PHY10 CO1 - Vectors

8/20/2019 PHY10 CO1 - Vectors

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Vectors

Engr. E. D. Dimaunahan


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#. $ifferentiate a scalar andvector quantities.

%. &ther name for vector sum.

'. Give eam(les of scalar and

vector quantities.


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VECTORS

resultant vector – the vector sum of two or more

concurrent vectors.

equilibrant vector – a single vector that balances twoor more concurrent vectors.

equivalent vectors – two vectors that have the samemagnitude and direction.

parallel vectors – two vectors that have the samedirection.

anti-parallel vectors – two vectors that have oppositedirections.

negative of a vector – a vector that has the samemagnitude as but opposite the direction of anothervector.


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Vectors and Their Components

The sim(lest eam(le is a displacement vector 0f a (article changes (osition from * to " we re(resent

this by a vector arrow (ointing from * to

0n a3 we see that all three arrows havethe same magnitude and direction- they

are identical dis(lacement vectors.

0n b3 we see that all three (aths

corres(ond to the same dis(lacementvector. The vector tells us nothing about

the actual (ath that was taken between

* and .


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The vector sum" or resultanto 0s the result of (erforming vector addition

o 4e(resents the net dis(lacement of two or more

dis(lacement vectors

o 5an be added gra(hically as shown-


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62.% m" #2 degrees east of north'6.# m" #7 degrees south of

west

16.8 m straight south


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ector addition is commutativeo We can add vectors in any order

Eq. (3-2)

igure (3-3)


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ector addition is associativeo We can grou( vector addition however we like

igure (3-!)


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* negative sign reverses vectordirection

We use this to define vector

subtraction

Eq. (3-!)


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These rules hold for all vectors" whether they re(resentdis(lacement" velocity" etc.

&nly vectors of the same kind can be added

o

distance3 9 distance3 makes senseo distance3 9 velocity 3 does not

*nswer-

a3 # m 9 % m : 6 m b3 % m ; # m : 1 m


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4ather than using a gra(hical method" vectors can beadded by components

o * com(onent is the (ro+ection of a vector on an ais

The (rocess of finding com(onents is called resolving

the vector

The com(onents of a vector

can be (ositive or negative.

They are unchanged if thevector is shifted in any

direction but not rotated3.


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5om(onents in two dimensions can be found by-

Where θ is the angle the vector makes with the (ositive x ais" and a is the vector length

The length and angle can also be found if the

com(onents are known

Therefore" com(onents fully define a vector


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*ngles may be measured in degrees or radians 4ecall that a full circle is #7


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#nit Vectors$ %dding Vectors &y Components

* unit vector o Aas magnitude 1

o Aas a (articular direction

o

Backs both dimension and unito 0s labeled with a hat- C

We use a right-handed coordinate system

o 4emains right;handed when rotated


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The quantities a x i and ay ' are vector components

The quantities a x and a

y alone are scalar components

o &r +ust Dcom(onentsE as before

ectors can be added using com(onents

Eq. (3-)


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To subtract two vectors" we subtract com(onents

Fnit ectors" *dding ectors by 5om(onents

*nswer- a3 (ositive" (ositive b3 (ositive" negative

c3 (ositive" (ositive


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ectors are inde(endent of thecoordinate system used to measure

them

We can rotate the coordinate system"

without rotating the vector" and thevector remains the same

*ll such coordinate systems are

equally valid


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*ultiplying Vectors

!ulti(lying a vector z by a scalar c o 4esults in a new vector

o 0ts magnitude is the magnitude of vector z times |c

o 0ts direction is the same as vector z " or o((osite if c is

negative

o To achieve this" we can sim(ly multi(ly each of the

com(onents of vector z by c

To divide a vector by a scalar we multi(ly by 1Hc E+ample !ulti(ly vector z by '

o

z : ;# i 9 ' '

o ' z : ;1' i 9 2' '


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*ultiplying Vectors

!ulti(lying two vectors- the scalar producto *lso called the dot product

o 4esults in a scalar" where a and b are magnitudes and φ isthe angle between the directions of the two vectors-

The commutative law a((lies" and we can do the dot

(roduct in com(onent form


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*ultiplying Vectors

* dot (roduct is- the (roduct of the magnitude of onevector times the scalar com(onent of the other vector in

the direction of the first vector

Eq. (3-2")

,ither (ro+ection of one

vector onto the other can

be used

To multi(ly a vector by the

(ro+ection" multi(ly the

magnitudes

> 2


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*ultiplying Vectors

*nswer- a3 I< degrees b3 < degrees c3 18< degrees


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*ultiplying Vectors

!ulti(lying two vectors- the vector producto The cross product of two vectors with magnitudes a @ b,

se(arated by angle φ, (roduces a vector with magnitude-

o *nd a direction (er(endicular to both original vectors

$irection is determined by the right-hand rule

Place vectors tail;to;tail" swee( fingers from the first to

the second" and thumb (oints in the direction of the

resultant vector

Eq. (3-2!)


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*ultiplying Vectors

The u((er shows vector a cross vector b" the lower shows vector b cross vector a


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*ultiplying Vectors

*nswer- a3 < degrees b3 I< degrees


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UNIT VECTOR

A unit vector is a vector with a magnitude of exactly

1 and points in a particular direction.

