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== 2.2#h(0.6em)Coordinate Systems and Components of a Vector


  Vectors are usually described in terms of their components in a #strong[coordinate system]. Even in everyday life we naturally invoke the concept of vector components in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction #math.equation(block: false, alt: "37 °")[$37 "°"$] north of east.


  In a rectangular (Cartesian) #emph[xy]-coordinate system in a plane, a point in a plane is described by a pair of coordinates (#emph[x], #emph[y]). In a similar fashion, a vector #math.equation(block: false, alt: "A →")[$arrow(A)$] in a plane is described by a pair of its #emph[vector] coordinates. The #emph[x]-coordinate of vector #math.equation(block: false, alt: "A →")[$arrow(A)$] is called its #emph[x]-component and the #emph[y]-coordinate of vector #math.equation(block: false, alt: "A →")[$arrow(A)$] is called its #emph[y]-component. The vector #emph[x]-component is a vector denoted by #math.equation(block: false, alt: "A → sub x")[$arrow(A)_(x)$]. The vector #emph[y]-component is a vector denoted by #math.equation(block: false, alt: "A → sub y")[$arrow(A)_(y)$]. In the Cartesian system, the #emph[x] and #emph[y] #strong[vector components] of a vector are the orthogonal projections, as illustrated in , of this vector onto the #emph[x]- and #emph[y]-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:


  #math.equation(block: true, alt: "A → equals A → sub x plus A → sub y .")[$arrow(A) = arrow(A)_(x) + arrow(A)_(y) .$]
  As illustrated in , vector #math.equation(block: false, alt: "A →")[$arrow(A)$] is the diagonal of the rectangle where the #emph[x]-component #math.equation(block: false, alt: "A → sub x")[$arrow(A)_(x)$] is the side parallel to the #emph[x]-axis and the #emph[y]-component #math.equation(block: false, alt: "A → sub y")[$arrow(A)_(y)$] is the side parallel to the #emph[y]-axis. Vector component #math.equation(block: false, alt: "A → sub x")[$arrow(A)_(x)$] is orthogonal to vector component #math.equation(block: false, alt: "A → sub y")[$arrow(A)_(y)$].


  #figure(figph[Vector A is shown in the x y coordinate system and extends from point b at A’s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.], alt: "Vector A is shown in the x y coordinate system and extends from point b at A’s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.", caption: [Vector #math.equation(block: false, alt: "A →")[$arrow(A)$] in a plane in the Cartesian coordinate system is the vector sum of its vector #emph[x]- and #emph[y]-components. The #emph[x]-vector component #math.equation(block: false, alt: "A → sub x")[$arrow(A)_(x)$] is the orthogonal projection of vector #math.equation(block: false, alt: "A →")[$arrow(A)$] onto the #emph[x]-axis. The #emph[y]-vector component #math.equation(block: false, alt: "A → sub y")[$arrow(A)_(y)$] is the orthogonal projection of vector #math.equation(block: false, alt: "A →")[$arrow(A)$] onto the #emph[y]-axis. The numbers #math.equation(block: false, alt: "A sub x")[$A_(x)$] and #math.equation(block: false, alt: "A sub y")[$A_(y)$] that multiply the unit vectors are the scalar components of the vector.])


  It is customary to denote the positive direction on the #emph[x]-axis by the unit vector #math.equation(block: false, alt: "i ^")[$hat(i)$] and the positive direction on the #emph[y]-axis by the unit vector #math.equation(block: false, alt: "j ^")[$hat(j)$]. #strong[Unit vectors of the axes], #math.equation(block: false, alt: "i ^")[$hat(i)$] and #math.equation(block: false, alt: "j ^")[$hat(j)$], define two orthogonal directions in the plane. As shown in , the #emph[x]- and #emph[y]- components of a vector can now be written in terms of the unit vectors of the axes:


  #math.equation(block: true, alt: "{ A → sub x equals A sub x i ^; A → sub y equals A sub y j ^ .")[$\{ arrow(A)_(x) = A_(x) hat(i) \ arrow(A)_(y) = A_(y) hat(j) .$]
  The vectors #math.equation(block: false, alt: "A → sub x")[$arrow(A)_(x)$] and #math.equation(block: false, alt: "A → sub y")[$arrow(A)_(y)$] defined by  are the #emph[vector components] of vector #math.equation(block: false, alt: "A →")[$arrow(A)$]. The numbers #math.equation(block: false, alt: "A sub x")[$A_(x)$] and #math.equation(block: false, alt: "A sub y")[$A_(y)$] that define the vector components in  are the #strong[scalar component]#strong[s] of vector #math.equation(block: false, alt: "A →")[$arrow(A)$]. Combining  with , we obtain #strong[the component form of a vector]:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A → equals A sub x i ^ plus A sub y j ^ .")[$arrow(A) = A_(x) hat(i) + A_(y) hat(j) .$]
]


  If we know the coordinates #math.equation(block: false, alt: "b open parenthesis x sub b , y sub b close parenthesis")[$"b" ( x_(b) , y_(b) )$] of the origin point of a vector (where #emph[b] stands for “beginning”) and the coordinates #math.equation(block: false, alt: "e open parenthesis x sub e , y sub e close parenthesis")[$"e" ( x_(e) , y_(e) )$] of the end point of a vector (where #emph[e] stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "{ A sub x equals x sub e minus x sub b; A sub y equals y sub e minus y sub b .")[$\{ A_(x) = x_(e) − x_(b) \ A_(y) = y_(e) − y_(b) .$]
]


  #examplebox("Example 1")[Displacement of a Mouse Pointer][
A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?


    Strategy
The origin of the #emph[xy]-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector #math.equation(block: false, alt: "i ^")[$hat(i)$] on the #emph[x]-axis points horizontally to the right and the unit vector
#math.equation(block: false, alt: "j ^")[$hat(j)$] on the #emph[y]-axis points vertically upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vector is located at point e(2.0, 4.5). Substitute the coordinates of these points into  to find the scalar components #math.equation(block: false, alt: "D sub x")[$D_(x)$] and #math.equation(block: false, alt: "D sub y")[$D_(y)$] of the displacement vector #math.equation(block: false, alt: "D →")[$arrow(D)$]. Finally, substitute the coordinates into  to write the displacement vector in the vector component form.


