2.4 Products of vectors  (Page 2/16)

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, [link] gives the following identities:

The scalar product $\overrightarrow{A}·\overrightarrow{B}$ can also be interpreted as either the product of B with the orthogonal projection $A_{\perp }$ of vector $\overrightarrow{A}$ onto the direction of vector $\overrightarrow{B}$ ( [link] (b)) or the product of A with the orthogonal projection $B_{\perp }$ of vector $\overrightarrow{B}$ onto the direction of vector $\overrightarrow{A}$ ( [link] (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 [link] 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 [link] and [link] ), there are only three nonzero terms in this expression. Thus, the scalar product simplifies to

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

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

Angle between two forces

Three dogs are pulling on a stick in different directions, as shown in [link] . 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 ${\overrightarrow{F}}_2$ ?

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 [link] 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 [link] 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 [link] 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.

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