3.3: Vectors
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The Use of Vectors in Physics
From the last section we have three important ideas about vectors:
- vectors can exist at any point P in space,
- vectors have direction and magnitude, and
- any two vectors that have the same direction and magnitude are equal no matter where in space they are located.
When we apply vectors to physical quantities it’s nice to keep in the back of our minds all these formal properties. However, from the physicist’s point of view, we are interested in representing physical quantities such as displacement, velocity, acceleration, force, impulse, and momentum as vectors. We can’t add force to velocity or subtract momentum from force. We must always understand the physical context for the vector quantity. Thus, instead of approaching vectors as formal mathematical objects we shall instead consider the following essential properties that enable us to represent physical quantities as vectors.
Vectors in Cartesian Coordinates
Vector Decomposition
Choose a coordinate system with an origin, axes, and unit vectors. We can decompose a vector into component vectors along each coordinate axis (Figure 3.14).
A vector →A at P can be decomposed into the vector sum,
→A=→Ax+→Ay+→Az
where →Ax is the x-component vector pointing in the positive or negative x-direction, →Ay is the y-component vector pointing in the positive or negative y-direction, and →Az is the z-component vector pointing in the positive or negative z-direction.
Vector Components
Once we have defined unit vectors (ˆi,ˆj,ˆk), we then define the components of a vector. Recall our vector decomposition, →A=→Ax+→Ay+→Az. We define the x-component vector, →Ax, as →Ax=Axˆi.In this expression the term Ax (without the arrow above) is called the x-component of the vector →Ax. The x-component Ax can be positive, zero, or negative. It is not the magnitude of →Ax which is given by (A2x)1/2. The x-component Ax is a scalar quantity and the x-component vector, →Ax is a vector. In a similar fashion we define the y-component, Ay , and the z-component, Az , of the vector →A according to
→Ay=Ayˆj,→Az=Azˆk.
A vector →A is represented by its three components (Ax,Ay,Az). Thus we need three numbers to describe a vector in three-dimensional space. We write the vector →A as
→A=Axˆi+Ayˆj+Azˆk
Magnitude
Using the Pythagorean theorem, the magnitude of →A is,A=√A2x+A2y+A2z
Direction
Let’s consider a vector →A=(Ax,Ay,0). Because the z-component is zero, the vector →A lies in the x−y plane. Let θ denote the angle that the vector →A makes in the counterclockwise direction with the positive x-axis (Figure 3.15).
Then the x -component and y -component are Ar=Acos(θ),Ay=Asin(θ)We now write a vector in the xy -plane as
→A=Acos(θ)ˆi+Asin(θ)ˆj
Once the components of a vector are known, the tangent of the angle θ can be determined by
AyAx=Asin(θ)Acos(θ)=tan(θ)and hence the angle θ is given by θ=tan−1(AyAx)
Clearly, the direction of the vector depends on the sign of Ax and Ay. For example, if both Ax>0 and Ay>0, then 0<θ<π/2. If Ax<0 and Ay>0 then π/2<θ<π. If Ax<0 and Ay<0 then π<θ<3π/2. If Ax>0 and Ay<0, then 3π/2<θ<2. Note that tanθ is a double valued function because
−Ay−Ax=AyAx, and Ay−Ax=−AyAx
Unit Vectors
Unit vector in the direction of →A: Let →A=Axˆi+Avˆj+A7ˆk. Let →A denote a unit vector in the direction of →A. Then, ˆA=→A|→A|=Axˆi+Ayˆj+Azˆk(A2x+A2y+A2z)1/2
Vector Addition
Let →A and →B be two vectors in the x−y plane. Let θA and θB denote the angles that the vectors →A and →B make (in the counterclockwise direction) with the positive x -axis. Then →A=Acos(θA)ˆi+Asin(θA)ˆj →B=Bcos(θB)ˆi+Bsin(θB)ˆj In Figure 3.16, the vector addition →C=→A+→B is shown. Let θC denote the angle that the vector →C makes with the positive x -axis.
From Figure 3.16, the components of →C are Cx=Ax+Bx,Cy=Ay+By In terms of magnitudes and angles, we have
Cx=Ccos(θC)=Acos(θA)+Bcos(θB)Cy=Csin(θC)=Asin(θA)+Bsin(θB) We can write the vector →C as →C=(Ax+Bx)ˆi+(Ay+By)ˆj=Ccos(θC)ˆi+Csin(θC)ˆj
Example 3.1: Vector Addition
Given two vectors, →A=2ˆi+−3ˆj+7ˆkand→B=5ˆi+ˆj+2ˆk, find: (a) |→A|; (b) |→B|; (c) →A+→B; (d) →A−→B; (e) a unit vector ˆA pointing in the direction of →A; (f) a unit vector ˆB pointing in the direction of →B;
Solution
(a)
|→A|=(22+(−3)2+72)1/2=√62=7.87.
