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== 2.3#h(0.6em)Algebra of Vectors


  Vectors can be added together and multiplied by scalars. Vector addition is associative  and commutative , and vector multiplication by a sum of scalars is distributive . Also, scalar multiplication by a sum of vectors is distributive:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "α open parenthesis A → plus B → close parenthesis equals α A → plus α B → .")[$α ( arrow(A) + arrow(B) ) = α arrow(A) + α arrow(B) .$]
]


  In this equation, #math.equation(block: false, alt: "α")[$α$] is any number (a scalar). For example, a vector antiparallel to vector #math.equation(block: false, alt: "A → equals A sub x i ^ plus A sub y j ^ plus A sub z k ^")[$arrow(A) = A_(x) hat(i) + A_(y) hat(j) + A_(z) hat(k)$] can be expressed simply by multiplying #math.equation(block: false, alt: "A →")[$arrow(A)$] by the scalar #math.equation(block: false, alt: "α equals −1")[$α = −1$]:


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


  #examplebox("Example 1")[Direction of Motion][
In a Cartesian coordinate system where #math.equation(block: false, alt: "i ^")[$hat(i)$] denotes geographic east, #math.equation(block: false, alt: "j ^")[$hat(j)$] denotes geographic north, and #math.equation(block: false, alt: "k ^")[$hat(k)$] denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity #math.equation(block: false, alt: "v → equals open parenthesis 4.0 i ^ plus 3.0 j ^ plus 0.1 k ^ close parenthesis km / h")[$arrow(v) = ( 4.0 hat(i) + 3.0 hat(j) + 0.1 hat(k) ) "km" "/" "h"$]. If the convoy had to retreat, in what geographic direction would it be moving?


    Solution
The velocity vector has the third component #math.equation(block: false, alt: "v → sub z equals open parenthesis plus 0.1 km / h close parenthesis k ^")[$arrow(v)_(z) = ( + 0.1 "km" "/" "h" ) hat(k)$], which says the convoy is climbing at a rate of 100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to the north, so it moves on the ground in direction #math.equation(block: false, alt: "tan to the power −1 open parenthesis 3 / 4 close parenthesis approximately equals 37 °")[$"tan"^(−1) ( 3 #h(0.2em) "/" 4 ) ≈ 37 "°"$] north of east. If the convoy had to retreat, its new velocity vector #math.equation(block: false, alt: "u →")[$arrow(u)$] would have to be antiparallel to #math.equation(block: false, alt: "v →")[$arrow(v)$] and be in the form #math.equation(block: false, alt: "u → equals − α v →")[$arrow(u) = "−" α arrow(v)$], where #math.equation(block: false, alt: "α")[$α$] is a positive number. Thus, the velocity of the retreat would be #math.equation(block: false, alt: "u → equals α open parenthesis −4.0 i ^ minus 3.0 j ^ minus 0.1 k ^ close parenthesis km / h")[$arrow(u) = α ( −4.0 hat(i) − 3.0 hat(j) − 0.1 hat(k) ) "km" "/" "h"$]. The negative sign of the third component indicates the convoy would be descending. The direction angle of the retreat velocity is #math.equation(block: false, alt: "tan to the power −1 open parenthesis −3 α / minus 4 α close parenthesis approximately equals 37 °")[$"tan"^(−1) ( −3 α "/" − 4 α ) ≈ 37 "°"$] south of west. Therefore, the convoy would be moving on the ground in direction #math.equation(block: false, alt: "37 °")[$37 "°"$] south of west while descending on its way back.
]


  The generalization of the number zero to vector algebra is called the #strong[null vector], denoted by #math.equation(block: false, alt: "0 →")[$arrow(0)$]. All components of the null vector are zero, #math.equation(block: false, alt: "0 → equals 0 i ^ plus 0 j ^ plus 0 k ^")[$arrow(0) = 0 hat(i) + 0 hat(j) + 0 hat(k)$], so the null vector has no length and no direction.


  Two vectors #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$] are #strong[equal vectors] if and only if their difference is the null vector:


  #math.equation(block: true, alt: "0 → equals A → minus B → equals open parenthesis A sub x i ^ plus A sub y j ^ plus A sub z k ^ close parenthesis minus open parenthesis B sub x i ^ plus B sub y j ^ plus B sub z k ^ close parenthesis equals open parenthesis A sub x minus B sub x close parenthesis i ^ plus open parenthesis A sub y minus B sub y close parenthesis j ^ plus open parenthesis A sub z minus B sub z close parenthesis k ^ .")[$arrow(0) = arrow(A) − arrow(B) = ( A_(x) hat(i) + A_(y) hat(j) + A_(z) hat(k) ) − ( B_(x) hat(i) + B_(y) hat(j) + B_(z) hat(k) ) = ( A_(x) − B_(x) ) hat(i) + ( A_(y) − B_(y) ) hat(j) + ( A_(z) − B_(z) ) hat(k) .$]
  This vector equation means we must have simultaneously #math.equation(block: false, alt: "A sub x minus B sub x equals 0")[$A_(x) − B_(x) = 0$], #math.equation(block: false, alt: "A sub y minus B sub y equals 0")[$A_(y) − B_(y) = 0$], and #math.equation(block: false, alt: "A sub z minus B sub z equals 0")[$A_(z) − B_(z) = 0$]. Hence, we can write #math.equation(block: false, alt: "A → equals B →")[$arrow(A) = arrow(B)$] if and only if the corresponding components of vectors #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$] are equal:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A → equals B → ⇔ { A sub x equals B sub x; A sub y equals B sub y; A sub z equals B sub z .")[$arrow(A) = arrow(B) #h(0.4em) ⇔ #h(0.4em) \{ A_(x) = B_(x) \ A_(y) = B_(y) \ A_(z) = B_(z) .$]
]


  Two vectors are equal when their corresponding scalar components are equal.


  Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form (given by ) allows us to use vector algebra to find sums or differences of many vectors #emph[analytically] (i.e., without using graphical methods). For example, to find the resultant of two vectors #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$], we simply add them component by component, as follows:


  #math.equation(block: true, alt: "R → equals A → plus B → equals open parenthesis A sub x i ^ plus A sub y j ^ plus A sub z k ^ close parenthesis plus open parenthesis B sub x i ^ plus B sub y j ^ plus B sub z k ^ close parenthesis equals open parenthesis A sub x plus B sub x close parenthesis i ^ plus open parenthesis A sub y plus B sub y close parenthesis j ^ plus open parenthesis A sub z plus B sub z close parenthesis k ^ .")[$arrow(R) = arrow(A) + arrow(B) = ( A_(x) hat(i) + A_(y) hat(j) + A_(z) hat(k) ) + ( B_(x) hat(i) + B_(y) hat(j) + B_(z) hat(k) ) = ( A_(x) + B_(x) ) hat(i) + ( A_(y) + B_(y) ) hat(j) + ( A_(z) + B_(z) ) hat(k) .$]
  In this way, using , scalar components of the resultant vector #math.equation(block: false, alt: "R → equals R sub x i ^ plus R sub y j ^ plus R sub z k ^")[$arrow(R) = R_(x) hat(i) + R_(y) hat(j) + R_(z) hat(k)$] are the sums of corresponding scalar components of vectors #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$]:


  #math.equation(block: true, alt: "{ R sub x equals A sub x plus B sub x ,; R sub y equals A sub y plus B sub y ,; R sub z equals A sub z plus B sub z .")[$\{ R_(x) = A_(x) + B_(x) , \ R_(y) = A_(y) + B_(y) , \ R_(z) = A_(z) + B_(z) .$]
  Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up #math.equation(block: false, alt: "N")[$N$] vectors #math.equation(block: false, alt: "F → sub 1 , F → sub 2 , F → sub 3 , and so on , F → sub N")[$arrow(F)_(1) , arrow(F)_(2) , arrow(F)_(3) , … , arrow(F)_(N)$], where each vector is #math.equation(block: false, alt: "F → sub k equals F sub k x i ^ plus F sub k y j ^ plus F sub k z k ^")[$arrow(F)_(k) = F_(k x) hat(i) + F_(k y) hat(j) + F_(k z) hat(k)$], the resultant vector #math.equation(block: false, alt: "F → sub R")[$arrow(F)_(R)$] is