    Solution
We identify #math.equation(block: false, alt: "x sub b equals 6.0")[$x_(b) = 6.0$], #math.equation(block: false, alt: "x sub e equals 2.0")[$x_(e) = 2.0$], #math.equation(block: false, alt: "y sub b equals 1.6")[$y_(b) = 1.6$], and #math.equation(block: false, alt: "y sub e equals 4.5")[$y_(e) = 4.5$], where the physical unit is 1 cm. The scalar #emph[x]- and #emph[y]-components of the displacement vector are


    #math.equation(block: true, alt: "D sub x, equals x sub e minus x sub b equals open parenthesis 2.0 minus 6.0 close parenthesis cm equals −4.0 cm ,; D sub y, equals y sub e minus y sub b equals open parenthesis 4.5 minus 1.6 close parenthesis cm equals plus 2.9 cm .")[$D_(x) & = x_(e) − x_(b) = ( 2.0 − 6.0 ) "cm" = −4.0 #h(0.2em) "cm" , \ D_(y) & = y_(e) − y_(b) = ( 4.5 − 1.6 ) "cm" = + 2.9 #h(0.2em) "cm" .$]
    The vector component form of the displacement vector is


    #math.equation(block: true, alt: "D → equals D sub x i ^ plus D sub y j ^ equals open parenthesis −4.0 cm close parenthesis i ^ plus open parenthesis 2.9 cm close parenthesis j ^ equals open parenthesis −4.0 i ^ plus 2.9 j ^ close parenthesis cm .")[$arrow(D) = D_(x) hat(i) + D_(y) hat(j) = ( −4.0 #h(0.2em) "cm" ) hat(i) + ( 2.9 #h(0.2em) "cm" ) hat(j) = ( −4.0 hat(i) + 2.9 hat(j) ) "cm" .$]
    This solution is shown.


    #figure(figph[Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 − 1.6.], alt: "Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 − 1.6.", caption: [The graph of the displacement vector. The vector points from the origin point at #emph[b] to the end point at #emph[e].])


    Significance
Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit vector or globally for both components, as in . Often, the latter way is more convenient because it is simpler.


    The vector #emph[x]-component #math.equation(block: false, alt: "D → sub x equals −4.0 i ^ equals 4.0 open parenthesis − i ^ close parenthesis")[$arrow(D)_(x) = −4.0 hat(i) = 4.0 ( "−" hat(i) )$] of the displacement vector has the magnitude #math.equation(block: false, alt: "| D → sub x | equals | minus 4.0 | | i ^ | equals 4.0")[$| arrow(D)_(x) | = | − 4.0 | | hat(i) | = 4.0$] because the magnitude of the unit vector is #math.equation(block: false, alt: "| i ^ | equals 1")[$| hat(i) | = 1$]. Notice, too, that the direction of the #emph[x]-component is #math.equation(block: false, alt: "− i ^")[$"−" hat(i)$], which is antiparallel to the direction of the +#emph[x]-axis; hence, the #emph[x]-component vector #math.equation(block: false, alt: "D → sub x")[$arrow(D)_(x)$] points to the left, as shown. The scalar #emph[x]-component of vector #math.equation(block: false, alt: "D →")[$arrow(D)$] is #math.equation(block: false, alt: "D sub x equals −4.0")[$D_(x) = −4.0$].


    Similarly, the vector #emph[y]-component #math.equation(block: false, alt: "D → sub y equals plus 2.9 j ^")[$arrow(D)_(y) = + 2.9 hat(j)$] of the displacement vector has magnitude #math.equation(block: false, alt: "| D → sub y | equals | 2.9 | | j ^ | equals 2.9")[$| arrow(D)_(y) | = | 2.9 | | hat(j) | = #h(0.2em) 2.9$] because the magnitude of the unit vector is #math.equation(block: false, alt: "| j ^ | equals 1")[$| hat(j) | = 1$]. The direction of the #emph[y]-component is #math.equation(block: false, alt: "plus j ^")[$+ hat(j)$], which is parallel to the direction of the +#emph[y]-axis. Therefore, the #emph[y]-component vector #math.equation(block: false, alt: "D → sub y")[$arrow(D)_(y)$] points up, as seen in . The scalar #emph[y]-component of vector #math.equation(block: false, alt: "D →")[$arrow(D)$] is #math.equation(block: false, alt: "D sub y equals plus 2.9")[$D_(y) = + 2.9$]. The displacement vector #math.equation(block: false, alt: "D →")[$arrow(D)$] is the resultant of its two #emph[vector] components.


    The vector component form of the displacement vector  tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.


    
    #solutionbox[
#math.equation(block: false, alt: "D → equals open parenthesis −5.0 i ^ minus 3.0 j ^ close parenthesis cm")[$arrow(D) = ( −5.0 hat(i) − 3.0 hat(j) ) "cm"$]; the fly moved 5.0 cm to the left and 3.0 cm down from its landing site.
]
]


  When we know the scalar components #math.equation(block: false, alt: "A sub x")[$A_(x)$] and #math.equation(block: false, alt: "A sub y")[$A_(y)$] of a vector #math.equation(block: false, alt: "A →")[$arrow(A)$], we can find its magnitude #emph[A] and its direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$]. The #strong[direction angle]—or direction, for short—is the angle the vector forms with the positive direction on the #emph[x]-axis. The angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] is measured in the #emph[counterclockwise direction] from the +#emph[x]-axis to the vector . Because the lengths #emph[A], #math.equation(block: false, alt: "A sub x")[$A_(x)$], and #math.equation(block: false, alt: "A sub y")[$A_(y)$] form a right triangle, they are related by the Pythagorean theorem:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A squared equals A x 2 plus A y 2 ⇔ A equals the square root of A x 2 plus A y 2 .")[$A^(2) = A_(x)^(2) + A_(y)^(2) #h(0.4em) ⇔ #h(0.4em) A = sqrt(A_(x)^(2) + A_(y)^(2)) .$]
]


  This equation works even if the scalar components of a vector are negative. The direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] of a vector is defined via the tangent function of angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] in the triangle shown in :


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "tan θ equals the fraction A sub y over A sub x")[$"tan" #h(0.2em) θ = frac(A_(y), A_(x))$]
]


  #figure(figph[Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A.], alt: "Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A.", caption: [When the vector lies either in the first quadrant or in the fourth quadrant, where component #math.equation(block: false, alt: "A sub x")[$A_(x)$]is positive (Figure 2.19), the direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] in Equation 2.16) is identical to the angle #math.equation(block: false, alt: "θ")[$θ$].])


  When the vector lies either in the first quadrant or in the fourth quadrant, where component #math.equation(block: false, alt: "A sub x")[$A_(x)$] is positive , the angle #math.equation(block: false, alt: "θ")[$θ$] in  is identical to the direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$]. For vectors in the fourth quadrant, angle #math.equation(block: false, alt: "θ")[$θ$] is negative, which means that for these vectors, direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] is measured #emph[clockwise] from the positive #emph[x]-axis. Similarly, for vectors in the second quadrant, angle #math.equation(block: false, alt: "θ")[$θ$] is negative. When the vector lies in either the second or third quadrant, where component #math.equation(block: false, alt: "A sub x")[$A_(x)$] is negative, the direction angle is #math.equation(block: false, alt: "θ sub A equals θ plus 180 °")[$θ_(A) = θ + 180 "°"$] .