(b)
|→B|=(52+12+22)1/2=√30=5.48.
(c)
→A+→B=(Ax+Bx)ˆi+(Ay+By)ˆj+(Az+Bz)ˆk=(2+5)ˆi+(−3+1)ˆj+(7+2)ˆk=7ˆi−2ˆj+9ˆk
(d)
→A−→B=(Ax−Bx)ˆi+(Ay−By)ˆj+(Az−Bz)ˆk=(2−5)ˆi+(−3−1)ˆj+(7−2)ˆk=−3ˆi−4ˆj+5ˆk
(e)
A unit vector ˆA in the direction of →A can be found by dividing the vector →A by the magnitude of →A. Therfore ˆA=→A/|→A|=(2ˆi+−3ˆj+7ˆk)/√62
(f)
In a similar fashion, ˆB=→B/|→B|=(5ˆi+ˆj+2ˆk)/√30
Example 3.2 Sinking Sailboat
A Coast Guard ship is located 35 km away from a checkpoint in a direction 52∘ north of west. A distressed sailboat located in still water 24 km from the same checkpoint in a direction 18∘ south of east is about to sink. Draw a diagram indicating the position of both ships. In what direction and how far must the Coast Guard ship travel to reach the sailboat?
Solution
The diagram of the set-up is Figure 3.17.
Choose the checkpoint as the origin of a Cartesian coordinate system with the positive x -axis in the East direction and the positive y –axis in the North direction. Choose the corresponding unit vectors ˆi andˆj as shown in Figure 3.18. The Coast Guard ship is then a distance r=35 km at an angle θ1=180∘−52∘=128∘ from the positive x -axis, The position of the Coast Guard ship is then
→r1=r1(cosθ1ˆi+sinθ1ˆj)→r=−21.5kmˆi+27.6kmˆj and the position of the sailboat is →r2=r2(cosθ2ˆi+sinθ2ˆj)→r2=22.8kmˆi−7.4kmˆj
The relative position vector from the Coast Guard ship to the sailboat is (Figure 3.19) →r2−→r1=(22.8kmˆi−7.4kmˆj)−(−21.5kmˆi+27.6kmˆj)→r2−→r1=44.4kmˆi−35.0kmˆj The distance between the ship and the sailboat is |→r2−→r1|=((44.4km)2+(−35.0km)2)1/2=56.5km The rescue ship’s heading would be the inverse tangent of the ratio of the y - and x - components of the relative position vector, θ21=tan−1(−35.0km/44.4km)=−38.3∘ or 38.3∘ South of East.
Example 3.3: Vector Addition
Two vectors →A and →B, such that |→B|=2|→A|, have a resultant →C=→A+→B of magnitude 26.5. The vector →C makes an angle θc=41∘ with respect to vector →A. Find the magnitude of each vector and the angle between vectors →A and →B.
Solution: We begin by making a sketch of the three vectors, choosing →A to point in the positive x-direction (Figure 3.20).
Denote the magnitude of →C by C≡|→C|=√(Cx)2+(Cy)2=26.5. The components of →C=→A+→B are given by
Cx=Ax+Bx=CcosθC=(26.5)cos(41∘)=20
Cy=By=CsinθC=(26.5)sin(41∘)=17.4.
From the condition that |→B|=2|→A|, the square of their magnitudes satisfy
(Bx)2+(By)2=4(Ax)2.
Using Equations (3.3.17) and (3.3.18), Equation (3.3.19) becomes (Cx−Ax)2+(Cy)2=4(Ax)2
(Cx)2−2CxAx+(Ax)2+(Cy)2=4(Ax)2
This is a quadratic equation
0=3(Ax)2+2CxAx−C2
which we solve for the component Ax:
Ax=−2Cx±√(2Cx)2+(4)(3)(C2)6=−2(20)±√(40))2+(4)(3)(26.5)26=10.0
where we choose the positive square root because we originally chose Ax>0. The components of →B are then given by Equations (3.3.17) and (3.3.18):
Bx=Cx−Ax=20.0−10.0=10.0B.=17.4
The magnitude of |→B|=√(Bx)2+(By)2=20.0 which is equal to two times the magnitude of |→A|=10.0. The angle between →A and →B is given by
θ=sin−1(By/|→B|)=sin−1(17.4/20.0N)=60∘
Example 3.4 Vector Description of a Point on a Line
Consider two points, P1 with coordinates (x1,y1) and P2 with coordinates (x1,y1) are separated by distance d . Find a vector →A from the origin to the point on the line connecting P1 and P2 that is located a distance a from the point P2 (Figure 3.21).