  #math.equation(block: true, alt: "F → sub R equals F → sub 1 plus F → sub 2 plus F → sub 3 plus … plus F → sub N, equals ∑ k equals 1 N F → sub k equals ∑ k equals 1 N open parenthesis F sub k x i ^ plus F sub k y j ^ plus F sub k z k ^ close parenthesis; equals open parenthesis ∑ k equals 1 N F sub k x close parenthesis i ^ plus open parenthesis ∑ k equals 1 N F sub k y close parenthesis j ^ plus open parenthesis ∑ k equals 1 N F sub k z close parenthesis k ^ .")[$arrow(F)_(R) = arrow(F)_(1) + arrow(F)_(2) + arrow(F)_(3) + "…" + arrow(F)_(N) & = display(limits(∑)^(N)_(k = 1) arrow(F)_(k)) = display(limits(∑)^(N)_(k = 1) ( F_(k x) hat(i) + F_(k y) hat(j) + F_(k z) hat(k) )) \  & = ( display(limits(∑)^(N)_(k = 1) F_(k x)) ) hat(i) + ( display(limits(∑)^(N)_(k = 1) F_(k y)) ) hat(j) + ( display(limits(∑)^(N)_(k = 1) F_(k z)) ) hat(k) .$]
  Therefore, scalar components of the resultant vector are


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "{ F sub R x equals ∑ k equals 1 N F sub k x equals F sub 1 x plus F sub 2 x plus … plus F sub N x; F sub R y equals ∑ k equals 1 N F sub k y equals F sub 1 y plus F sub 2 y plus … plus F sub N y; F sub R z equals ∑ k equals 1 N F sub k z equals F sub 1 z plus F sub 2 z plus … plus F sub N z .")[$\{ F_(R x) = display(limits(∑)^(N)_(k = 1) F_(k x)) = F_(1 x) + F_(2 x) + "…" + F_(N x) \ F_(R y) = display(limits(∑)^(N)_(k = 1) F_(k y)) = F_(1 y) + F_(2 y) + "…" + F_(N y) \ F_(R z) = display(limits(∑)^(N)_(k = 1) F_(k z)) = F_(1 z) + F_(2 z) + "…" + F_(N z) .$]
]


  Having found the scalar components, we can write the resultant in vector component form:


  #math.equation(block: true, alt: "F → sub R equals F sub R x i ^ plus F sub R y j ^ plus F sub R z k ^ .")[$arrow(F)_(R) = F_(R x) hat(i) + F_(R y) hat(j) + F_(R z) hat(k) .$]
  Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.


  #examplebox("Example 2")[Analytical Computation of a Resultant][
Three displacement vectors #math.equation(block: false, alt: "A →")[$arrow(A)$], #math.equation(block: false, alt: "B →")[$arrow(B)$], and #math.equation(block: false, alt: "C →")[$arrow(C)$] in a plane  are specified by their magnitudes #emph[A] = 10.0, #emph[B] = 7.0, and #emph[C] = 8.0, respectively, and by their respective direction angles with the horizontal direction #math.equation(block: false, alt: "α equals 35 ° ,")[$α = 35 "°" ,$] #math.equation(block: false, alt: "β equals −110 °")[$β = −110 "°"$], and #math.equation(block: false, alt: "γ equals 30 °")[$γ = 30 "°"$]. The physical units of the magnitudes are centimeters. Resolve the vectors to their scalar components and find the following vector sums: (a) #math.equation(block: false, alt: "R → equals A → plus B → plus C →")[$arrow(R) = arrow(A) + arrow(B) + arrow(C)$], (b) #math.equation(block: false, alt: "D → equals A → minus B →")[$arrow(D) = arrow(A) − arrow(B)$], and (c) #math.equation(block: false, alt: "S → equals A → minus 3 B → plus C →")[$arrow(S) = arrow(A) − 3 arrow(B) + arrow(C)$].


    Strategy
First, we use  to find the scalar components of each vector and then we express each vector in its vector component form given by . Then, we use analytical methods of vector algebra to find the resultants.


    Solution
We resolve the given vectors to their scalar components:


    #math.equation(block: true, alt: "{ A sub x equals A cos α equals open parenthesis 10.0 cm close parenthesis cos 35 ° equals 8.19 cm; A sub y equals A sin α equals open parenthesis 10.0 cm close parenthesis sin 35 ° equals 5.73 cm; { B sub x equals B cos β equals open parenthesis 7.0 cm close parenthesis cos open parenthesis −110 ° close parenthesis equals −2.39 cm; B sub y equals B sin β equals open parenthesis 7.0 cm close parenthesis sin open parenthesis −110 ° close parenthesis equals −6.58 cm; { C sub x equals C cos γ equals open parenthesis 8.0 cm close parenthesis cos 30 ° equals 6.93 cm; C sub y equals C sin γ equals open parenthesis 8.0 cm close parenthesis sin 30 ° equals 4.00 cm .")[$\{ A_(x) = A #h(0.2em) "cos" #h(0.2em) α = ( 10.0 #h(0.2em) "cm" ) #h(0.2em) "cos" #h(0.2em) 35 "°" = 8.19 #h(0.2em) "cm" \ A_(y) = A #h(0.2em) "sin" #h(0.2em) α = ( 10.0 #h(0.2em) "cm" ) #h(0.2em) "sin" #h(0.2em) 35 "°" = 5.73 #h(0.2em) "cm" \ \{ B_(x) = B #h(0.2em) "cos" #h(0.2em) β = ( 7.0 #h(0.2em) "cm" ) #h(0.2em) "cos" #h(0.2em) ( −110 "°" ) = −2.39 #h(0.2em) "cm" \ B_(y) = B #h(0.2em) "sin" #h(0.2em) β = ( 7.0 #h(0.2em) "cm" ) #h(0.2em) "sin" #h(0.2em) ( −110 "°" ) = −6.58 #h(0.2em) "cm" \ \{ C_(x) = C #h(0.2em) "cos" #h(0.2em) γ = ( 8.0 #h(0.2em) "cm" ) #h(0.2em) "cos" #h(0.2em) 30 "°" = 6.93 #h(0.2em) "cm" \ C_(y) = C #h(0.2em) "sin" #h(0.2em) γ = ( 8.0 #h(0.2em) "cm" ) #h(0.2em) "sin" #h(0.2em) 30 "°" = 4.00 #h(0.2em) "cm" .$]
    For (a) we may substitute directly into  to find the scalar components of the resultant:


    #math.equation(block: true, alt: "{ R sub x equals A sub x plus B sub x plus C sub x equals 8.19 cm minus 2.39 cm plus 6.93 cm equals 12.73 cm; R sub y equals A sub y plus B sub y plus C sub y equals 5.73 cm minus 6.58 cm plus 4.00 cm equals 3.15 cm .")[$\{ R_(x) = A_(x) + B_(x) + C_(x) = 8.19 #h(0.2em) "cm" − 2.39 #h(0.2em) "cm" + 6.93 #h(0.2em) "cm" = 12.73 #h(0.2em) "cm" \ R_(y) = A_(y) + B_(y) + C_(y) = 5.73 #h(0.2em) "cm" − 6.58 #h(0.2em) "cm" + 4.00 #h(0.2em) "cm" = 3.15 #h(0.2em) "cm" .$]
    Therefore, the resultant vector is #math.equation(block: false, alt: "R → equals R sub x i ^ plus R sub y j ^ equals open parenthesis 12.7 i ^ plus 3.1 j ^ close parenthesis cm")[$arrow(R) = R_(x) hat(i) + R_(y) hat(j) = ( 12.7 hat(i) + 3.1 hat(j) ) "cm"$].


    For (b), we may want to write the vector difference as


    #math.equation(block: true, alt: "D → equals A → minus B → equals open parenthesis A sub x i ^ plus A sub y j ^ close parenthesis minus open parenthesis B sub x i ^ plus B sub y j ^ close parenthesis equals open parenthesis A sub x minus B sub x close parenthesis i ^ plus open parenthesis A sub y minus B sub y close parenthesis j ^ .")[$arrow(D) = arrow(A) − arrow(B) = ( A_(x) hat(i) + A_(y) hat(j) ) − ( B_(x) hat(i) + B_(y) hat(j) ) = ( A_(x) − B_(x) ) hat(i) + ( A_(y) − B_(y) ) hat(j) .$]
    Then, the scalar components of the vector difference are


    #math.equation(block: true, alt: "{ D sub x equals A sub x minus B sub x equals 8.19 cm minus open parenthesis −2.39 cm close parenthesis equals 10.58 cm; D sub y equals A sub y minus B sub y equals 5.73 cm minus open parenthesis −6.58 cm close parenthesis equals 12.31 cm .")[$\{ D_(x) = A_(x) − B_(x) = 8.19 #h(0.2em) "cm" − ( −2.39 #h(0.2em) "cm" ) = 10.58 #h(0.2em) "cm" \ D_(y) = A_(y) − B_(y) = 5.73 #h(0.2em) "cm" − ( −6.58 #h(0.2em) "cm" ) = 12.31 #h(0.2em) "cm" .$]
    Hence, the difference vector is #math.equation(block: false, alt: "D → equals D sub x i ^ plus D sub y j ^ equals open parenthesis 10.6 i ^ plus 12.3 j ^ close parenthesis cm")[$arrow(D) = D_(x) hat(i) + D_(y) hat(j) = ( 10.6 hat(i) + 12.3 hat(j) ) "cm"$].


