2.4 Products of vectors  (Page 5/16)

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

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 [link] 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 ( [link] ), the cross product has the following distributive property:

The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectors of Cartesian axes.

When we apply the definition of the cross product, [link] , to unit vectors $\widehat{i}$ , $\widehat{j}$ , and $\widehat{k}$ that define the positive x -, y -, and z -directions in space, we find that

All other cross products of these three unit vectors must be vectors of unit magnitudes because $\widehat{i}$ , $\widehat{j}$ , and $\widehat{k}$ are orthogonal. For example, for the pair $\widehat{i}$ and $\widehat{j}$ , the magnitude is $|\widehat{i}\times \widehat{j}|=ij\text{sin}90\text{°}=(1)(1)(1)=1$ . The direction of the vector product $\widehat{i}\times \widehat{j}$ must be orthogonal to the xy -plane, which means it must be along the z -axis. The only unit vectors along the z -axis are $\text{−}\widehat{k}$ or $+\widehat{k}$ . By the corkscrew rule, the direction of vector $\widehat{i}\times \widehat{j}$ must be parallel to the positive z -axis. Therefore, the result of the multiplication $\widehat{i}\times \widehat{j}$ is identical to $+\widehat{k}$ . We can repeat similar reasoning for the remaining pairs of unit vectors. The results of these multiplications are

Notice that in [link] , the three unit vectors $\widehat{i}$ , $\widehat{j}$ , and $\widehat{k}$ appear in the cyclic order shown in a diagram in [link] (a). The cyclic order means that in the product formula, $\widehat{i}$ follows $\widehat{k}$ and comes before $\widehat{j}$ , or $\widehat{k}$ follows $\widehat{j}$ and comes before $\widehat{i}$ , or $\widehat{j}$ follows $\widehat{i}$ and comes before $\widehat{k}$ . The cross product of two different unit vectors is always a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplication is the remaining unit vector, as illustrated in [link] (b). When unit vectors in the cross product appear in a different order, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, as shown by the examples in [link] (c) and [link] (d). In practice, when the task is to find cross products of vectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.