Solution
Let →r1=x1ˆi+y1ˆj be the position vector of P1 and →r2=x2ˆi+y2ˆj the position vector of P2. Let →r1−→r2 be the vector from P2 to P1 (Figure 3.22a). The unit vector pointing from P2 to P1 is given by
ˆr21=(→r1−→r2)/|→r1−→r2|=(→r1−→r2)/d, where d=((x2−x1)2+(y2−y1)2)1/2
The vector →s in Figure 3.22b connects →A to the point at →r1, points in the direction of →r12 and has length a. Therefore →s=aˆr21=a(→r1−→r2)/d. The vector →r1=→A+→s. Therefore →A=→r1−→s=→r1−a(→r1−→r2)/d=(1−a/d)→r1+(a/d)→r2→A=(1−a/d)(x1ˆi+y1ˆj)+(a/d)(x2ˆi+y2ˆj)→A=(x1+a(x2−x1)((x2−x1)2+(y2−y1)2)1/2)ˆi+(y1+a(y2−y1)((x2−x1)2+(y2−y1)2)1/2)ˆj
Transformation of Vectors in Rotated Coordinate Systems
Consider two Cartesian coordinate systems S and S′ such that the (x′,y′) coordinate axes in S′ are rotated by an angle θ with respect to the (x,y) coordinate axes in S, (Figure 3.23).
The components of the unit vector ^i′ in the ˆi and ˆj direction are given by
i′x=|ˆi′|cosθ=cosθ
and
i′y=|ˆi′|sinθ=sinθ.
Therefore
ˆi′=i′xˆi+i′yˆj=ˆicosθ+ˆjsinθ
A similar argument holds for the components of the unit vector \hat{\mathbf{j^{'}}}. The components of \hat{\mathbf{j^{'}}} in the \hat{\mathbf{i}} and \hat{\mathbf{j}} direction are given by
j_{x}^{\prime}=-\left|\hat{\mathbf{j}}^{\prime}\right| \sin \theta=-\sin \theta \nonumber
and
j_{y}^{\prime}=\left|\hat{\mathbf{j}}^{\prime}\right| \cos \theta=\cos \theta. \nonumber
Therefore
\hat{\mathbf{j}^{\prime}}=j_{x}^{\prime} \hat{\mathbf{i}}+j_{y}^{\prime} \hat{\mathbf{j}}=\hat{\mathbf{j}} \cos \theta-\hat{\mathbf{i}} \sin \theta \nonumber
Conversely, from Figure 3.23 and similar vector decomposition arguments, the components of (\hat{\mathbf{i}}\) and (\hat{\mathbf{j}}\) in S' are given by
\hat{\mathbf{i}}=\hat{\mathbf{i}}^{\prime} \cos \theta-\hat{\mathbf{j}}^{\prime} \sin \theta \nonumber \hat{\mathbf{j}}=\hat{\mathbf{i}}^{\prime} \sin \theta+\hat{\mathbf{j}}^{\prime} \cos \theta \nonumber
Consider a fixed vector \overrightarrow{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}} with components (x, y) in coordinate system S. In coordinate system S', the vector is given by \overrightarrow{\mathbf{r}}=x^{\prime} \hat{\mathbf{i}}^{\prime}+y^{\hat{\prime}} \hat{\mathbf{j}}^{\prime}, \text { where }\left(x^{\prime}, y^{\prime}\right), where (x',y') are the components in S', (Figure 3.24).
Using the Equations (3.3.20) and (3.3.21), we have that
\begin{array}{l} \overrightarrow{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}=x\left(\hat{\mathbf{i}}^{\prime} \cos \theta-\hat{\mathbf{j}}^{\prime} \sin \theta\right)+y\left(\hat{\mathbf{j}}^{\prime} \cos \theta+\hat{\mathbf{i}}^{\prime} \sin \theta\right) \\ \overrightarrow{\mathbf{r}}=(x \cos \theta+y \sin \theta) \hat{\mathbf{i}}^{\prime}+(x \sin \theta-y \cos \theta) \hat{\mathbf{j}}^{\prime} \end{array} \nonumber
Therefore the components of the vector transform according to
x^{\prime}=x \cos \theta+y \sin \theta \nonumber
y^{\prime}=x \sin \theta-y \cos \theta \nonumber
We now consider an alternate approach to understanding the transformation laws for the components of the position vector of a fixed point in space. In coordinate system S, suppose the position vector \overrightarrow{\mathbf{r}} has length r=|\overrightarrow{\mathbf{r}}| and makes an angle \phi with respect to the positive x-axis (Figure 3.25).
