Add the vector AA size 12{A} {} to the vector BB size 12{B} {} shown in Figure 8, using perpendicular components along the *x*- and *y*-axes. The *x*- and *y*-axes are along the east–west and north–south directions, respectively. Vector AA size 12{A} {} represents the first leg of a walk in which a person walks 53.0 m53.0 m size 12{"53" "." "0 m"} {} in a direction 20.0º20.0º size 12{"20" "." 0º } {} north of east. Vector BB size 12{B} {} represents the second leg, a displacement of 34.0 m34.0 m size 12{"34" "." "0 m"} {} in a direction 63.0º63.0º size 12{"63" "." 0º } {} north of east.

*Strategy*

The components of AA size 12{A} {} and BB size 12{B} {} along the *x*- and *y*-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

*Solution*

Following the method outlined above, we first find the components of
AA size 12{A} {} and
BB size 12{B} {} along the *x*- and *y*-axes. Note that
A=53.0 mA=53.0 m size 12{"A" "=" "53.0 m"} {},
θA=20.0ºθA=20.0º size 12{"θ" "subA" "=" "20.0°" } {},
B=34.0 mB=34.0 m size 12{"B" "=" "34.0" "m"} {}, and θB=63.0ºθB=63.0º size 12{θ rSub { size 8{B} } } {}.
We find the *x*-components by using Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}, which gives

A
x
=
A
cos
θ
A
=
(
53.
0 m
)
(
cos 20.0º
)
=
(
53.
0 m
)
(
0
.940
)
=
49.
8 m
A
x
=
A
cos
θ
A
=
(
53.
0 m
)
(
cos 20.0º
)
=
(
53.
0 m
)
(
0
.940
)
=
49.
8 m
alignl { stack {
size 12{A rSub { size 8{x} } =A"cos"θ rSub { size 8{A} } = \( "53" "." 0" m" \) \( "cos""20" "." 0 { size 12{ circ } } \) } {} #
" "= \( "53" "." 0" m" \) \( 0 "." "940" \) ="49" "." 8" m" {}
} } {}

(14)and

B
x
=
B
cos
θ
B
=
(
34
.
0 m
)
(
cos 63.0º
)
=
(
34
.
0 m
)
(
0
.
454
)
=
15
.
4 m
.
B
x
=
B
cos
θ
B
=
(
34
.
0 m
)
(
cos 63.0º
)
=
(
34
.
0 m
)
(
0
.
454
)
=
15
.
4 m
.
alignl { stack {
size 12{B rSub { size 8{x} } =B"cos"θ rSub { size 8{B} } = \( "34" "." 0" m" \) \( "cos""63" "." 0 { size 12{ circ } } \) } {} #
" "= \( "34" "." 0" m" \) \( 0 "." "454" \) ="15" "." 4" m" {}
} } {}

(15)Similarly, the *y*-components are found using Ay=AsinθAAy=AsinθA size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } } {}:

A
y
=
A
sin
θ
A
=
(
53
.
0 m
)
(
sin 20.0º
)
=
(
53
.
0 m
)
(
0
.
342
)
=
18
.
1 m
A
y
=
A
sin
θ
A
=
(
53
.
0 m
)
(
sin 20.0º
)
=
(
53
.
0 m
)
(
0
.
342
)
=
18
.
1 m
alignl { stack {
size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } = \( "53" "." 0" m" \) \( "sin""20" "." 0 { size 12{ circ } } \) } {} #
" "= \( "53" "." 0" m" \) \( 0 "." "342" \) ="18" "." 1" m" {}
} } {}

(16)and

B
y
=
B
sin
θ
B
=
(
34
.
0 m
)
(
sin 63
.
0
º
)
=
(
34
.
0 m
)
(
0
.
891
)
=
30
.
3 m
.
B
y
=
B
sin
θ
B
=
(
34
.
0 m
)
(
sin 63
.
0
º
)
=
(
34
.
0 m
)
(
0
.
891
)
=
30
.
3 m
.
alignl { stack {
size 12{B rSub { size 8{y} } =B"sin"θ rSub { size 8{B} } = \( "34" "." 0" m" \) \( "sin""63" "." 0 { size 12{ circ } } \) } {} #
" "= \( "34" "." 0" m" \) \( 0 "." "891" \) ="30" "." 3" m" "." {}
} } {}

(17)The *x*- and *y*-components of the resultant are thus

R
x
=
A
x
+
B
x
=
49
.
8 m
+
15
.
4 m
=
65
.
2 m
R
x
=
A
x
+
B
x
=
49
.
8 m
+
15
.
4 m
=
65
.
2 m
size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } ="49" "." 8" m"+"15" "." 4" m"="65" "." 2" m"} {}

(18)and

Ry=Ay+By=18.1 m+30.3 m=48.4 m.Ry=Ay+By=18.1 m+30.3 m=48.4 m. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } ="18" "." 1" m"+"30" "." 3" m"="48" "." 4" m."} {}

(19)Now we can find the magnitude of the resultant by using the Pythagorean theorem:

R
=
R
x
2
+
R
y
2
=
(
65
.
2
)
2
+
(
48
.
4
)
2
m
R
=
R
x
2
+
R
y
2
=
(
65
.
2
)
2
+
(
48
.
4
)
2
m
size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } = sqrt { \( "65" "." 2 \) rSup { size 8{2} } + \( "48" "." 4 \) rSup { size 8{2} } } " m"} {}

(20)so that

R
=
81.2 m.
R
=
81.2 m.
size 12{R ="81.2" "m."} {}

(21)Finally, we find the direction of the resultant:

θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2).θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2). size 12{θ="tan" rSup { size 8{ - 1} } \( R rSub { size 8{y} } /R rSub { size 8{x} } \) "=+""tan" rSup { size 8{ - 1} } \( "48" "." 4/"65" "." 2 \) "."} {}

(22)Thus,

θ=tan−1(0.742)=36.6º.θ=tan−1(0.742)=36.6º. size 12{θ="tan" rSup { size 8{ - 1} } \( 0 "." "742" \) ="36" "." 6 { size 12{ circ } } "."} {}

(23)
*Discussion*

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, A−B≡A+(–B)A−B≡A+(–B) size 12{A – B equiv A+ \( - B \) } {}. Thus, *the method for the subtraction of vectors using perpendicular components is identical to that for addition*. The components of
–B–B are the negatives of the components of
BB size 12{B} {}. The *x*- and *y*-components of the resultant A−B = RA−B = R size 12{A- bold "B = R"} {} are thus

R
x
=
A
x
+
(–
B x
)
R
x
=
A
x
+
(–
B x
)
size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +-B rSub { size 8{x} } } {}

(24)and

R
y
=
A
y
+
(–
B y
)
R
y
=
A
y
+
(–
B y
)
size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +-B rSub { size 8{y} } } {}

(25)and the rest of the method outlined above is identical to that for addition. (See Figure 10.)