  #figure(figph[Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees.], alt: "Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees.", caption: [Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is #math.equation(block: false, alt: "θ sub A equals θ plus 180 °")[$θ_(A) = θ + 180 "°"$].])


  #examplebox("Example 2")[Magnitude and Direction of the Displacement Vector][
You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer?


    Strategy
In , we found the displacement vector #math.equation(block: false, alt: "D →")[$arrow(D)$] of the mouse pointer. We identify its scalar components #math.equation(block: false, alt: "D sub x equals −4.0 cm")[$D_(x) = −4.0 #h(0.2em) "cm"$] and #math.equation(block: false, alt: "D sub y equals plus 2.9 cm")[$D_(y) = + 2.9 #h(0.2em) "cm"$] and substitute into  and  to find the magnitude #emph[D] and direction #math.equation(block: false, alt: "θ sub D")[$θ_(D)$], respectively.


    Solution
The magnitude of vector #math.equation(block: false, alt: "D →")[$arrow(D)$] is


    #math.equation(block: true, alt: "D equals the square root of D x 2 plus D y 2 equals the square root of open parenthesis −4.0 cm close parenthesis squared plus open parenthesis 2.9 cm close parenthesis squared equals the square root of open parenthesis 4.0 close parenthesis squared plus open parenthesis 2.9 close parenthesis squared cm equals 4.9 cm .")[$D = sqrt(D_(x)^(2) + D_(y)^(2)) = sqrt(attach(( −4.0 #h(0.2em) "cm" ), t: 2) + attach(( 2.9 #h(0.2em) "cm" ), t: 2)) = sqrt(attach(( 4.0 ), t: 2) + attach(( 2.9 ), t: 2)) #h(0.2em) "cm" = 4.9 #h(0.2em) "cm" .$]
    The direction angle is


    #math.equation(block: true, alt: "tan θ equals the fraction D sub y over D sub x equals the fraction plus 2.9 cm over −4.0 cm equals −0.725 ⇒ θ equals tan to the power −1 open parenthesis −0.725 close parenthesis equals −35.9 ° .")[$"tan" #h(0.2em) θ = frac(D_(y), D_(x)) = frac(+ 2.9 #h(0.2em) "cm", −4.0 #h(0.2em) "cm") = −0.725 #h(0.6em) ⇒ #h(0.4em) θ = "tan"^(−1) ( −0.725 ) = −35.9 "°" .$]
    Vector #math.equation(block: false, alt: "D →")[$arrow(D)$] lies in the second quadrant, so its direction angle is


    #math.equation(block: true, alt: "θ sub D equals θ plus 180 ° equals −35.9 ° plus 180 ° equals 144.1 ° .")[$θ_(D) = θ + 180 "°" = −35.9 "°" + 180 "°" = 144.1 "°" .$]
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
If the displacement vector of a blue fly walking on a sheet of graph paper is #math.equation(block: false, alt: "D → equals open parenthesis −5.00 i ^ minus 3.00 j ^ close parenthesis cm")[$arrow(D) = ( −5.00 hat(i) − 3.00 hat(j) ) "cm"$], find its magnitude and direction.


    
    #solutionbox[
5.83 cm, #math.equation(block: false, alt: "211 °")[$211 "°"$]
]
]


  In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.


  Let us return to the right triangle in . The quotient of the adjacent side #math.equation(block: false, alt: "A sub x")[$A_(x)$] to the hypotenuse #emph[A] is the cosine function of direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$], #math.equation(block: false, alt: "A sub x / A equals cos θ sub A")[$A_(x) "/" A = "cos" #h(0.2em) θ_(A)$], and the quotient of the opposite side #math.equation(block: false, alt: "A sub y")[$A_(y)$] to the hypotenuse #emph[A] is the sine function of #math.equation(block: false, alt: "θ sub A")[$θ_(A)$], #math.equation(block: false, alt: "A sub y / A equals sin θ sub A")[$A_(y) "/" A = "sin" #h(0.2em) θ_(A)$]. When magnitude #emph[A] and direction #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] are known, we can solve these relations for the scalar components:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "{ A sub x equals A cos θ sub A; A sub y equals A sin θ sub A .")[$\{ A_(x) = A #h(0.2em) "cos" #h(0.2em) θ_(A) \ A_(y) = A #h(0.2em) "sin" #h(0.2em) θ_(A) .$]
]


  When calculating vector components with , care must be taken with the angle. The direction angle #math.equation(block: false, alt: "θ sub A")[$θ_(A)$] of a vector is the angle measured #emph[counterclockwise] from the positive direction on the #emph[x]-axis to the vector. The clockwise measurement gives a negative angle.


  #examplebox("Example 3")[Components of Displacement Vectors][
A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction #math.equation(block: false, alt: "30 °")[$30 "°"$] west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns #math.equation(block: false, alt: "23 °")[$23 "°"$] west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg.


    Strategy
Let’s adopt a rectangular coordinate system with the positive #emph[x]-axis in the direction of geographic east, with the positive #emph[y]-direction pointed to geographic north. Explicitly, the unit vector #math.equation(block: false, alt: "i ^")[$hat(i)$] of the #emph[x]-axis points east and the unit vector #math.equation(block: false, alt: "j ^")[$hat(j)$] of the #emph[y]-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use  to find the scalar components of the displacements and  for the displacement vectors.


    Solution
On the first leg, the displacement magnitude is #math.equation(block: false, alt: "L sub 1 equals 200.0 m")[$L_(1) = 200.0 #h(0.2em) "m"$] and the direction is southeast. For direction angle #math.equation(block: false, alt: "θ sub 1")[$θ_(1)$] we can take either #math.equation(block: false, alt: "45 °")[$45 "°"$] measured clockwise from the east direction or #math.equation(block: false, alt: "45 ° plus 270 °")[$45 "°" + 270 "°"$] measured counterclockwise from the east direction. With the first choice, #math.equation(block: false, alt: "θ sub 1 equals −45 °")[$θ_(1) = −45 "°"$]. With the second choice, #math.equation(block: false, alt: "θ sub 1 equals plus 315 °")[$θ_(1) = + 315 "°"$]. We can use either one of these two angles. The components are