    For (c), we can write vector #math.equation(block: false, alt: "S →")[$arrow(S)$] in the following explicit form:


    #math.equation(block: true, alt: "S →, equals A → minus 3 B → plus C → equals open parenthesis A sub x i ^ plus A sub y j ^ close parenthesis minus 3 open parenthesis B sub x i ^ plus B sub y j ^ close parenthesis plus open parenthesis C sub x i ^ plus C sub y j ^ close parenthesis; equals open parenthesis A sub x minus 3 B sub x plus C sub x close parenthesis i ^ plus open parenthesis A sub y minus 3 B sub y plus C sub y close parenthesis j ^ .")[$arrow(S) & = arrow(A) − 3 arrow(B) + arrow(C) = ( A_(x) hat(i) + A_(y) hat(j) ) − 3 ( B_(x) hat(i) + B_(y) hat(j) ) + ( C_(x) hat(i) + C_(y) hat(j) ) \  & = ( A_(x) − 3 B_(x) + C_(x) ) hat(i) + ( A_(y) − 3 B_(y) + C_(y) ) hat(j) .$]
    Then, the scalar components of #math.equation(block: false, alt: "S →")[$arrow(S)$] are


    #math.equation(block: true, alt: "{ S sub x equals A sub x minus 3 B sub x plus C sub x equals 8.19 cm minus 3 open parenthesis −2.39 cm close parenthesis plus 6.93 cm equals 22.29 cm; S sub y equals A sub y minus 3 B sub y plus C sub y equals 5.73 cm minus 3 open parenthesis −6.58 cm close parenthesis plus 4.00 cm equals 29.47 cm .")[$\{ S_(x) = A_(x) − 3 B_(x) + C_(x) = 8.19 #h(0.2em) "cm" − 3 ( −2.39 #h(0.2em) "cm" ) + 6.93 #h(0.2em) "cm" = 22.29 #h(0.2em) "cm" \ S_(y) = A_(y) − 3 B_(y) + C_(y) = 5.73 #h(0.2em) "cm" − 3 ( −6.58 #h(0.2em) "cm" ) + 4.00 #h(0.2em) "cm" = 29.47 #h(0.2em) "cm" .$]
    The vector is #math.equation(block: false, alt: "S → equals S sub x i ^ plus S sub y j ^ equals open parenthesis 22.3 i ^ plus 29.5 j ^ close parenthesis cm")[$arrow(S) = S_(x) hat(i) + S_(y) hat(j) = ( 22.3 hat(i) + 29.5 hat(j) ) "cm"$].


    Significance
Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes and direction angles, as shown. Results for the magnitudes in (b) and (c) can be compared with results for the same problems obtained with the graphical method, shown in  and . Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler or a protractor, as it was with the graphical method used in  for finding this same resultant.


    #figure(figph[Vector R has magnitude 13.11. The angle between R and the positive x direction is theta sub R equals 13.9 degrees. The components of R are R sub x on the x axis and R sub y on the y axis. Vector D has magnitude 16.23. The angle between D and the positive x direction is theta sub D equals 49.3 degrees. The components of D are D sub x on the x axis and D sub y on the y axis. Vector S has magnitude 36.95. The angle between S and the positive x direction is theta sub S equals 52.9 degrees. The components of S are S sub x on the x axis and S sub y on the y axis.], alt: "Vector R has magnitude 13.11. The angle between R and the positive x direction is theta sub R equals 13.9 degrees. The components of R are R sub x on the x axis and R sub y on the y axis. Vector D has magnitude 16.23. The angle between D and the positive x direction is theta sub D equals 49.3 degrees. The components of D are D sub x on the x axis and D sub y on the y axis. Vector S has magnitude 36.95. The angle between S and the positive x direction is theta sub S equals 52.9 degrees. The components of S are S sub x on the x axis and S sub y on the y axis.", caption: [Graphical illustration of the solutions obtained analytically in .])
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Three displacement vectors #math.equation(block: false, alt: "A →")[$arrow(A)$], #math.equation(block: false, alt: "B →")[$arrow(B)$], and #math.equation(block: false, alt: "F →")[$arrow(F)$]  are specified by their magnitudes #emph[A] = 10.00, #emph[B] = 7.00, and #emph[F] = 20.00, respectively, and by their respective direction angles with the horizontal direction #math.equation(block: false, alt: "α equals 35 °")[$α = 35 "°"$], #math.equation(block: false, alt: "β equals −110 °")[$β = −110 "°"$], and #math.equation(block: false, alt: "φ equals 110 °")[$φ = 110 "°"$]. The physical units of the magnitudes are centimeters. Use the analytical method to find vector #math.equation(block: false, alt: "G → equals A → plus 2 B → minus F →")[$arrow(G) = arrow(A) + 2 arrow(B) − arrow(F)$]. Verify that #linebreak() #emph[G] = 28.15 cm and that #math.equation(block: false, alt: "θ sub G equals −68.65 °")[$θ_(G) = −68.65 "°"$].


    
    #solutionbox[
#math.equation(block: true, alt: "G → equals open parenthesis 10.25 i ^ minus 26.22 j ^ close parenthesis cm")[$arrow(G) = ( 10.25 hat(i) − 26.22 hat(j) ) "cm"$]
]
]


  #examplebox("Example 3")[The Tug-of-War Game][
Four dogs named Astro, Balto, Clifford, and Dug play a tug-of-war game with a toy . Astro pulls on the toy in direction #math.equation(block: false, alt: "α equals 55 °")[$α = 55 "°"$] south of east, Balto pulls in direction #math.equation(block: false, alt: "β equals 60 °")[$β = 60 "°"$] east of north, and Clifford pulls in direction #math.equation(block: false, alt: "γ equals 55 °")[$γ = 55 "°"$] west of north. Astro pulls strongly with 160.0 units of force (N), which we abbreviate as #emph[A] = 160.0 N. Balto pulls even stronger than Astro with a force of magnitude #emph[B] = 200.0 N, and Clifford pulls with a force of magnitude #emph[C] = 140.0 N. When Dug pulls on the toy in such a way that his force balances out the resultant of the other three forces, the toy does not move in any direction. With how big a force and in what direction must Dug pull on the toy for this to happen?


    #figure(figph[Illustration of 4 dogs pulling on a toy. The toy is at the origin of a coordinate system, with plus x aligned with east and plus y with north. Ang is pulling at an angle alpha which is 55 degrees clockwise from the plus x (east) direction. Bing is pulling at an angle beta which is 60 degrees clockwise from the plus y (north) direction. Chang is pulling at an angle gamma which is 55 degrees counterclockwise from the plus y (north) direction. Dong is pulling in an unspecified direction in the third quadrant.], alt: "Illustration of 4 dogs pulling on a toy. The toy is at the origin of a coordinate system, with plus x aligned with east and plus y with north. Ang is pulling at an angle alpha which is 55 degrees clockwise from the plus x (east) direction. Bing is pulling at an angle beta which is 60 degrees clockwise from the plus y (north) direction. Chang is pulling at an angle gamma which is 55 degrees counterclockwise from the plus y (north) direction. Dong is pulling in an unspecified direction in the third quadrant.", caption: [Four dogs play a tug-of-war game with a toy.])


    Strategy We assume that east is the direction of the positive #emph[x]-axis and north is the direction of the positive #emph[y]-axis. As in , we have to resolve the three given forces— #math.equation(block: false, alt: "A →")[$arrow(A)$] (the pull from Astro), #math.equation(block: false, alt: "B →")[$arrow(B)$] (the pull from Balto), and #math.equation(block: false, alt: "C →")[$arrow(C)$] (the pull from Clifford)—into their scalar components and then find the scalar components of the resultant vector #math.equation(block: false, alt: "R → equals A → plus B → plus C →")[$arrow(R) = arrow(A) + arrow(B) + arrow(C)$]. When the pulling force #math.equation(block: false, alt: "D →")[$arrow(D)$] from Dug balances out this resultant, the sum of #math.equation(block: false, alt: "D →")[$arrow(D)$] and #math.equation(block: false, alt: "R →")[$arrow(R)$] must give the null vector #math.equation(block: false, alt: "D → plus R → equals 0 →")[$arrow(D) + arrow(R) = arrow(0)$]. This means that #math.equation(block: false, alt: "D → equals − R →")[$arrow(D) = "−" arrow(R)$], so the pull from Dug must be antiparallel to #math.equation(block: false, alt: "R →")[$arrow(R)$].


