## Vector Components

We can describe any vector

Vin terms of its magnitude V and its direction . Here, we are measuring angle from the positive x-axis. And angle is positive for counter clockwise directions. Other coordinate systems may also be used; it is also common to measure angles from North.We can also describe a vector

Vin terms of itscomponentsalong the x- and y-axes,

V = V_{x}+ V_{y}How are these two descriptions related?

How are these two descriptions related? We have already seen that

V

_{x}= V cosV

_{y}= V sinfor this coordinate system. We will use those relationships shortly.

Now consider a second vector

U. It, too, may be expressed in terms of its components,

U = U_{x}+ U_{y}How can we add vectors

UandVin terms of their components? That is, we want a resultant vectorRthat is

R = U + V

R = (U_{x}+ U_{y}) + (V_{x}+ V_{y})Since

U = Uand_{x}+ U_{y}V = V_{x}+ V, we may eliminate the vectors_{y}UandVfrom our diagram.Now let us continue with our vector addition. Add the x-components together by themselves and then add the y-components together by themselves.

R = (U_{x}+ V_{x }) + ( U_{y }+ V_{y})

R_{x}= U_{x}+ V_{x}and

R_{y}= U_{y}+ V_{y}But look at these diagrams. While we have written

Rand_{x}= U_{x}+ V_{x}Ras_{y}= U_{y}+ V_{y}vector addition equations, they are rather unusual vector equations. All the x-componets lie along the x-axis and all the y-components lie along they y-axis. So these particular equations are also true asscalar addition equations!

That is,

R

_{x}= U_{x}+ V_{x}and

R

_{y}= U_{y}+ V_{y}Finding the components of the resultant, then, has been reduced tor ordinary, commonplace, scalar addition!

Now that we know the

componentsof the resultant, we can use them to reconstruct the resultant itself.This will involve trig functions and the Pythagorean theorem.

Now that we have the

idea, let's apply this numerical method of vector addition to the vector addition problem of the treasure map:Vector A has only an x-component, so we can write

A

_{x}= 5 pacesA

_{y}= 0"Northeast" means = 45

^{o}, soB

_{x}= (7 paces) (cos 45^{o}) = 5 pacesB

_{y}= (7 paces) (sin 45^{o}) = 5 paces"West" means "to the left" or that the x-component is

negative. The y-component is zero, so we haveC

_{x}= - 10 pacesC

_{y}= 0"Northwest" means = 135

^{o}, soD

_{x}= (7 paces) (cos 135^{o}) = - 5 pacesD

_{y}= (7 paces) (sin 135^{o}) = + 5 pacesAs always, watch the signs! They're important.

Now we're ready to add the component to find the components of the resultant. Vector notation is elegant shorthand notation. We write

R = A + B + C + Dbut this

meansthe followingtwoordinary, scalar equations:R

_{x}= A_{x}+ B_{x}+ C_{x}+ D_{x}

andR

_{y}= A_{y}+ B_{y}+ C_{y}+ D_{y}For our example, these equations are

R

_{x}= A_{x}+ B_{x}+ C_{x}+ D_{x}R

_{x}= ( 5 + 5 - 10 - 5 ) paces = - 5 pacesor 5 paces to the

westor to the "left" of the old oak tree (our origin).

andR

_{y}= A_{y}+ B_{y}+ C_{y}+ D_{y}R

_{y}= ( 0 + 5 + 0 + 5) paces = 10 pacesor 10 paces

northor "up" from the old oak tree ("up" meaning north, tho').Now we can reconstruct the resultant R and its direction.

R = SQRT [ R

_{x}^{2}+ R_{y}^{2}]R = SQRT[ (- 5)

^{2}+ (10)^{2}] pacesR = 11.2 paces

tan =

^{opp}/_{adj}= R_{y}/ R_{x}tan = ( 10 ) / ( - 5 ) = - 2

= 116.6

^{o}Remember, with our present coordinate system, this angle is measured from the positive x-axis. Let's talk about this angle itself. If you ask your calculator for the

inverse tangent of - 2, it will probably tell you = - 63.4. But look at the diagram. You know that cannotbe right! What happened? There are two possible solutions,= - 63.4

^{o}and

= - 63.4

^{o}+ 180^{o}= 116.6

^{o}You and your diagram must decide which to use. Calculators are great; just be sure you control the calculator rather than the other way round (and that is even more true -- and often more difficult -- with computers!)!

(c) Doug Davis, 2001; all rights reserved