    #math.equation(block: true, alt: "L sub 1 x equals L sub 1 cos θ sub 1 equals open parenthesis 200.0 m close parenthesis cos 315 ° equals 141.4 m,; L sub 1 y equals L sub 1 sin θ sub 1 equals open parenthesis 200.0 m close parenthesis sin 315 ° equals −141.4 m .")[$L_(1 x) = L_(1) #h(0.2em) "cos" #h(0.2em) θ_(1) = ( 200.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) 315 "°" = 141.4 #h(0.2em) "m," \ L_(1 y) = L_(1) #h(0.2em) "sin" #h(0.2em) θ_(1) = ( 200.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) 315 "°" = −141.4 #h(0.2em) "m" .$]
    The displacement vector of the first leg is


    #math.equation(block: true, alt: "L → sub 1 equals L sub 1 x i ^ plus L sub 1 y j ^ equals open parenthesis 141.4 i ^ minus 141.4 j ^ close parenthesis m .")[$arrow(L)_(1) = L_(1 x) hat(i) + L_(1 y) hat(j) = ( 141.4 hat(i) − 141.4 hat(j) ) #h(0.2em) "m" .$]
    On the second leg of Trooper’s wanderings, the magnitude of the displacement is #math.equation(block: false, alt: "L sub 2 equals 300.0 m")[$L_(2) = 300.0 #h(0.2em) "m"$] and the direction is north. The direction angle is #math.equation(block: false, alt: "θ sub 2 equals plus 90 °")[$θ_(2) = + 90 "°"$]. We obtain the following results:


    #math.equation(block: true, alt: "L sub 2 x, equals, L sub 2 cos θ sub 2 equals open parenthesis 300.0 m close parenthesis cos 90 ° equals 0.0 ,; L sub 2 y, equals, L sub 2 sin θ sub 2 equals open parenthesis 300.0 m close parenthesis sin 90 ° equals 300.0 m,; L → sub 2, equals, L sub 2 x i ^ plus L sub 2 y j ^ equals open parenthesis 300.0 m close parenthesis j ^ .")[$L_(2 x) & = & L_(2) #h(0.2em) "cos" #h(0.2em) θ_(2) = ( 300.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) 90 "°" = 0.0 #h(0.2em) , \ L_(2 y) & = & L_(2) #h(0.2em) "sin" #h(0.2em) θ_(2) = ( 300.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) 90 "°" = 300.0 #h(0.2em) "m," \ arrow(L)_(2) & = & L_(2 x) hat(i) + L_(2 y) hat(j) = ( 300.0 #h(0.2em) "m" ) hat(j) .$]
    On the third leg, the displacement magnitude is #math.equation(block: false, alt: "L sub 3 equals 50.0 m")[$L_(3) = 50.0 #h(0.2em) "m"$] and the direction is #math.equation(block: false, alt: "30 °")[$30 "°"$] west of north. The direction angle measured counterclockwise from the eastern direction is #math.equation(block: false, alt: "θ sub 3 equals 30 ° plus 90 ° equals plus 120 °")[$θ_(3) = 30 "°" + 90 "°" = + 120 "°"$]. This gives the following answers:


    #math.equation(block: true, alt: "L sub 3 x, equals, L sub 3 cos θ sub 3 equals open parenthesis 50.0 m close parenthesis cos 120 ° equals −25.0 m,; L sub 3 y, equals, L sub 3 sin θ sub 3 equals open parenthesis 50.0 m close parenthesis sin 120 ° equals plus 43.3 m,; L → sub 3, equals, L sub 3 x i ^ plus L sub 3 y j ^ equals open parenthesis −25.0 i ^ plus 43.3 j ^ close parenthesis m .")[$L_(3 x) & = & L_(3) #h(0.2em) "cos" #h(0.2em) θ_(3) = ( 50.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) 120 "°" = −25.0 #h(0.2em) "m," \ L_(3 y) & = & L_(3) #h(0.2em) "sin" #h(0.2em) θ_(3) = ( 50.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) 120 "°" = + 43.3 #h(0.2em) "m," \ arrow(L)_(3) & = & L_(3 x) hat(i) + L_(3 y) hat(j) = ( −25.0 hat(i) + 43.3 hat(j) ) "m" .$]
    On the fourth leg of the excursion, the displacement magnitude is #math.equation(block: false, alt: "L sub 4 equals 80.0 m")[$L_(4) = 80.0 #h(0.2em) "m"$] and the direction is south. The direction angle can be taken as either #math.equation(block: false, alt: "θ sub 4 equals −90 °")[$θ_(4) = −90 "°"$] or #math.equation(block: false, alt: "θ sub 4 equals plus 270 °")[$θ_(4) = + 270 "°"$]. We obtain


    #math.equation(block: true, alt: "L sub 4 x, equals, L sub 4 cos θ sub 4 equals open parenthesis 80.0 m close parenthesis cos open parenthesis −90 ° close parenthesis equals 0 ,; L sub 4 y, equals, L sub 4 sin θ sub 4 equals open parenthesis 80.0 m close parenthesis sin open parenthesis −90 ° close parenthesis equals −80.0 m,; L → sub 4, equals, L sub 4 x i ^ plus L sub 4 y j ^ equals open parenthesis −80.0 m close parenthesis j ^ .")[$L_(4 x) & = & L_(4) #h(0.2em) "cos" #h(0.2em) θ_(4) = ( 80.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) ( −90 "°" ) = 0 #h(0.2em) , \ L_(4 y) & = & L_(4) #h(0.2em) "sin" #h(0.2em) θ_(4) = ( 80.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) ( −90 "°" ) = −80.0 #h(0.2em) "m," \ arrow(L)_(4) & = & L_(4 x) hat(i) + L_(4 y) hat(j) = ( −80.0 #h(0.2em) "m" ) hat(j) .$]
    On the last leg, the magnitude is #math.equation(block: false, alt: "L sub 5 equals 150.0 m")[$L_(5) = 150.0 #h(0.2em) "m"$] and the angle is #math.equation(block: false, alt: "θ sub 5 equals −23 ° plus 270 ° equals plus 247 °")[$θ_(5) = −23 "°" + 270 "°" = + 247 "°"$] #math.equation(block: false, alt: "open parenthesis 23 °")[$\( 23 "°"$] west of south), which gives


    #math.equation(block: true, alt: "L sub 5 x, equals, L sub 5 cos θ sub 5 equals open parenthesis 150.0 m close parenthesis cos 247 ° equals −58.6 m,; L sub 5 y, equals, L sub 5 sin θ sub 5 equals open parenthesis 150.0 m close parenthesis sin 247 ° equals −138.1 m,; L → sub 5, equals, L sub 5 x i ^ plus L sub 5 y j ^ equals open parenthesis −58.6 i ^ minus 138.1 j ^ close parenthesis m .")[$L_(5 x) & = & L_(5) #h(0.2em) "cos" #h(0.2em) θ_(5) = ( 150.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) 247 "°" = −58.6 #h(0.2em) "m," \ L_(5 y) & = & L_(5) #h(0.2em) "sin" #h(0.2em) θ_(5) = ( 150.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) 247 "°" = −138.1 #h(0.2em) "m," \ arrow(L)_(5) & = & L_(5 x) hat(i) + L_(5 y) hat(j) = ( −58.6 hat(i) − 138.1 hat(j) ) "m" .$]
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
If Trooper runs 20 m west before taking a rest, what is his displacement vector?