    Solution
The direction angles are #math.equation(block: false, alt: "θ sub A equals − α equals −55 °")[$θ_(A) = "−" α = −55 "°"$], #math.equation(block: false, alt: "θ sub B equals 90 ° minus β equals 30 °")[$θ_(B) = 90 "°" − β = 30 "°"$], and #math.equation(block: false, alt: "θ sub C equals 90 ° plus γ equals 145 °")[$θ_(C) = 90 "°" + γ = 145 "°"$], and substituting them into  gives the scalar components of the three given forces:


    #math.equation(block: true, alt: "{ A sub x equals A cos θ sub A equals open parenthesis 160.0 N close parenthesis cos open parenthesis −55 ° close parenthesis equals plus 91.8 N; A sub y equals A sin θ sub A equals open parenthesis 160.0 N close parenthesis sin open parenthesis −55 ° close parenthesis equals −131.1 N; { B sub x equals B cos θ sub B equals open parenthesis 200.0 N close parenthesis cos 30 ° equals plus 173.2 N; B sub y equals B sin θ sub B equals open parenthesis 200.0 N close parenthesis sin 30 ° equals plus 100.0 N; { C sub x equals C cos θ sub C equals open parenthesis 140.0 N close parenthesis cos 145 ° equals −114.7 N; C sub y equals C sin θ sub C equals open parenthesis 140.0 N close parenthesis sin 145 ° equals plus 80.3 N .")[$\{ A_(x) = A #h(0.2em) "cos" #h(0.2em) θ_(A) = ( 160.0 #h(0.2em) "N" ) #h(0.2em) "cos" #h(0.2em) ( −55 "°" ) = + 91.8 #h(0.2em) "N" \ A_(y) = A #h(0.2em) "sin" #h(0.2em) θ_(A) = ( 160.0 #h(0.2em) "N" ) #h(0.2em) "sin" #h(0.2em) ( −55 "°" ) = −131.1 #h(0.2em) "N" \ \{ B_(x) = B #h(0.2em) "cos" #h(0.2em) θ_(B) = ( 200.0 #h(0.2em) "N" ) #h(0.2em) "cos" #h(0.2em) 30 "°" = + 173.2 #h(0.2em) "N" \ B_(y) = B #h(0.2em) "sin" #h(0.2em) θ_(B) = ( 200.0 #h(0.2em) "N" ) #h(0.2em) "sin" #h(0.2em) 30 "°" = + 100.0 #h(0.2em) "N" \ \{ C_(x) = C #h(0.2em) "cos" #h(0.2em) θ_(C) = ( 140.0 #h(0.2em) "N" ) #h(0.2em) "cos" #h(0.2em) 145 "°" = −114.7 #h(0.2em) "N" \ C_(y) = C #h(0.2em) "sin" #h(0.2em) θ_(C) = ( 140.0 #h(0.2em) "N" ) #h(0.2em) "sin" #h(0.2em) 145 "°" = + 80.3 #h(0.2em) "N" .$]
    Now we compute scalar components of the resultant vector #math.equation(block: false, alt: "R → equals A → plus B → plus C →")[$arrow(R) = arrow(A) + arrow(B) + arrow(C)$]:


    #math.equation(block: true, alt: "{ R sub x equals A sub x plus B sub x plus C sub x equals plus 91.8 N plus 173.2 N minus 114.7 N equals plus 150.3 N; R sub y equals A sub y plus B sub y plus C sub y equals −131.1 N plus 100.0 N plus 80.3 N equals plus 49.2 N .")[$\{ R_(x) = A_(x) + B_(x) + C_(x) = + 91.8 #h(0.2em) "N" + 173.2 #h(0.2em) "N" − 114.7 #h(0.2em) "N" = + 150.3 #h(0.2em) "N" \ R_(y) = A_(y) + B_(y) + C_(y) = −131.1 #h(0.2em) "N" + 100.0 #h(0.2em) "N" + 80.3 #h(0.2em) "N" = + 49.2 #h(0.2em) "N" .$]
    The antiparallel vector to the resultant #math.equation(block: false, alt: "R →")[$arrow(R)$] is


    #math.equation(block: true, alt: "D → equals − R → equals − R sub x i ^ minus R sub y j ^ equals open parenthesis −150.3 i ^ minus 49.2 j ^ close parenthesis N .")[$arrow(D) = "−" arrow(R) = "−" R_(x) hat(i) − R_(y) hat(j) = ( −150.3 hat(i) − 49.2 hat(j) ) #h(0.2em) "N" .$]
    The magnitude of Dug’s pulling force 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 −150.3 close parenthesis squared plus open parenthesis −49.2 close parenthesis squared N equals 158.1 N .")[$D = sqrt(D_(x)^(2) + D_(y)^(2)) = sqrt(attach(( −150.3 ), t: 2) + attach(( −49.2 ), t: 2)) #h(0.2em) "N" = 158.1 #h(0.2em) "N" .$]
    The direction of Dug’s pulling force is


    #math.equation(block: true, alt: "θ equals tan to the power −1 open parenthesis the fraction D sub y over D sub x close parenthesis equals tan to the power −1 open parenthesis the fraction −49.2 N over −150.3 N close parenthesis equals tan to the power −1 open parenthesis the fraction 49.2 over 150.3 close parenthesis equals 18.1 ° .")[$θ = "tan"^(−1) ( frac(D_(y), D_(x)) ) = "tan"^(−1) ( frac(−49.2 #h(0.2em) "N", −150.3 #h(0.2em) "N") ) = "tan"^(−1) ( frac(49.2, 150.3) ) = 18.1 "°" .$]
    Dug pulls in the direction #math.equation(block: false, alt: "18.1 °")[$18.1 "°"$] south of west because both components are negative, which means the pull vector lies in the third quadrant .
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that Balto in  leaves the game to attend to more important matters, but Astro, Clifford, and Dug continue playing. Astro and Clifford's pull on the toy does not change, but Dug runs around and bites on the toy in a different place. With how big a force and in what direction must Dug pull on the toy now to balance out the combined pulls from Clifford and Astro? Illustrate this situation by drawing a vector diagram indicating all forces involved.


    
    #solutionbox[
#emph[D] = 55.7 N; direction #math.equation(block: false, alt: "65.7 °")[$65.7 "°"$] north of east
]
]


  #examplebox("Example 4")[Vector Algebra][
Find the magnitude of the vector #math.equation(block: false, alt: "C →")[$arrow(C)$] that satisfies the equation #math.equation(block: false, alt: "2 A → minus 6 B → plus 3 C → equals 2 j ^")[$2 arrow(A) − 6 arrow(B) + 3 arrow(C) = 2 hat(j)$], where #math.equation(block: false, alt: "A → equals i ^ minus 2 k ^")[$arrow(A) = hat(i) − 2 hat(k)$] and #math.equation(block: false, alt: "B → equals − j ^ plus k ^ / 2")[$arrow(B) = "−" hat(j) + hat(k) "/" 2$].


    Strategy
We first solve the given equation for the unknown vector #math.equation(block: false, alt: "C →")[$arrow(C)$]. Then we substitute #math.equation(block: false, alt: "A →")[$arrow(A)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$]; group the terms along each of the three directions #math.equation(block: false, alt: "i ^")[$hat(i)$], #math.equation(block: false, alt: "j ^")[$hat(j)$], and #math.equation(block: false, alt: "k ^")[$hat(k)$]; and identify the scalar components #math.equation(block: false, alt: "C sub x")[$C_(x)$], #math.equation(block: false, alt: "C sub y")[$C_(y)$], and #math.equation(block: false, alt: "C sub z")[$C_(z)$]. Finally, we substitute into  to find magnitude #emph[C].


    Solution


    #math.equation(block: true, alt: "2 A → minus 6 B → plus 3 C →, equals, 2 j ^; 3 C →, equals, 2 j ^ minus 2 A → plus 6 B →; C →, equals, the fraction 2 over 3 j ^ minus the fraction 2 over 3 A → plus 2 B →; equals, the fraction 2 over 3 j ^ minus the fraction 2 over 3 open parenthesis i ^ minus 2 k ^ close parenthesis plus 2 open parenthesis − j ^ plus the fraction k ^ over 2 close parenthesis equals the fraction 2 over 3 j ^ minus the fraction 2 over 3 i ^ plus the fraction 4 over 3 k ^ minus 2 j ^ plus k ^; equals, minus the fraction 2 over 3 i ^ plus open parenthesis the fraction 2 over 3 minus 2 close parenthesis j ^ plus open parenthesis the fraction 4 over 3 plus 1 close parenthesis k ^; equals, minus the fraction 2 over 3 i ^ minus the fraction 4 over 3 j ^ plus the fraction 7 over 3 k ^ .")[$2 arrow(A) − 6 arrow(B) + 3 arrow(C) & = & 2 hat(j) \ 3 arrow(C) & = & 2 hat(j) − 2 arrow(A) + 6 arrow(B) \ arrow(C) & = & frac(2, 3) hat(j) − frac(2, 3) arrow(A) + 2 arrow(B) \  & = & frac(2, 3) hat(j) − frac(2, 3) ( hat(i) − 2 hat(k) ) + 2 ( "−" hat(j) + frac(hat(k), 2) ) = frac(2, 3) hat(j) − frac(2, 3) hat(i) + frac(4, 3) hat(k) − 2 hat(j) + hat(k) \  & = & − frac(2, 3) hat(i) + ( frac(2, 3) − 2 ) hat(j) + ( frac(4, 3) + 1 ) hat(k) \  & = & − frac(2, 3) hat(i) − frac(4, 3) hat(j) + frac(7, 3) hat(k) .$]
    The components are #math.equation(block: false, alt: "C sub x equals − 2 / 3")[$C_(x) = "−" 2 #h(0.2em) "/" 3$], #math.equation(block: false, alt: "C sub y equals −4 / 3")[$C_(y) = −4 "/" 3$], and #math.equation(block: false, alt: "C sub z equals 7 / 3")[$C_(z) = 7 #h(0.2em) "/" 3$], and substituting into  gives