Then the components of \overrightarrow{\mathbf{r}} in S are given by x=r \cos \phi \nonumber y=r \sin \phi \nonumber . In coordinate system S', the components of \overrightarrow{\mathbf{r}} are given by
x^{\prime}=r \cos (\phi-\theta) \nonumber y^{\prime}=r \sin (\phi-\theta) \nonumber
Apply the addition of angle trigonometric identities to Equations (3.3.29) and (3.3.30) yielding
x^{\prime}=r \cos (\phi-\theta)=r \cos \phi \cos \theta+r \sin \phi \sin \theta=x \cos \theta+y \sin \theta \nonumber
y^{\prime}=r \sin (\phi-\theta)=r \sin \phi \cos \theta-r \cos \phi \sin \theta=y \cos \theta-x \sin \theta \nonumber
in agreement with Equations (3.3.25) and (3.3.26).
Example 3.5 Vector Decomposition in Rotated Coordinate Systems
With respect to a given Cartesian coordinate system S, a vector \overrightarrow{\mathbf{A}} has components A_{x} = 5, A_{y} = -3, A_{z} = 0. Consider a second coordinate system S' such that the (x',y' coordinate axes in S' are rotated by an angle \theta = 60^{\circ} with respect to the (x, y) coordinate axes in S, (Figure 3.26).
- What are the components A_{x} and A_{y} of vector \overrightarrow{\mathbf{A}} in coordinate system S'?
- Calculate the magnitude of the vector using the (A_{x}, A_{y}) components and using the (A_{x}, A_{y}) components. Does your result agree with what you expect?
Solution:
We begin by considering the vector decomposition of \overrightarrow{\mathbf{A}} with respect to the coordinate system S, \overrightarrow{\mathbf{A}}=A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}} \nonumber Now we can use our results for the transformation of unit vectors \overrightarrow{\mathbf{i}} and \overrightarrow{\mathbf{j}} in terms of \overrightarrow{\mathbf{i^{'}}} and \overrightarrow{\mathbf{j^{'}}}, (Equations (3.3.22) and (3.3.23)) in order decompose the vector \overrightarrow{\mathbf{A}} in coordinate system S'
\begin{array}{l} \overrightarrow{\mathbf{A}}=A \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}=A_{x}\left(\cos \theta \hat{\mathbf{i}}^{\prime}-\sin \theta \hat{\mathbf{j}}^{\prime}\right)+A_{y}\left(\sin \theta \hat{\mathbf{i}}^{\prime}+\cos \theta \hat{\mathbf{j}}^{\prime}\right) \\ =\left(A_{x} \cos \theta+A_{y} \sin \theta\right) \hat{\mathbf{i}}^{\prime}+\left(-A_{x} \sin \theta+A_{y} \cos \theta\right) \hat{\mathbf{j}}^{\prime} \\ =A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}} \end{array} \nonumber
where
A_{x^{\prime}}=A_{x} \cos \theta+A_{y} \sin \theta \nonumber A_{y^{\prime}}=-A_{x} \sin \theta+A_{y} \cos \theta \nonumber
We now use the given information that A_{x} = 5, A_{y} = -3, and \theta = 60^{\circ} to solve for the components of \overrightarrow{\mathbf{A}} in coordinate system S'
A_{x^{\prime}}=A_{x} \cos \theta+A_{y} \sin \theta=(1 / 2)(5-3 \sqrt{3}) \nonumber
A_{y^{\prime}}=-A_{x} \sin \theta+A_{y} \cos \theta=(1 / 2)(-5 \sqrt{3}-3) \nonumber
b) The magnitude can be calculated in either coordinate system
|\overrightarrow{\mathbf{A}}|=\sqrt{\left(A_{x}\right)^{2}+\left(A_{y}\right)^{2}}=\sqrt{(5)^{2}+(-3)^{2}}=\sqrt{34} \nonumber
|\overrightarrow{\mathbf{A}}|=\sqrt{\left(A_{x^{\prime}}\right)^{2}+\left(A_{y^{\prime}}\right)^{2}}=\sqrt{((1 / 2)(5-3 \sqrt{3}))^{2}+((1 / 2)(-5 \sqrt{3}-3))^{2}}=\sqrt{34} \nonumber
This result agrees with what I expect because the length of vector \overrightarrow{\mathbf{A}} independent of the choice of coordinate system.