    
    #solutionbox[
#math.equation(block: true, alt: "D → equals open parenthesis −20 m close parenthesis i ^")[$arrow(D) = ( −20 #h(0.2em) "m" ) hat(i)$]
]
]


  
    
=== Polar Coordinates


    To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors #math.equation(block: false, alt: "i ^")[$hat(i)$] and #math.equation(block: false, alt: "j ^")[$hat(j)$] along the #emph[x]-axis and the #emph[y]-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the #strong[polar coordinate system].


    In the polar coordinate system, the location of point #emph[P] in a plane is given by two #strong[polar coordinates] . The first polar coordinate is the #strong[radial coordinate] #emph[r], which is the distance of point #emph[P] from the origin. The second polar coordinate is an angle #math.equation(block: false, alt: "φ")[$φ$] that the radial vector makes with some chosen direction, usually the positive #emph[x]-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point #emph[P.] This radial direction is described by a unit radial vector #math.equation(block: false, alt: "r ^")[$hat(r)$]. The second unit vector #math.equation(block: false, alt: "t ^")[$hat(t)$] is a vector orthogonal to the radial direction #math.equation(block: false, alt: "r ^")[$hat(r)$]. The positive #math.equation(block: false, alt: "plus t ^")[$+ hat(t)$] direction indicates how the angle #math.equation(block: false, alt: "φ")[$φ$] changes in the counterclockwise direction. In this way, a point #emph[P] that has coordinates (#emph[x], #emph[y]) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates #math.equation(block: false, alt: "open parenthesis r , φ close parenthesis")[$( r , φ )$].  is valid for any vector, so we can use it to express the #emph[x]- and #emph[y]-coordinates of vector #math.equation(block: false, alt: "r →")[$arrow(r)$]. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point #emph[P]:


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "{ x equals r cos φ; y equals r sin φ .")[$\{ x = r #h(0.2em) "cos" #h(0.2em) φ \ y = r #h(0.2em) "sin" #h(0.2em) φ .$]
]


    #figure(figph[Vector r points from the origin of the x y coordinate system to point P. The angle between the vector r and the positive x direction is phi. X equals r cosine phi and y equals r sine phi. Extending a line in the direction of r vector past point P, a unit vector r hat is drawn in the same direction as r. A unit vector t hat is perpendicular to r hat, pointing 90 degrees counterclockwise to r hat.], alt: "Vector r points from the origin of the x y coordinate system to point P. The angle between the vector r and the positive x direction is phi. X equals r cosine phi and y equals r sine phi. Extending a line in the direction of r vector past point P, a unit vector r hat is drawn in the same direction as r. A unit vector t hat is perpendicular to r hat, pointing 90 degrees counterclockwise to r hat.", caption: [Using polar coordinates, the unit vector #math.equation(block: false, alt: "r ^")[$hat(r)$] defines the positive direction along the radius #emph[r] (radial direction) and, orthogonal to it, the unit vector #math.equation(block: false, alt: "t ^")[$hat(t)$] defines the positive direction of rotation by the angle #math.equation(block: false, alt: "φ")[$φ$].])


    #examplebox("Example 4")[Polar Coordinates][
A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction #math.equation(block: false, alt: "20 °")[$20 "°"$] north of east and finds one gold coin at a location 10.0 m away from the well in the direction #math.equation(block: false, alt: "20 °")[$20 "°"$] north of west. What are the polar and rectangular coordinates of these findings with respect to the well?


      Strategy
The well marks the origin of the coordinate system and east is the +#emph[x]-direction. We identify radial distances from the locations to the origin, which are #math.equation(block: false, alt: "r sub S equals 20.0 m")[$r_(S) = 20.0 #h(0.2em) "m"$] (for the silver coin) and #math.equation(block: false, alt: "r sub G equals 10.0 m")[$r_(G) = 10.0 #h(0.2em) "m"$] (for the gold coin). To find the angular coordinates, we convert #math.equation(block: false, alt: "20 °")[$20 "°"$] to radians: #math.equation(block: false, alt: "20 ° equals π 20 / 180 equals π / 9")[$20 "°" = π 20 "/" 180 = π "/" 9$]. We use  to find the #emph[x]- and #emph[y]-coordinates of the coins.


      Solution
The angular coordinate of the silver coin is #math.equation(block: false, alt: "φ sub S equals π / 9")[$φ_(S) = π "/" 9$], whereas the angular coordinate of the gold coin is #math.equation(block: false, alt: "φ sub G equals π minus π / 9 equals 8 π / 9")[$φ_(G) = π − π "/" 9 = 8 π "/" 9$]. Hence, the polar coordinates of the silver coin are #math.equation(block: false, alt: "open parenthesis r sub S , φ sub S close parenthesis equals open parenthesis 20.0 m , π / 9 close parenthesis")[$( r_(S) , φ_(S) ) = ( 20.0 #h(0.2em) "m" , π "/" 9 )$] and those of the gold coin are #math.equation(block: false, alt: "open parenthesis r sub G , φ sub G close parenthesis equals open parenthesis 10.0 m , 8 π / 9 close parenthesis")[$( r_(G) , φ_(G) ) = ( 10.0 #h(0.2em) "m" , 8 π "/" 9 )$]. We substitute these coordinates into  to obtain rectangular coordinates. For the gold coin, the coordinates are


      #math.equation(block: true, alt: "{ x sub G equals r sub G cos φ sub G equals open parenthesis 10.0 m close parenthesis cos 8 π / 9 equals −9.4 m; y sub G equals r sub G sin φ sub G equals open parenthesis 10.0 m close parenthesis sin 8 π / 9 equals 3.4 m ⇒ open parenthesis x sub G , y sub G close parenthesis equals open parenthesis −9.4 m , 3.4 m close parenthesis .")[$\{ x_(G) = r_(G) #h(0.2em) "cos" #h(0.2em) φ_(G) = ( 10.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) 8 π "/" 9 = −9.4 #h(0.2em) "m" \ y_(G) = r_(G) #h(0.2em) "sin" #h(0.2em) φ_(G) = ( 10.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) 8 π "/" 9 = 3.4 #h(0.2em) "m" #h(0.4em) ⇒ #h(0.4em) ( x_(G) , y_(G) ) = ( −9.4 #h(0.2em) "m" , 3.4 #h(0.2em) "m" ) .$]
      For the silver coin, the coordinates are