    #math.equation(block: true, alt: "C equals the square root of C x 2 plus C y 2 plus C z 2 equals the square root of open parenthesis −2 / 3 close parenthesis squared plus open parenthesis − 4 / 3 close parenthesis squared plus open parenthesis 7 / 3 close parenthesis squared equals the square root of 23 / 3 .")[$C = sqrt(C_(x)^(2) + C_(y)^(2) + C_(z)^(2)) = sqrt(attach(( −2 "/" 3 ), t: 2) + attach(( "−" 4 #h(0.2em) "/" 3 ), t: 2) + attach(( 7 #h(0.2em) "/" 3 ), t: 2)) = sqrt(23 "/" 3) .$]
]


  #examplebox("Example 5")[Displacement of a Skier][
Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest before taking a rest. Find their total displacement vector relative to the lodge when they are at the rest point. How far and in what direction must they ski from the rest point to return directly to the lodge?


    Strategy We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector #math.equation(block: false, alt: "i ^")[$hat(i)$] pointing east and the unit vector #math.equation(block: false, alt: "j ^")[$hat(j)$] pointing north. There are three displacements: #math.equation(block: false, alt: "D → sub 1")[$arrow(D)_(1)$], #math.equation(block: false, alt: "D → sub 2")[$arrow(D)_(2)$], and #math.equation(block: false, alt: "D → sub 3")[$arrow(D)_(3)$]. We identify their magnitudes as #math.equation(block: false, alt: "D sub 1 equals 5.0 km")[$D_(1) = 5.0 #h(0.2em) "km"$], #math.equation(block: false, alt: "D sub 2 equals 3.0 km")[$D_(2) = 3.0 #h(0.2em) "km"$], and #math.equation(block: false, alt: "D sub 3 equals 4.0 km")[$D_(3) = 4.0 #h(0.2em) "km"$]. We identify their directions are the angles #math.equation(block: false, alt: "θ sub 1 equals 90 °")[$θ_(1) = 90 "°"$], #math.equation(block: false, alt: "θ sub 2 equals 180 °")[$θ_(2) = 180 "°"$], and #math.equation(block: false, alt: "θ sub 3 equals 180 ° plus 45 ° equals 225 °")[$θ_(3) = 180 "°" + 45 "°" = 225 "°"$]. We resolve each displacement vector to its scalar components and substitute the components into  to obtain the scalar components of the resultant displacement #math.equation(block: false, alt: "D →")[$arrow(D)$] from the lodge to the rest point. On the way back from the rest point to the lodge, the displacement is #math.equation(block: false, alt: "B → equals − D →")[$arrow(B) = "−" arrow(D)$]. Finally, we find the magnitude and direction of #math.equation(block: false, alt: "B →")[$arrow(B)$].


    Solution
Scalar components of the displacement vectors are


    #math.equation(block: true, alt: "{ D sub 1 x equals D sub 1 cos θ sub 1 equals open parenthesis 5.0 km close parenthesis cos 90 ° equals 0; D sub 1 y equals D sub 1 sin θ sub 1 equals open parenthesis 5.0 km close parenthesis sin 90 ° equals 5.0 km; { D sub 2 x equals D sub 2 cos θ sub 2 equals open parenthesis 3.0 km close parenthesis cos 180 ° equals −3.0 km; D sub 2 y equals D sub 2 sin θ sub 2 equals open parenthesis 3.0 km close parenthesis sin 180 ° equals 0; { D sub 3 x equals D sub 3 cos θ sub 3 equals open parenthesis 4.0 km close parenthesis cos 225 ° equals −2.8 km; D sub 3 y equals D sub 3 sin θ sub 3 equals open parenthesis 4.0 km close parenthesis sin 225 ° equals −2.8 km .")[$\{ D_(1 x) = D_(1) #h(0.2em) "cos" #h(0.2em) θ_(1) = ( 5.0 #h(0.2em) "km" ) #h(0.2em) "cos" #h(0.2em) 90 "°" = 0 \ D_(1 y) = D_(1) #h(0.2em) "sin" #h(0.2em) θ_(1) = ( 5.0 #h(0.2em) "km" ) #h(0.2em) "sin" #h(0.2em) 90 "°" = 5.0 #h(0.2em) "km" \ \{ D_(2 x) = D_(2) #h(0.2em) "cos" #h(0.2em) θ_(2) = ( 3.0 #h(0.2em) "km" ) #h(0.2em) "cos" #h(0.2em) 180 "°" = −3.0 #h(0.2em) "km" \ D_(2 y) = D_(2) #h(0.2em) "sin" #h(0.2em) θ_(2) = ( 3.0 #h(0.2em) "km" ) #h(0.2em) "sin" #h(0.2em) 180 "°" = 0 \ \{ D_(3 x) = D_(3) #h(0.2em) "cos" #h(0.2em) θ_(3) = ( 4.0 #h(0.2em) "km" ) #h(0.2em) "cos" #h(0.2em) 225 "°" = −2.8 #h(0.2em) "km" \ D_(3 y) = D_(3) #h(0.2em) "sin" #h(0.2em) θ_(3) = ( 4.0 #h(0.2em) "km" ) #h(0.2em) "sin" #h(0.2em) 225 "°" = −2.8 #h(0.2em) "km" .$]
    Scalar components of the net displacement vector are


    #math.equation(block: true, alt: "{ D sub x equals D sub 1 x plus D sub 2 x plus D sub 3 x equals open parenthesis 0 minus 3.0 minus 2.8 close parenthesis km equals −5.8 km; D sub y equals D sub 1 y plus D sub 2 y plus D sub 3 y equals open parenthesis 5.0 plus 0 minus 2.8 close parenthesis km equals plus 2.2 km .")[$\{ D_(x) = D_(1 x) + D_(2 x) + D_(3 x) = ( 0 − 3.0 − 2.8 ) "km" = −5.8 #h(0.2em) "km" \ D_(y) = D_(1 y) + D_(2 y) + D_(3 y) = ( 5.0 + 0 − 2.8 ) "km" = + 2.2 #h(0.2em) "km" .$]
    Hence, the skier’s net displacement vector is #math.equation(block: false, alt: "D → equals D sub x i ^ plus D sub y j ^ equals open parenthesis −5.8 i ^ plus 2.2 j ^ close parenthesis km")[$arrow(D) = D_(x) hat(i) + D_(y) hat(j) = ( −5.8 hat(i) + 2.2 hat(j) ) "km"$]. On the way back to the lodge, their displacement is #math.equation(block: false, alt: "B → equals − D → equals − open parenthesis −5.8 i ^ plus 2.2 j ^ close parenthesis km equals open parenthesis 5.8 i ^ minus 2.2 j ^ close parenthesis km")[$arrow(B) = "−" arrow(D) = "−" ( −5.8 hat(i) + 2.2 hat(j) ) "km" = ( 5.8 hat(i) − 2.2 hat(j) ) "km"$]. Its magnitude is #math.equation(block: false, alt: "B equals the square root of B x 2 plus B y 2 equals the square root of open parenthesis 5.8 close parenthesis squared plus open parenthesis −2.2 close parenthesis squared km equals 6.2 km")[$B = sqrt(B_(x)^(2) + B_(y)^(2)) = sqrt(attach(( 5.8 ), t: 2) + attach(( −2.2 ), t: 2)) #h(0.2em) "km" = 6.2 #h(0.2em) "km"$] and its direction angle is #math.equation(block: false, alt: "θ equals tan to the power −1 open parenthesis −2.2 / 5.8 close parenthesis equals −20.8 °")[$θ = "tan"^(−1) ( −2.2 "/" 5.8 ) = −20.8 "°"$]. Therefore, to return to the lodge, they must go 6.2 km in a direction about #math.equation(block: false, alt: "21 °")[$21 "°"$] south of east.


    Significance
Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step in solving any vector problem.
]


  #examplebox("Example 6")[Displacement of a Jogger][
A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m before he stops at a drinking fountain . His displacement vector from point #emph[A] at the bottom of the steps to point #emph[B] at the fountain is #math.equation(block: false, alt: "D → sub A B equals open parenthesis −90.0 i ^ plus 30.0 j ^ close parenthesis m")[$arrow(D)_(A B) = ( −90.0 hat(i) + 30.0 hat(j) ) "m"$]. What is the height and width of each step in the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point #emph[A], what is his net displacement vector?


