\right)\text{N}[/latex]. What is the angle between forces [latex]{\stackrel{\to }{F}}_{1}[/latex] and [latex]{\stackrel{\to }{F}}_{2}[/latex]?

Strategy
The components of force vector [latex]{\stackrel{\to }{F}}_{1}[/latex] are [latex]{F}_{1x}=10.0\phantom{\rule{0.2em}{0ex}}\text{N}[/latex], [latex]{F}_{1y}=-20.4\phantom{\rule{0.2em}{0ex}}\text{N}[/latex], and [latex]{F}_{1z}=2.0\phantom{\rule{0.2em}{0ex}}\text{N}[/latex], whereas those of force vector [latex]{\stackrel{\to }{F}}_{2}[/latex] are [latex]{F}_{2x}=-15.0\phantom{\rule{0.2em}{0ex}}\text{N}[/latex], [latex]{F}_{2y}=0.0\phantom{\rule{0.2em}{0ex}}\text{N}[/latex], and [latex]{F}_{2z}=-6.2\phantom{\rule{0.2em}{0ex}}\text{N}[/latex]. Computing the scalar product of these vectors and their magnitudes, and substituting into (Figure) gives the angle of interest.

Solution
The magnitudes of forces [latex]{\stackrel{\to }{F}}_{1}[/latex] and [latex]{\stackrel{\to }{F}}_{2}[/latex] are

and

Substituting the scalar components into (Figure) yields the scalar product

Finally, substituting everything into (Figure) gives the angle

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 Work of a Force
When force [latex]\stackrel{\to }{F}[/latex] pulls on an object and when it causes its displacement [latex]\stackrel{\to }{D}[/latex], we say the force performs work. The amount of work the force does is the scalar product [latex]\stackrel{\to }{F}·\stackrel{\to }{D}[/latex]. If the stick in (Figure) moves momentarily and gets displaced by vector [latex]\stackrel{\to }{D}=\left(-7.9\stackrel{^}{j}-4.2\stackrel{^}{k}\right)\phantom{\rule{0.2em}{0ex}}\text{cm}[/latex], how much work is done by the third dog in (Figure)?

Strategy
We compute the scalar product of displacement vector [latex]\stackrel{\to }{D}[/latex] with force vector [latex]{\stackrel{\to }{F}}_{3}=\left(5.0\stackrel{^}{i}+12.5\stackrel{^}{j}\right)\text{N}[/latex], which is the pull from the third dog. Let’s use [latex]{W}_{3}[/latex] to denote the work done by force [latex]{\stackrel{\to }{F}}_{3}[/latex] on displacement [latex]\stackrel{\to }{D}[/latex].

Solution
Calculating the work is a straightforward application of the dot product:

Significance
The SI unit of work is called the joule [latex]\left(\text{J}\right)[/latex], where 1 J = 1 [latex]\text{N}·\text{m}[/latex]. The unit [latex]\text{cm}·\text{N}[/latex] can be written as [latex]{10}^{-2}\text{m}·\text{N}={10}^{-2}\text{J}[/latex], so the answer can be expressed as [latex]{W}_{3}=-0.9875\phantom{\rule{0.2em}{0ex}}\text{J}\approx -1.0\phantom{\rule{0.2em}{0ex}}\text{J}[/latex].

The Torque of a ForceThe mechanical advantage that a familiar tool called a wrench provides ((Figure)) 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 [latex]\stackrel{\to }{F}[/latex] is attached and is represented by the radial vector [latex]\stackrel{\to }{R}[/latex]. The physical vector quantity that makes the nut turn is called torque (denoted by [latex]\stackrel{\to }{\tau }\right)[/latex], and it is the vector product of the distance between the pivot to force with the force: [latex]\stackrel{\to }{\tau }=\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex].

To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle [latex]\phi =40\text{°}[/latex] and at a distance of 0.25 m from the nut, as shown in (Figure)(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 [latex]\phi =45\text{°}[/latex], as shown in (Figure)(b)? For what value of angle [latex]\phi[/latex] does the torque have the largest magnitude?

Strategy
We adopt the frame of reference shown in (Figure), where vectors [latex]\stackrel{\to }{R}[/latex] and [latex]\stackrel{\to }{F}[/latex] lie in the xy-plane and the origin is at the position of the nut. The radial direction along vector [latex]\stackrel{\to }{R}[/latex] (pointing away from the origin) is the reference direction for measuring the angle [latex]\phi[/latex] because [latex]\stackrel{\to }{R}[/latex] is the first vector in the vector product [latex]\stackrel{\to }{\tau }=\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex]. Vector [latex]\stackrel{\to }{\tau }[/latex] must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both [latex]\stackrel{\to }{R}[/latex] and [latex]\stackrel{\to }{F}[/latex] lie. To compute the magnitude [latex]\tau[/latex], we use (Figure). To find the direction of [latex]\stackrel{\to }{\tau }[/latex], we use the corkscrew right-hand rule ((Figure)).

Solution
For the situation in (a), the corkscrew rule gives the direction of [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex] in the positive direction of the z-axis. Physically, it means the torque vector [latex]\stackrel{\to }{\tau }[/latex] 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 (Figure):

For the situation in (b), the corkscrew rule gives the direction of [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex] in the negative direction of the z-axis. Physically, it means the vector [latex]\stackrel{\to }{\tau }[/latex] points into the page, perpendicular to the wrench handle. The magnitude of this torque is

The torque has the largest value when [latex]\text{sin}\phantom{\rule{0.2em}{0ex}}\phi =1[/latex], which happens when [latex]\phi =90\text{°}[/latex]. 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 [latex]{\tau }_{\text{best}}=RF=\left(0.25\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(20.00\phantom{\rule{0.2em}{0ex}}\text{N}\right)=5.00\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}[/latex].