      #math.equation(block: true, alt: "{ x sub S equals r sub S cos φ sub S equals open parenthesis 20.0 m close parenthesis cos π / 9 equals 18.9 m; y sub S equals r sub S sin φ sub S equals open parenthesis 20.0 m close parenthesis sin π / 9 equals 6.8 m ⇒ open parenthesis x sub S , y sub S close parenthesis equals open parenthesis 18.9 m , 6.8 m close parenthesis .")[$\{ x_(S) = r_(S) #h(0.2em) "cos" #h(0.2em) φ_(S) = ( 20.0 #h(0.2em) "m" ) #h(0.2em) "cos" #h(0.2em) π "/" 9 = 18.9 #h(0.2em) "m" \ y_(S) = r_(S) #h(0.2em) "sin" #h(0.2em) φ_(S) = ( 20.0 #h(0.2em) "m" ) #h(0.2em) "sin" #h(0.2em) π "/" 9 = 6.8 #h(0.2em) "m" #h(0.4em) ⇒ #h(0.4em) ( x_(S) , y_(S) ) = ( 18.9 #h(0.2em) "m" , 6.8 #h(0.2em) "m" ) .$]
]


  
  
    
=== Vectors in Three Dimensions


    To specify the location of a point in space, we need three coordinates (#emph[x], #emph[y], #emph[z]), where coordinates #emph[x] and #emph[y] specify locations in a plane, and coordinate #emph[z] gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but #emph[three] unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the #emph[x]-axis #math.equation(block: false, alt: "i ^")[$hat(i)$] and the unit vector of the #emph[y]-axis #math.equation(block: false, alt: "j ^")[$hat(j)$]. The third unit vector #math.equation(block: false, alt: "k ^")[$hat(k)$] is the direction of the #emph[z]-axis . The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order #emph[x]-#emph[y]-#emph[z], which is equivalent to the order #math.equation(block: false, alt: "i ^")[$hat(i)$] - #math.equation(block: false, alt: "j ^")[$hat(j)$] - #math.equation(block: false, alt: "k ^")[$hat(k)$], defines the standard right-handed coordinate system (positive orientation). Notice that while previously the #emph[y]-axis was used for the upward direction, in three dimensions we usually choose #emph[z] to represent the vertical. A right-handed coordinate system with #emph[z] up and with one of the remaining axes pointing to the right and the other out of the page must have the #emph[y]-axis pointing right and #emph[x] out of the page, as illustrated here.


    #figure(figph[The x y z coordinate system, with unit vectors I hat, j hat and k hat respectively. I hat points out at us, j hat points to the right, and k hat points up the page. The unit vectors form the sides of a cube.], alt: "The x y z coordinate system, with unit vectors I hat, j hat and k hat respectively. I hat points out at us, j hat points to the right, and k hat points up the page. The unit vectors form the sides of a cube.", caption: [Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.])


    In three-dimensional space, vector #math.equation(block: false, alt: "A →")[$arrow(A)$] has three vector components: the #emph[x]-component #math.equation(block: false, alt: "A → sub x equals A sub x i ^")[$arrow(A)_(x) = A_(x) hat(i)$], which is the part of vector #math.equation(block: false, alt: "A →")[$arrow(A)$] along the #emph[x]-axis; the #emph[y]-component #math.equation(block: false, alt: "A → sub y equals A sub y j ^")[$arrow(A)_(y) = A_(y) hat(j)$], which is the part of #math.equation(block: false, alt: "A →")[$arrow(A)$] along the #emph[y]-axis; and the #emph[z]-component #math.equation(block: false, alt: "A → sub z equals A sub z k ^")[$arrow(A)_(z) = A_(z) hat(k)$], which is the part of the vector along the #emph[z]-axis. A vector in three-dimensional space is the vector sum of its three vector components :


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A → equals A sub x i ^ plus A sub y j ^ plus A sub z k ^ .")[$arrow(A) = A_(x) hat(i) + A_(y) hat(j) + A_(z) hat(k) .$]
]


    If we know the coordinates of its origin #math.equation(block: false, alt: "b open parenthesis x sub b , y sub b , z sub b close parenthesis")[$b ( x_(b) , y_(b) , z_(b) )$] and of its end #math.equation(block: false, alt: "e open parenthesis x sub e , y sub e , z sub e close parenthesis")[$e ( x_(e) , y_(e) , z_(e) )$], its scalar components are obtained by taking their differences: #math.equation(block: false, alt: "A sub x")[$A_(x)$] and #math.equation(block: false, alt: "A sub y")[$A_(y)$] are given by  and the #emph[z]-component is given by


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A sub z equals z sub e minus z sub b .")[$A_(z) = z_(e) − z_(b) .$]
]


    Magnitude #emph[A] is obtained by generalizing  to three dimensions:


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A equals the square root of A x 2 plus A y 2 plus A z 2 .")[$A = sqrt(A_(x)^(2) + A_(y)^(2) + A_(z)^(2)) .$]
]


    This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in , the diagonal in the #emph[xy]-plane has length #math.equation(block: false, alt: "the square root of A x 2 plus A y 2")[$sqrt(A_(x)^(2) + A_(y)^(2))$] and its square adds to the square #math.equation(block: false, alt: "A z 2")[$A_(z)^(2)$] to give #math.equation(block: false, alt: "A squared")[$A^(2)$]. Note that when the #emph[z]-component is zero, the vector lies entirely in the #emph[xy]-plane and its description is reduced to two dimensions.


    #figure(figph[Vector A in the x y z coordinate system extends from the origin. Vector A equals the sum of vectors A sub x, A sub y and A sub z. Vector A sub x is the x component along the x axis and has length A sub x I hat. Vector A sub y is the y component along the y axis and has length A sub y j hat. Vector A sub z is the z component along the z axis and has length A sub x k hat. The components form the sides of a rectangular box with sides length A sub x, A sub y, and A sub z.], alt: "Vector A in the x y z coordinate system extends from the origin. Vector A equals the sum of vectors A sub x, A sub y and A sub z. Vector A sub x is the x component along the x axis and has length A sub x I hat. Vector A sub y is the y component along the y axis and has length A sub y j hat. Vector A sub z is the z component along the z axis and has length A sub x k hat. The components form the sides of a rectangular box with sides length A sub x, A sub y, and A sub z.", caption: [A vector in three-dimensional space is the vector sum of its three vector components.])


    #examplebox("Example 5")[Takeoff of a Drone][
During a takeoff of IAI Heron , its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?


      #figure(figph[A photo of a drone plane.], alt: "A photo of a drone plane.", caption: [The drone IAI Heron in flight.])