    #figure(figph[A coordinate system is shown with positive x to the right and positive y up. A jogger is at point A at the bottom of steps which lead up and to the left. The top of the steps is labeled as point T. At the top of the steps is a flat section extending from point T to the fountain at point B. The distance between T and B is 50 meters.], alt: "A coordinate system is shown with positive x to the right and positive y up. A jogger is at point A at the bottom of steps which lead up and to the left. The top of the steps is labeled as point T. At the top of the steps is a flat section extending from point T to the fountain at point B. The distance between T and B is 50 meters.", caption: [A jogger runs up a flight of steps.])


    Strategy The displacement vector #math.equation(block: false, alt: "D → sub A B")[$arrow(D)_(A B)$] is the vector sum of the jogger’s displacement vector #math.equation(block: false, alt: "D → sub A T")[$arrow(D)_(A T)$] along the stairs (from point #emph[A] at the bottom of the stairs to point #emph[T] at the top of the stairs) and his displacement vector #math.equation(block: false, alt: "D → sub T B")[$arrow(D)_(T B)$] on the top of the hill (from point #emph[T] at the top of the stairs to the fountain at point #emph[B]). We must find the horizontal and the vertical components of #math.equation(block: false, alt: "D → sub A T")[$arrow(D)_(A T)$]. If each step has width #emph[w] and height #emph[h], the horizontal component of #math.equation(block: false, alt: "D → sub A T")[$arrow(D)_(A T)$] must have a length of 200#emph[w] and the vertical component must have a length of 200#emph[h.] The actual distance the jogger covers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of the hill.


    Solution
In the coordinate system indicated in , the jogger’s displacement vector on the top of the hill is #math.equation(block: false, alt: "D → sub T B equals open parenthesis −50.0 m close parenthesis i ^")[$arrow(D)_(T B) = ( −50.0 #h(0.2em) "m" ) hat(i)$]. His net displacement vector is


    #math.equation(block: true, alt: "D → sub A B equals D → sub A T plus D → sub T B .")[$arrow(D)_(A B) = arrow(D)_(A T) + arrow(D)_(T B) .$]
    Therefore, his displacement vector #math.equation(block: false, alt: "D → sub A T")[$arrow(D)_(A T)$] along the stairs is


    #math.equation(block: true, alt: "D → sub A T, equals D → sub A B minus D → sub T B equals open parenthesis −90.0 i ^ plus 30.0 j ^ close parenthesis m minus open parenthesis −50.0 m close parenthesis i ^ equals [ open parenthesis −90.0 plus 50.0 close parenthesis i ^ plus 30.0 j ^ ] m; equals open parenthesis −40.0 i ^ plus 30.0 j ^ close parenthesis m .")[$arrow(D)_(A T) & = arrow(D)_(A B) − arrow(D)_(T B) = ( −90.0 hat(i) + 30.0 hat(j) ) "m" − ( −50.0 #h(0.2em) "m" ) hat(i) = [ ( −90.0 + 50.0 ) hat(i) + 30.0 hat(j) ] "m" \  & = ( −40.0 hat(i) + 30.0 hat(j) ) "m" .$]
    Its scalar components are #math.equation(block: false, alt: "D sub A T x equals −40.0 m")[$D_(A T x) = −40.0 #h(0.2em) "m"$] and #math.equation(block: false, alt: "D sub A T y equals 30.0 m")[$D_(A T y) = 30.0 #h(0.2em) "m"$]. Therefore, we must have


    #math.equation(block: true, alt: "200 w equals | minus 40.0 | m and 200 h equals 30.0 m .")[$200 w = | − 40.0 | "m" #h(0.2em) "and" #h(0.2em) 200 h = 30.0 #h(0.2em) "m" .$]
    Hence, the step width is #emph[w] = 40.0 m/200 = 0.2 m = 20 cm, and the step height is #emph[h] = 30.0 m/200 = 0.15 m = 15 cm. The distance that the jogger covers along the stairs is


    #math.equation(block: true, alt: "D sub A T equals the square root of D A T x 2 plus D A T y 2 equals the square root of open parenthesis −40.0 close parenthesis squared plus open parenthesis 30.0 close parenthesis squared m equals 50.0 m .")[$D_(A T) = sqrt(D_(A T x)^(2) + D_(A T y)^(2)) = sqrt(attach(( −40.0 ), t: 2) + attach(( 30.0 ), t: 2)) #h(0.2em) "m" = 50.0 #h(0.2em) "m" .$]
    Thus, the actual distance he runs is #math.equation(block: false, alt: "D sub A T plus D sub T B equals 50.0 m plus 50.0 m equals 100.0 m")[$D_(A T) + D_(T B) = 50.0 #h(0.2em) "m" + 50.0 #h(0.2em) "m" = 100.0 #h(0.2em) "m"$]. When he makes a loop and comes back from the fountain to his initial position at point #emph[A], the total distance he covers is twice this distance, or 200.0 m. However, his net displacement vector is zero, because when his final position is the same as his initial position, the scalar components of his net displacement vector are zero .
]


  In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector #math.equation(block: false, alt: "d → equals −5 m i ^")[$arrow(d) = −5 #h(0.2em) "m" hat(i)$] is unit vector #math.equation(block: false, alt: "d ^ equals − i ^")[$hat(d) = "−" hat(i)$]. The general rule of finding the unit vector #math.equation(block: false, alt: "V ^")[$hat(V)$] of direction for any vector #math.equation(block: false, alt: "V →")[$arrow(V)$] is to divide it by its magnitude #emph[V]:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "V ^ equals the fraction V → over V .")[$hat(V) = frac(arrow(V), V) .$]
]


  We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in  have the same physical unit. In this way,  allows us to express the unit vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.


  #examplebox("Example 7")[The Unit Vector of Direction][
If the velocity vector of the military convoy in  is #math.equation(block: false, alt: "v → equals open parenthesis 4.000 i ^ plus 3.000 j ^ plus 0.100 k ^ close parenthesis km / h")[$arrow(v) = ( 4.000 hat(i) + 3.000 hat(j) + 0.100 hat(k) ) "km" "/" "h"$], what is the unit vector of its direction of motion?


    Strategy
The unit vector of the convoy’s direction of motion is the unit vector #math.equation(block: false, alt: "v ^")[$hat(v)$] that is parallel to the velocity vector. The unit vector is obtained by dividing a vector by its magnitude, in accordance with .


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


    #math.equation(block: true, alt: "v equals the square root of v x 2 plus v y 2 plus v z 2 equals the square root of 4.000 squared plus 3.000 squared plus 0.100 squared km / h equals 5.001 km / h .")[$v = sqrt(v_(x)^(2) + v_(y)^(2) + v_(z)^(2)) = sqrt(4.000^(2) + 3.000^(2) + 0.100^(2)) "km" "/" "h" = 5.001 "km" "/" "h" .$]
    To obtain the unit vector #math.equation(block: false, alt: "v ^")[$hat(v)$], divide #math.equation(block: false, alt: "v →")[$arrow(v)$] by its magnitude:


    #math.equation(block: true, alt: "v ^, equals the fraction v → over v equals the fraction open parenthesis 4.000 i ^ plus 3.000 j ^ plus 0.100 k ^ close parenthesis km / h over 5.001 km / h; equals the fraction open parenthesis 4.000 i ^ plus 3.000 j ^ plus 0.100 k ^ close parenthesis over 5.001; equals the fraction 4.000 over 5.001 i ^ plus the fraction 3.000 over 5.001 j ^ plus the fraction 0.100 over 5.001 k ^; equals open parenthesis 79.98 i ^ plus 59.99 j ^ plus 2.00 k ^ close parenthesis times 10 to the power −2 .")[$hat(v) & = frac(arrow(v), v) = frac(( 4.000 hat(i) + 3.000 hat(j) + 0.100 hat(k) ) "km" "/" "h", 5.001 "km" "/" "h") \  & = frac(( 4.000 hat(i) + 3.000 hat(j) + 0.100 hat(k) ), 5.001) \  & = frac(4.000, 5.001) hat(i) + frac(3.000, 5.001) hat(j) + frac(0.100, 5.001) hat(k) \  & = ( 79.98 hat(i) + 59.99 hat(j) + 2.00 hat(k) ) #h(0.2em) × #h(0.2em) 10^(−2) .$]
    Significance
Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to at least three decimal places and then round off the final answer to the required number of significant figures, which is the way we performed calculations in this example. If you round off your partial answer too early, you risk your final answer having a huge numerical error, and it may be far off from the exact answer or from a value measured in an experiment.
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify that vector #math.equation(block: false, alt: "v ^")[$hat(v)$] obtained in  is indeed a unit vector by computing its magnitude. If the convoy in  was moving across a desert flatland—that is, if the third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic direction does it represent?