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 [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex] lies along the z-axis, we can write this vector in terms of the unit vector [latex]\stackrel{^}{k}[/latex] of the z-axis:

In this equation, the number that multiplies [latex]\stackrel{^}{k}[/latex] is the scalar z-component of the vector [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}[/latex]. In the computation of this component, care must be taken that the angle [latex]\phi[/latex] is measured counterclockwise from [latex]\stackrel{\to }{R}[/latex] (first vector) to [latex]\stackrel{\to }{F}[/latex] (second vector). Following this principle for the angles, we obtain [latex]RF\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\left(+40\text{°}\right)=+3.2\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}[/latex] for the situation in (a), and we obtain [latex]RF\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\left(-45\text{°}\right)=-3.5\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}[/latex] for the situation in (b). In the latter case, the angle is negative because the graph in (Figure) indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because [latex]+\left(360\text{°}-45\text{°}\right)=+315\text{°}[/latex] and [latex]\text{sin}\phantom{\rule{0.2em}{0ex}}\left(+315\text{°}\right)=\text{sin}\phantom{\rule{0.2em}{0ex}}\left(-45\text{°}\right)[/latex]. In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}=+3.2\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\stackrel{^}{k}[/latex]; for the situation in (b), the solution is [latex]\stackrel{\to }{R}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}=-3.5\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\stackrel{^}{k}[/latex].

A Particle in a Magnetic Field
When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field [latex]\stackrel{\to }{B}[/latex] is a vector, the magnetic force [latex]\stackrel{\to }{F}[/latex] is a vector, and the velocity [latex]\stackrel{\to }{u}[/latex] of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we express as [latex]\stackrel{\to }{F}=\zeta \stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}[/latex]. In this equation, a constant [latex]\zeta[/latex] takes care of the consistency in physical units, so we can omit physical units on vectors [latex]\stackrel{\to }{u}[/latex] and [latex]\stackrel{\to }{B}[/latex]. In this example, let’s assume the constant [latex]\zeta[/latex] is positive.

A particle moving in space with velocity vector [latex]\stackrel{\to }{u}=-5.0\stackrel{^}{i}-2.0\stackrel{^}{j}+3.5\stackrel{^}{k}[/latex] enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force [latex]\stackrel{\to }{F}[/latex] on this particle at the entry point to the region where the magnetic field vector is (a) [latex]\stackrel{\to }{B}=7.2\stackrel{^}{i}-\stackrel{^}{j}-2.4\stackrel{^}{k}[/latex] and (b) [latex]\stackrel{\to }{B}=4.5\stackrel{^}{k}[/latex]. In each case, find magnitude F of the magnetic force and angle [latex]\theta[/latex] the force vector [latex]\stackrel{\to }{F}[/latex] makes with the given magnetic field vector [latex]\stackrel{\to }{B}[/latex].

Strategy
First, we want to find the vector product [latex]\stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}[/latex], because then we can determine the magnetic force using [latex]\stackrel{\to }{F}=\zeta \stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}[/latex]. Magnitude F can be found either by using components, [latex]F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}+{F}_{z}^{2}}[/latex], or by computing the magnitude [latex]|\stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}|[/latex] directly using (Figure). In the latter approach, we would have to find the angle between vectors [latex]\stackrel{\to }{u}[/latex] and [latex]\stackrel{\to }{B}[/latex]. When we have [latex]\stackrel{\to }{F}[/latex], the general method for finding the direction angle [latex]\theta[/latex] involves the computation of the scalar product [latex]\stackrel{\to }{F}·\stackrel{\to }{B}[/latex] and substitution into (Figure). To compute the vector product we can either use (Figure) or compute the product directly, whichever way is simpler.

Solution
The components of the velocity vector are [latex]{u}_{x}=-5.0[/latex], [latex]{u}_{y}=-2.0[/latex], and [latex]{u}_{z}=3.5[/latex].

(a) The components of the magnetic field vector are [latex]{B}_{x}=7.2[/latex], [latex]{B}_{y}=-1.0[/latex], and [latex]{B}_{z}=-2.4[/latex]. Substituting them into (Figure) gives the scalar components of vector [latex]\stackrel{\to }{F}=\zeta \stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}[/latex]:

Thus, the magnetic force is [latex]\stackrel{\to }{F}=\zeta \left(8.3\stackrel{^}{i}+13.2\stackrel{^}{j}+19.4\stackrel{^}{k}\right)[/latex] and its magnitude is

To compute angle [latex]\theta[/latex], we may need to find the magnitude of the magnetic field vector,

and the scalar product [latex]\stackrel{\to }{F}·\stackrel{\to }{B}[/latex]:

Now, substituting into (Figure) gives angle [latex]\theta[/latex]:

Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some time if we had computed the scalar product earlier.)

(b) Because vector [latex]\stackrel{\to }{B}=4.5\stackrel{^}{k}[/latex] has only one component, we can perform the algebra quickly and find the vector product directly:

The magnitude of the magnetic force is

Because the scalar product is

the magnetic force vector [latex]\stackrel{\to }{F}[/latex] is perpendicular to the magnetic field vector [latex]\stackrel{\to }{B}[/latex].

Significance
Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic force vector is the vector product [latex]\stackrel{\to }{F}=\zeta \stackrel{\to }{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}[/latex] and, by the definition of the vector product (see (Figure)), vector [latex]\stackrel{\to }{F}[/latex] must be perpendicular to both vectors [latex]\stackrel{\to }{u}[/latex] and [latex]\stackrel{\to }{B}[/latex].