      Strategy
We take the origin of the Cartesian coordinate system as the control tower. The direction of the +#emph[x]-axis is given by unit vector #math.equation(block: false, alt: "i ^")[$hat(i)$] to the east, the direction of the +#emph[y]-axis is given by unit vector #math.equation(block: false, alt: "j ^")[$hat(j)$] to the north, and the direction of the +#emph[z]-axis is given by unit vector #math.equation(block: false, alt: "k ^")[$hat(k)$], which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.


      Solution We identify #emph[b] (300.0 m, 200.0 m, 100.0 m) and #emph[e] (1200 m, 2100 m, 250 m), and use  and  to find the scalar components of the drone’s displacement vector:


      #math.equation(block: true, alt: "{ D sub x equals x sub e minus x sub b equals 1200.0 m minus 300.0 m equals 900.0 m ,; D sub y equals y sub e minus y sub b equals 2100.0 m minus 200.0 m equals 1900.0 m,; D sub z equals z sub e minus z sub b equals 250.0 m minus 100.0 m equals 150.0 m .")[$\{ D_(x) = x_(e) − x_(b) = 1200.0 #h(0.2em) "m" − 300.0 #h(0.2em) "m" = 900.0 #h(0.2em) "m" , \ D_(y) = y_(e) − y_(b) = 2100.0 #h(0.2em) "m" − 200.0 #h(0.2em) "m" = 1900.0 #h(0.2em) "m," \ D_(z) = z_(e) − z_(b) = 250.0 #h(0.2em) "m" − 100.0 #h(0.2em) "m" = 150.0 #h(0.2em) "m" .$]
      We substitute these components into  to find the displacement vector:


      #math.equation(block: true, alt: "D → equals D sub x i ^ plus D sub y j ^ plus D sub z k ^ equals 900.0 m i ^ plus 1900.0 m j ^ plus 150.0 m k ^ equals open parenthesis 0.90 i ^ plus 1.90 j ^ plus 0.15 k ^ close parenthesis km .")[$arrow(D) = D_(x) hat(i) + D_(y) hat(j) + D_(z) hat(k) = 900.0 #h(0.2em) "m" hat(i) + 1900.0 #h(0.2em) "m" hat(j) + 150.0 #h(0.2em) "m" hat(k) = ( 0.90 hat(i) + 1.90 hat(j) + 0.15 hat(k) ) #h(0.2em) "km" .$]
      We substitute into  to find the magnitude of the displacement:


      #math.equation(block: true, alt: "D equals the square root of D x 2 plus D y 2 plus D z 2 equals the square root of open parenthesis 0.90 km close parenthesis squared plus open parenthesis 1.90 km close parenthesis squared plus open parenthesis 0.15 km close parenthesis squared equals 2.11 km .")[$D = sqrt(D_(x)^(2) + D_(y)^(2) + D_(z)^(2)) = sqrt(attach(( 0.90 #h(0.2em) "km" ), t: 2) + attach(( 1.90 #h(0.2em) "km" ), t: 2) + attach(( 0.15 #h(0.2em) "km" ), t: 2)) = 2.11 #h(0.2em) "km" .$]
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
If the average velocity vector of the drone in the displacement in  is #math.equation(block: false, alt: "u → equals open parenthesis 15.0 i ^ plus 31.7 j ^ plus 2.5 k ^ close parenthesis m / s")[$arrow(u) = ( 15.0 hat(i) + 31.7 hat(j) + 2.5 hat(k) ) "m" "/" "s"$], what is the magnitude of the drone’s velocity vector?


      
      #solutionbox[
35.2 m/s = 126.7 km/h
]
]


  
  
    
=== Summary


    - Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectors have two components. In three dimensions (in space), vectors have three components.
- A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vector of an axis with its scalar component along this axis. A vector is the resultant of its vector components.
- Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted from end point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum of the squares of its components.
- In a plane, the direction of a vector is given by an angle the vector has with the positive #emph[x]-axis. This direction angle is measured counterclockwise. The scalar #emph[x]-component of a vector can be expressed as the product of its magnitude with the cosine of its direction angle, and the scalar #emph[y]-component can be expressed as the product of its magnitude with the sine of its direction angle.
- In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors #math.equation(block: false, alt: "i ^")[$hat(i)$] and #math.equation(block: false, alt: "j ^")[$hat(j)$] along the #emph[x]-axis and the #emph[y]-axis, respectively. The polar coordinate system is defined by the radial unit vector #math.equation(block: false, alt: "r ^")[$hat(r)$], which gives the direction from the origin, and a unit vector #math.equation(block: false, alt: "t ^")[$hat(t)$], which is perpendicular (orthogonal) to the radial direction.


  
  
    
=== Conceptual Questions


    
      Give an example of a nonzero vector that has a component of zero.


    
    #solutionbox[
a unit vector of the #emph[x]-axis
]


    
      Explain why a vector cannot have a component greater than its own magnitude.


    
    
      If two vectors are equal, what can you say about their components?


    
    #solutionbox[
They are equal.
]


    
      If vectors #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$] are orthogonal, what is the component of #math.equation(block: false, alt: "B →")[$arrow(B)$] along the direction of #math.equation(block: false, alt: "A →")[$arrow(A)$]? What is the component of #math.equation(block: false, alt: "A →")[$arrow(A)$] along the direction of #math.equation(block: false, alt: "B →")[$arrow(B)$]?


    
    
      If one of the two components of a vector is not zero, can the magnitude of the other vector component of this vector be zero?


    
    #solutionbox[
yes
]


    
      If two vectors have the same magnitude, do their components have to be the same?


    
  
  
    
=== Problems


    
      Assuming the +#emph[x]-axis is horizontal and points to the right, resolve the vectors given in the following figure to their scalar components and express them in vector component form.