    
    #solutionbox[
#math.equation(block: false, alt: "v ^ equals 0.8 i ^ plus 0.6 j ^")[$hat(v) = 0.8 hat(i) + 0.6 hat(j)$], #math.equation(block: false, alt: "36.87 °")[$36.87 "°"$] north of east
]
]


  
    
=== Summary


    - Analytical methods of vector algebra allow us to find resultants of sums or differences of vectors without having to draw them. Analytical methods of vector addition are exact, contrary to graphical methods, which are approximate.
- Analytical methods of vector algebra are used routinely in mechanics, electricity, and magnetism. They are important mathematical tools of physics.


  
  
    
=== Problems


    
      For vectors #math.equation(block: false, alt: "B → equals − i ^ minus 4 j ^")[$arrow(B) = "−" hat(i) − 4 hat(j)$] and #math.equation(block: false, alt: "A → equals −3 i ^ minus 2 j ^")[$arrow(A) = −3 hat(i) − 2 hat(j)$], calculate (a) #math.equation(block: false, alt: "A → plus B →")[$arrow(A) + arrow(B)$] and its magnitude and direction angle, and (b) #math.equation(block: false, alt: "A → minus B →")[$arrow(A) − arrow(B)$] and its magnitude and direction angle.


    
    #solutionbox[
a. #math.equation(block: false, alt: "A → plus B → equals −4 i ^ minus 6 j ^")[$arrow(A) + arrow(B) = −4 hat(i) − 6 hat(j)$], #math.equation(block: false, alt: "| A → plus B → | equals 7.211 , θ equals 236 °")[$| arrow(A) + arrow(B) | = 7.211 , θ = 236 "°"$]; b. #math.equation(block: false, alt: "A → minus B → equals –2 i ^ plus 2 j ^")[$arrow(A) − arrow(B) = –2 hat(i) + 2 hat(j)$], #math.equation(block: false, alt: "| A → minus B → | equals 2 the square root of 2 , θ equals 135 °")[$| arrow(A) − arrow(B) | = 2 sqrt(2) , θ = 135 "°"$]
]


    
      A particle undergoes three consecutive displacements given by vectors #math.equation(block: false, alt: "D → sub 1 equals open parenthesis 3.0 i ^ minus 4.0 j ^ minus 2.0 k ^ close parenthesis mm")[$arrow(D)_(1) = ( 3.0 hat(i) − 4.0 hat(j) − 2.0 hat(k) ) "mm"$], #math.equation(block: false, alt: "D → sub 2 equals open parenthesis 1.0 i ^ minus 7.0 j ^ plus 4.0 k ^ close parenthesis mm")[$arrow(D)_(2) = ( 1.0 hat(i) − 7.0 hat(j) + 4.0 hat(k) ) "mm"$], and #math.equation(block: false, alt: "D → sub 3 equals open parenthesis −7.0 i ^ plus 4.0 j ^ plus 1.0 k ^ close parenthesis mm")[$arrow(D)_(3) = ( −7.0 hat(i) + 4.0 hat(j) + 1.0 hat(k) ) "mm"$]. (a) Find the resultant displacement vector of the particle. (b) What is the magnitude of the resultant displacement? (c) If all displacements were along one line, how far would the particle travel?


    
    
      Given two displacement vectors #math.equation(block: false, alt: "A → equals open parenthesis 3.00 i ^ minus 4.00 j ^ plus 4.00 k ^ close parenthesis m")[$arrow(A) = ( 3.00 hat(i) − 4.00 hat(j) + 4.00 hat(k) ) "m"$] and #math.equation(block: false, alt: "B → equals open parenthesis 2.00 i ^ plus 3.00 j ^ minus 7.00 k ^ close parenthesis m")[$arrow(B) = ( 2.00 hat(i) + 3.00 hat(j) − 7.00 hat(k) ) "m"$], find the displacements and their magnitudes for (a) #math.equation(block: false, alt: "C → equals A → plus B →")[$arrow(C) = arrow(A) + arrow(B)$] and (b) #math.equation(block: false, alt: "D → equals 2 A → minus B →")[$arrow(D) = 2 arrow(A) − arrow(B)$].


    
    #solutionbox[
a. #math.equation(block: false, alt: "C → equals open parenthesis 5.0 i ^ minus 1.0 j ^ minus 3.0 k ^ close parenthesis m , C equals 5.92 m")[$arrow(C) = ( 5.0 hat(i) − 1.0 hat(j) − 3.0 hat(k) ) "m" , C = 5.92 #h(0.2em) "m"$]; #linebreak() 
b. #math.equation(block: false, alt: "D → equals open parenthesis 4.0 i ^ minus 11.0 j ^ plus 15.0 k ^ close parenthesis m , D equals 19.03 m")[$arrow(D) = ( 4.0 hat(i) − 11.0 hat(j) + 15.0 hat(k) ) "m" , D = 19.03 #h(0.2em) "m"$]
]


    
      A small plane flies #math.equation(block: false, alt: "40.0 km")[$40.0 #h(0.2em) "km"$] in a direction #math.equation(block: false, alt: "60 °")[$60 "°"$] north of east and then flies #math.equation(block: false, alt: "30.0 km")[$30.0 #h(0.2em) "km"$] in a direction #math.equation(block: false, alt: "15 °")[$15 "°"$] north of east. Use the analytical method to find the total distance the plane covers from the starting point, and the geographic direction of its displacement vector. What is its displacement vector?


    
    
      In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day, and she is blown along the following straight lines: 2.50 km and #math.equation(block: false, alt: "45.0 °")[$45.0 "°"$] north of west, then 4.70 km and #math.equation(block: false, alt: "60.0 °")[$60.0 "°"$] south of east, then 1.30 km and #math.equation(block: false, alt: "25.0 °")[$25.0 "°"$] south of west, then 5.10 km due east, then 1.70 km and #math.equation(block: false, alt: "5.00 °")[$5.00 "°"$] east of north, then 7.20 km and #math.equation(block: false, alt: "55.0 °")[$55.0 "°"$] south of west, and finally 2.80 km and #math.equation(block: false, alt: "10.0 °")[$10.0 "°"$] north of east. Use the analytical method to find the resultant vector of all her displacement vectors. What is its magnitude and direction?


    
    #solutionbox[
#math.equation(block: false, alt: "D → equals open parenthesis 3.3 i ^ minus 6.6 j ^ close parenthesis km")[$arrow(D) = ( 3.3 hat(i) − 6.6 hat(j) ) "km"$], #math.equation(block: false, alt: "i ^")[$hat(i)$] is to the east, 7.34 km, #math.equation(block: false, alt: "−63.5 °")[$−63.5 "°"$]
]


    
      Assuming the +#emph[x]-axis is horizontal to the right for the vectors given in the following figure, use the analytical method to find the following resultants: (a) #math.equation(block: false, alt: "A → plus B →")[$arrow(A) + arrow(B)$], (b) #math.equation(block: false, alt: "C → plus B →")[$arrow(C) + arrow(B)$], (c) #math.equation(block: false, alt: "D → plus F →")[$arrow(D) + arrow(F)$], (d) #math.equation(block: false, alt: "A → minus B →")[$arrow(A) − arrow(B)$], (e) #math.equation(block: false, alt: "D → minus F →")[$arrow(D) − arrow(F)$], (f) #math.equation(block: false, alt: "A → plus 2 F →")[$arrow(A) + 2 arrow(F)$], (g) #math.equation(block: false, alt: "C → minus 2 D → plus 3 F →")[$arrow(C) − 2 arrow(D) + 3 arrow(F)$], and (h) #math.equation(block: false, alt: "A → minus 4 D → plus 2 F →")[$arrow(A) − 4 arrow(D) + 2 arrow(F)$].


      #figure(figph[The x y coordinate system has positive x to the right and positive y up. Vector A has magnitude 10.0 and points 30 degrees counterclockwise from the positive x direction. Vector B has magnitude 5.0 and points 53 degrees counterclockwise from the positive x direction. Vector C has magnitude 12.0 and points 60 degrees clockwise from the positive x direction. Vector D has magnitude 20.0 and points 37 degrees clockwise from the negative x direction. Vector F has magnitude 20.0 and points 30 degrees counterclockwise from the negative x direction.], alt: "The x y coordinate system has positive x to the right and positive y up. Vector A has magnitude 10.0 and points 30 degrees counterclockwise from the positive x direction. Vector B has magnitude 5.0 and points 53 degrees counterclockwise from the positive x direction. Vector C has magnitude 12.0 and points 60 degrees clockwise from the positive x direction. Vector D has magnitude 20.0 and points 37 degrees clockwise from the negative x direction. Vector F has magnitude 20.0 and points 30 degrees counterclockwise from the negative x direction.", caption: none)


    
    
      Given the vectors in the preceding figure, find vector #math.equation(block: false, alt: "R →")[$arrow(R)$] that solves equations (a) #math.equation(block: false, alt: "D → plus R → equals F →")[$arrow(D) + arrow(R) = arrow(F)$] and (b) #math.equation(block: false, alt: "C → minus 2 D → plus 5 R → equals 3 F →")[$arrow(C) − 2 arrow(D) + 5 arrow(R) = 3 arrow(F)$]. Assume the +#emph[x]-axis is horizontal to the right.


