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Components along the same axis, say the x -axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y -axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R size 12{R} {} are known, its magnitude and direction can be found.

Step 3. To get the magnitude R size 12{R } {} of the resultant, use the Pythagorean theorem:

R = R x 2 + R y 2 . size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } "."} {}

Step 4. To get the direction of the resultant:

θ = tan 1 ( R y / R x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( R rSub { size 8{y} } /R rSub { size 8{x} } \) "."} {}

The following example illustrates this technique for adding vectors using perpendicular components.

Adding vectors using analytical methods

Add the vector A size 12{A} {} to the vector B size 12{B} {} shown in [link] , 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 A size 12{A} {} represents the first leg of a walk in which a person walks 53 . 0 m size 12{"53" "." "0 m"} {} in a direction 20 . 0 º size 12{"20" "." 0º } {} north of east. Vector B size 12{B} {} represents the second leg, a displacement of 34 . 0 m size 12{"34" "." "0 m"} {} in a direction 63 . 0 º size 12{"63" "." 0º } {} north of east.

Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B.
Vector A size 12{A} {} has magnitude 53 . 0 m size 12{"53" "." "0 m"} {} and direction 20 . 0 º size 12{"20" "." 0 { size 12{ circ } } } {} north of the x -axis. Vector B size 12{B} {} has magnitude 34 . 0 m size 12{"34" "." "0 m"} {} and direction 63 . 0 º size 12{"63" "." 0° } {} north of the x -axis. You can use analytical methods to determine the magnitude and direction of R size 12{R} {} .

Strategy

The components of A size 12{A} {} and B 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 A size 12{A} {} and B size 12{B} {} along the x - and y -axes. Note that A = 53.0 m size 12{"A" "=" "53.0 m"} {} , θ A = 20.0º size 12{"θ" "subA" "=" "20.0°" } {} , B = 34.0 m size 12{"B" "=" "34.0" "m"} {} , and θ B = 63.0º size 12{θ rSub { size 8{B} } } {} . We find the x -components by using A x = A cos θ 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 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" {} } } {}

and

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" {} } } {}

Similarly, the y -components are found using A y = A sin θ A 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 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" {} } } {}

and

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" "." {} } } {}

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 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"} {}

and

R y = A y + B y = 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."} {}

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 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"} {}

so that

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

Finally, we find the direction of the resultant:

θ = tan 1 ( R y / R x ) =+ 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} } \( "48" "." 4/"65" "." 2 \) "."} {}

Thus,

θ = tan 1 ( 0 . 742 ) = 36 . 6 º . size 12{θ="tan" rSup { size 8{ - 1} } \( 0 "." "742" \) ="36" "." 6 { size 12{ circ } } "."} {}
The addition of two vectors A and B is shown. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. Vector B is of magnitude thirty four units and is inclined at angle sixty three degrees to the horizontal. The components of vector A are shown as dotted vectors A X is equal to forty nine point eight meter along x axis and A Y is equal to eighteen point one meter along Y axis. The components of vector B are also shown as dotted vectors B X is equal to fifteen point four meter and B Y is equal to thirty point three meter. The horizontal component of the resultant R X is equal to A X plus B X is equal to sixty five point two meter. The vertical component of the resultant R Y is equal to A Y plus B Y is equal to forty eight point four meter. The magnitude of the resultant of two vectors is eighty one point two meters. The direction of the resultant R is in thirty six point six degree from the vector A in anticlockwise direction.
Using analytical methods, we see that the magnitude of R size 12{R} {} is 81 . 2 m size 12{"81" "." "2 m"} {} and its direction is 36 . size 12{"36" "." 6°} {} north of east.

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 ) 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 are the negatives of the components of B size 12{B} {} . The x - and y -components of the resultant A B = R size 12{A- bold "B = R"} {} are thus

Practice Key Terms 1

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Source:  OpenStax, Sample chapters: openstax college physics for ap® courses. OpenStax CNX. Oct 23, 2015 Download for free at http://legacy.cnx.org/content/col11896/1.9
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