      #figure(figph[The figure shows an x-y axis in the upper left corner followed by 5 unique vectors drawn on the x-y plane. Vector A has a magnitude of 10.0 and points 30 degrees above the x-axis. Vector B has a magnitude of 5.0 and points 53 degrees above the x-axis. Vector C has a magnitude of 12.0 and points 60 degrees below the x-axis. Vector D has a magnitude of 20.0 and points 37 degrees above the negative x-axis. Vector E has a magnitude of 20.0 and points 30 degrees below the negative x-axis.], alt: "The figure shows an x-y axis in the upper left corner followed by 5 unique vectors drawn on the x-y plane. Vector A has a magnitude of 10.0 and points 30 degrees above the x-axis. Vector B has a magnitude of 5.0 and points 53 degrees above the x-axis. Vector C has a magnitude of 12.0 and points 60 degrees below the x-axis. Vector D has a magnitude of 20.0 and points 37 degrees above the negative x-axis. Vector E has a magnitude of 20.0 and points 30 degrees below the negative x-axis.", caption: none)


    
    #solutionbox[
a. #math.equation(block: false, alt: "A → equals plus 8.66 i ^ plus 5.00 j ^")[$arrow(A) = + 8.66 hat(i) + 5.00 hat(j)$], b. #math.equation(block: false, alt: "B → equals plus 3.01 i ^ plus 3.99 j ^")[$arrow(B) = + 3.01 hat(i) + 3.99 hat(j)$], c. #math.equation(block: false, alt: "C → equals plus 6.00 i ^ minus 10.39 j ^")[$arrow(C) = + 6.00 hat(i) − 10.39 hat(j)$], d. #math.equation(block: false, alt: "D → equals −15.97 i ^ plus 12.04 j ^")[$arrow(D) = −15.97 hat(i) + 12.04 hat(j)$], f. #math.equation(block: false, alt: "F → equals −17.32 i ^ minus 10.00 j ^")[$arrow(F) = −17.32 hat(i) − 10.00 hat(j)$] #linebreak() 
#figure(figph[The figure shows an x-y axis in the upper left corner followed by 5 unique vectors drawn on the x-y plane. Vector A has a magnitude of 10.0 and points 30 degrees above the x-axis. Vector B has a magnitude of 5.0 and points 53 degrees above the x-axis. Vector C has a magnitude of 12.0 and points 60 degrees below the x-axis. Vector D has a magnitude of 20.0 and points 37 degrees above the negative x-axis. Vector F has a magnitude of 20.0 and points 30 degrees below the negative x-axis.], alt: "The figure shows an x-y axis in the upper left corner followed by 5 unique vectors drawn on the x-y plane. Vector A has a magnitude of 10.0 and points 30 degrees above the x-axis. Vector B has a magnitude of 5.0 and points 53 degrees above the x-axis. Vector C has a magnitude of 12.0 and points 60 degrees below the x-axis. Vector D has a magnitude of 20.0 and points 37 degrees above the negative x-axis. Vector F has a magnitude of 20.0 and points 30 degrees below the negative x-axis.", caption: none)
]


    
      Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +#emph[x]-axis is to the east.


    
    
      You drive 7.50 km in a straight line in a direction #math.equation(block: false, alt: "15 °")[$15 "°"$] east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (b) Show that you still arrive at the same point if the east and north legs are reversed in order. Assume the +#emph[x]-axis is to the east.


    
    #solutionbox[
a. 1.94 km, 7.24 km; b. proof
]


    
      A sledge is being pulled by two horses on a flat terrain. The net force on the sledge can be expressed in the Cartesian coordinate system as vector #math.equation(block: false, alt: "F → equals open parenthesis −2980.0 i ^ plus 8200.0 j ^ close parenthesis N")[$arrow(F) = ( −2980.0 hat(i) + 8200.0 hat(j) ) "N"$], where #math.equation(block: false, alt: "i ^")[$hat(i)$] and #math.equation(block: false, alt: "j ^")[$hat(j)$] denote directions to the east and north, respectively. Find the magnitude and direction of the pull.


    
    
      A trapper walks a 5.0-km straight-line distance from her cabin to the lake, as shown in the following figure. Determine the east and north components of her displacement vector. How many more kilometers would she have to walk if she walked along the component displacements? What is her displacement vector?


      #figure(figph[The vector from the cabin to the lake is vector S, magnitude 5.0 kilometers and pointing 40 degrees north of east. Two additional meandering paths are shown and labeled path 1 and path 2.], alt: "The vector from the cabin to the lake is vector S, magnitude 5.0 kilometers and pointing 40 degrees north of east. Two additional meandering paths are shown and labeled path 1 and path 2.", caption: none)


    
    #solutionbox[
3.8 km east, 3.2 km north, 2.0 km, #math.equation(block: false, alt: "D → equals open parenthesis 3.8 i ^ plus 3.2 j ^ close parenthesis km")[$arrow(D) = ( 3.8 hat(i) + 3.2 hat(j) ) "km"$]
]


    
      The polar coordinates of a point are #math.equation(block: false, alt: "4 π / 3")[$4 π "/" 3$] and 5.50 m. What are its Cartesian coordinates?


    
    
      Two points in a plane have polar coordinates #math.equation(block: false, alt: "P sub 1 open parenthesis 2.500 m , π / 6 close parenthesis")[$P_(1) ( 2.500 #h(0.2em) "m" , π "/" 6 )$] and #math.equation(block: false, alt: "P sub 2 open parenthesis 3.800 m , 2 π / 3 close parenthesis")[$P_(2) ( 3.800 #h(0.2em) "m" , 2 π "/" 3 )$]. Determine their Cartesian coordinates and the distance between them in the Cartesian coordinate system. Round the distance to a nearest centimeter.


    
    #solutionbox[
#math.equation(block: false, alt: "P sub 1 open parenthesis 2.165 m , 1.250 m close parenthesis")[$P_(1) ( 2.165 #h(0.2em) "m" , 1.250 #h(0.2em) "m" )$], #math.equation(block: false, alt: "P sub 2 open parenthesis −1.900 m , 3.290 m close parenthesis")[$P_(2) ( −1.900 #h(0.2em) "m" , 3.290 #h(0.2em) "m" )$], 4.55 m
]


    
      A chameleon is resting quietly on a lanai screen, waiting for an insect to come by. Assume the origin of a Cartesian coordinate system at the lower left-hand corner of the screen and the horizontal direction to the right as the +#emph[x]-direction. If its coordinates are (2.000 m, 1.000 m), (a) how far is it from the corner of the screen? (b) What is its location in polar coordinates?


    
    
      Two points in the Cartesian plane are #emph[A] (2.00 m, −4.00 m) and #emph[B] (−3.00 m, 3.00 m). Find the distance between them and their polar coordinates.


    
    #solutionbox[
8.60 m, #math.equation(block: false, alt: "A open parenthesis 2 the square root of 5 m , 1.647 π close parenthesis")[$A ( 2 sqrt(5) #h(0.2em) "m" , 1.647 π )$], #math.equation(block: false, alt: "B open parenthesis 3 the square root of 2 m , 0.75 π close parenthesis")[$B ( 3 sqrt(2) #h(0.2em) "m" , 0.75 π )$]
]


    
      A fly enters through an open window and zooms around the room. In a Cartesian coordinate system with three axes along three edges of the room, the fly changes its position from point #emph[b] (4.0 m, 1.5 m, 2.5 m) to point #emph[e] (1.0 m, 4.5 m, 0.5 m). Find the scalar components of the fly’s displacement vector and express its displacement vector in vector component form. What is its magnitude?