    
    #solutionbox[
a. #math.equation(block: false, alt: "R → equals −1.35 i ^ minus 22.04 j ^")[$arrow(R) = −1.35 hat(i) − 22.04 hat(j)$], b. #math.equation(block: false, alt: "R → equals −17.98 i ^ plus 0.89 j ^")[$arrow(R) = −17.98 hat(i) + 0.89 hat(j)$]
]


    
      A delivery driver starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to stop for lunch. Use the analytical method to determine the following: (a) Find his net displacement vector. (b) How far is the restaurant from the post office? (c) If he returns directly from the restaurant to the post office, what is his displacement vector on the return trip? (d) What is his compass heading on the return trip? Assume the +#emph[x]-axis is to the east.


    
    
      An adventurous dog strays from home, runs three blocks east, two blocks north, and one block east, one block north, and two blocks west. Assuming that each block is about a 100 yd, use the analytical method to find the dog’s net displacement vector, its magnitude, and its direction. Assume the +#emph[x]-axis is to the east. How would your answer be affected if each block was about 100 m?


    
    #solutionbox[
#math.equation(block: false, alt: "D → equals open parenthesis 200 i ^ plus 300 j ^ close parenthesis yd")[$arrow(D) = ( 200 hat(i) + 300 hat(j) ) "yd"$], #emph[D] = 360.5 yd, #math.equation(block: false, alt: "56.3 °")[$56.3 "°"$] north of east; The numerical answers would stay the same but the physical unit would be meters. The physical meaning and distances would be about the same because 1 yd is comparable with 1 m.
]


    
      If #math.equation(block: false, alt: "D → equals open parenthesis 6.00 i ^ minus 8.00 j ^ close parenthesis m")[$arrow(D) = ( 6.00 hat(i) − 8.00 hat(j) ) "m"$], #math.equation(block: false, alt: "B → equals open parenthesis −8.00 i ^ plus 3.00 j ^ close parenthesis m")[$arrow(B) = ( −8.00 hat(i) + 3.00 hat(j) ) "m"$], and #math.equation(block: false, alt: "A → equals open parenthesis 26.0 i ^ plus 19.0 j ^ close parenthesis m")[$arrow(A) = ( 26.0 hat(i) + 19.0 hat(j) ) "m"$], find the unknown constants #emph[a] and #emph[b] such that #math.equation(block: false, alt: "a D → plus b B → plus A → equals 0 →")[$a arrow(D) + b arrow(B) + arrow(A) = arrow(0)$].


    
    
      Given the displacement vector #math.equation(block: false, alt: "D → equals open parenthesis 3 i ^ minus 4 j ^ close parenthesis m,")[$arrow(D) = ( 3 hat(i) − 4 hat(j) ) "m,"$] find the displacement vector #math.equation(block: false, alt: "R →")[$arrow(R)$] so that #math.equation(block: false, alt: "D → plus R → equals −4 D j ^")[$arrow(D) + arrow(R) = −4 D hat(j)$].


    
    #solutionbox[
#math.equation(block: true, alt: "R → equals open parenthesis −3 i ^ minus 16 j ^ close parenthesis m")[$arrow(R) = ( −3 hat(i) − 16 hat(j) ) "m"$]
]


    
      Find the unit vector of direction for the following vector quantities: (a) Force #math.equation(block: false, alt: "F → equals open parenthesis 3.0 i ^ minus 2.0 j ^ close parenthesis N")[$arrow(F) = ( 3.0 hat(i) − 2.0 hat(j) ) "N"$], (b) displacement #math.equation(block: false, alt: "D → equals open parenthesis −3.0 i ^ minus 4.0 j ^ close parenthesis m")[$arrow(D) = ( −3.0 hat(i) − 4.0 hat(j) ) "m"$], and (c) velocity #math.equation(block: false, alt: "v → equals open parenthesis −5.00 i ^ plus 4.00 j ^ close parenthesis m/s")[$arrow(v) = ( −5.00 hat(i) + 4.00 hat(j) ) "m/s"$].


    
    
      At one point in space, the direction of the electric field vector is given in the Cartesian system by the unit vector #math.equation(block: false, alt: "E ^ equals 1 / the square root of 5 i ^ minus 2 / the square root of 5 j ^")[$hat(E) = 1 #h(0.2em) "/" sqrt(5) hat(i) − 2 #h(0.2em) "/" sqrt(5) hat(j)$]. If the magnitude of the electric field vector is #emph[E] = 400.0 V/m, what are the scalar components #math.equation(block: false, alt: "E sub x")[$E_(x)$], #math.equation(block: false, alt: "E sub y")[$E_(y)$], and #math.equation(block: false, alt: "E sub z")[$E_(z)$] of the electric field vector #math.equation(block: false, alt: "E →")[$arrow(E)$] at this point? What is the direction angle #math.equation(block: false, alt: "θ sub E")[$θ_(E)$] of the electric field vector at this point?


    
    #solutionbox[
#math.equation(block: true, alt: "E → equals E E ^")[$arrow(E) = E hat(E)$], #math.equation(block: true, alt: "E sub x equals plus 178.9 V / m")[$E_(x) = + 178.9 "V" "/" "m"$], #math.equation(block: true, alt: "E sub y equals −357.8 V / m")[$E_(y) = −357.8 "V" "/" "m"$], #math.equation(block: true, alt: "E sub z equals 0.0 V / m")[$E_(z) = 0.0 "V" "/" "m"$], #math.equation(block: true, alt: "θ sub E equals − tan to the power −1 open parenthesis 2 close parenthesis")[$θ_(E) = "−" "tan"^(−1) ( 2 )$]
]


    
      A barge is pulled by the two tugboats shown in the following figure. One tugboat pulls on the barge with a force of magnitude 4000 units of force at #math.equation(block: false, alt: "15 °")[$15 "°"$] above the line AB (see the figure) and the other tugboat pulls on the barge with a force of magnitude 5000 units of force at #math.equation(block: false, alt: "12 °")[$12 "°"$] below the line AB. Resolve the pulling forces to their scalar components and find the components of the resultant force pulling on the barge. What is the magnitude of the resultant pull? What is its direction relative to the line AB?


      #figure(figph[The situation in the problem is illustrated as viewed from above. Line A B is vertical on the page, with A at the top and B at the bottom. Two tugboats above the barge are pulling it. The one on the right with 5000 units at an angle of 12 degrees counterclockwise from the line A B and the one on the right with 4000 units at an angle of 15 degrees.], alt: "The situation in the problem is illustrated as viewed from above. Line A B is vertical on the page, with A at the top and B at the bottom. Two tugboats above the barge are pulling it. The one on the right with 5000 units at an angle of 12 degrees counterclockwise from the line A B and the one on the right with 4000 units at an angle of 15 degrees.", caption: none)


    
    
      In the control tower at a regional airport, an air traffic controller monitors two aircraft that travel in straight paths directly away from the control tower. One plane is a cargo carrier Boeing 747 climbing at #math.equation(block: false, alt: "10 °")[$10 "°"$] above the horizontal, and moving #math.equation(block: false, alt: "30 °")[$30 "°"$] north of west.  At some moment in time, the controller notes that the Boeing is at an altitude of 2500 m and the DC-3 is at an altitude of 3000 m. The other plane is a Douglas DC-3 climbing at #math.equation(block: false, alt: "5 °")[$5 "°"$] above the horizontal, and cruising directly west. (a) Find the position vectors of the planes relative to the control tower at this time. (b) What is the distance between the planes at this time?


    
    #solutionbox[
a. #math.equation(block: false, alt: "R → sub B equals open parenthesis 12.278 i ^ plus 7.089 j ^ plus 2.500 k ^ close parenthesis km")[$arrow(R)_(B) = ( 12.278 hat(i) + 7.089 hat(j) + 2.500 hat(k) ) "km"$], #math.equation(block: false, alt: "R → sub D equals open parenthesis −34.290 i ^ plus 3.000 k ^ close parenthesis km")[$arrow(R)_(D) = ( −34.290 hat(i) + 3.000 hat(k) ) "km"$]; b. #math.equation(block: false, alt: "| R → sub B minus R → sub D | equals 23.131 km")[$| arrow(R)_(B) − arrow(R)_(D) | = 23.131 #h(0.2em) "km"$]
]