9

9y

9)

A vector in its “regular”form or “magnitude-angle”form, such as

can be expressed in “unitvector” form or “rectangular”

form.

î

jˆ

xabovem A o += 30,00.10

jmim A ˆ)00.5(ˆ)66.8( +=


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9

9y

9)

VECTOR AITION

9

9y

9)

i A x ˆ

k A zˆ

j A y

ˆ

A

B

i B x

ˆ

k B Z ˆ

j B yˆ

k m jmim A ˆ)6(ˆ)8(ˆ)10( ++= k m jmim B

ˆ

)4(ˆ

)12(ˆ

)8( ++−=


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Find the resultant of the following vectors.

k m jmim A ˆ)6(ˆ)8(ˆ)10( ++=

k m jmim B ˆ)4(ˆ)12(ˆ)8( ++−=

k m jmimC ˆ)10(ˆ)20(ˆ)2( ++=

B AC +=k m jmim A ˆ)6(ˆ)8(ˆ)10( ++=

k m jmim B ˆ)4(ˆ)12(ˆ)8( ++−=


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VECTOR !ROUCTS" Scalar#ot !ro$uct

iven the following vectors!

"he scalar #or dot$ product of these vectors is defined as!

%here! A is the magnitude of

B is the magnitude of

φ is the angle between and .

k A j Ai A A z y xˆˆˆ ++= k B j Bi B B z y x

ˆˆˆ ++=

φ cos AB B A =⋅

A

B

A B


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φ cos AB B A =⋅

10cos)1)(1(ˆˆ ==⋅ii

090cos)1)(1(ˆˆ ==⋅ ji090cos)1)(1(ˆˆ ==⋅k i

090cos)1)(1(ˆˆ ==⋅i j

10cos)1)(1(ˆˆ ==⋅ j j

090cos)1)(1(ˆˆ ==⋅k j

090cos)1)(1(ˆˆ ==⋅ik

090cos)1)(1(ˆˆ ==⋅ jk

10cos)1)(1(ˆˆ ==⋅k k


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k B j Bi Bk A j Ai A B A z y x z y xˆˆˆˆˆˆ ++⋅++=⋅

k B j Bi B B z y xˆˆˆ ++=k A j Ai A A z y x

ˆˆˆ ++=

( ) ( ) ( ) z z y y x x B A B A B A B A ++=⋅

φ cos AB B A =⋅


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Find the dot product of the following vectors!


ˆ

)4(ˆ

)12(ˆ

)8( ++−=

( ) ( ) ( ) z z y y x x B A B A B A B A ++=⋅

)]4)(6[()]12)(8[()]8)(10[( mmmmmm B A ++−=⋅

222 249680 mmm B A ++−=⋅

240m B A =⋅


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Find the angle between the following vectors!


ˆ

)4(ˆ

)12(ˆ

)8( ++−=

φ cos AB B A =⋅

( ) ( ) φ cos22420040 =

224200

40cos 1−=φ

o11.79=φ


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VECTOR !ROUCTS" vector#Cross !ro$uct

iven the following vectors!

"he vector #or cross$ product of these vectors is defined

as!

%here! A is the magnitude of

B is the magnitude of

φ is the angle between and .

k A j Ai A A z y xˆˆˆ ++= k B j Bi B B z y x

ˆˆˆ ++=

φ sin AB B A =×

A

B

A B


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VECTOR !ROUCTS" Vector#Cross !ro$uct

φ sin AB B A =×

00sin)1)(1(ˆˆ ==× ii

k ji ˆ

90sin)1)(1(ˆˆ ==×

jk i ˆ90sin)1)(1(ˆˆ −==×

k i j ˆ

90sin)1)(1(ˆˆ −==×

00sin)1)(1(ˆˆ ==× j j

ik j ˆ90sin)1)(1(ˆˆ ==×

jik ˆ90sin)1)(1(ˆˆ ==×

i jk ˆ90sin)1)(1(ˆˆ −==×

00sin)1)(1(ˆˆ ==× k k


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VECTOR !ROUCTS" Vector#Cross !ro$uct

k B j Bi Bk A j Ai A B A z y x z y x ˆˆˆˆˆˆ ++×++=×

k B j Bi B B z y xˆˆˆ ++=k A j Ai A A z y x

ˆˆˆ ++=

φ sin AB B A =×

( ) ( ) ( ) k B A B A j B A B Ai B A B A B A x y y x z x x z y z z y

ˆˆˆ −+−+−=×


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Find the cross product of the following vectors!

k m jmim A ˆ)6(ˆ)8(ˆ)10( ++= k m jmim B ˆ)4(ˆ)12(ˆ)8( ++−=

( ) ( ) ( ) k B A B A j B A B Ai B A B A B A x y y x z x x z y z z y ˆˆˆ −+−+−=×

( ) ( ) ( ) k ji B A ˆ)64(120ˆ4048ˆ7232 −−+−−+−=×

( ) ( ) ( ) k m jmim B A ˆ184ˆ88ˆ40 222 +−−=×


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!RO%&E'"


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!RO%&E'"

&n the figure at right, a cube isplaced so that one corner #point'$ is at the origin and threeedges are along the (x (y, and() axes of a coordinate system.*se unit vectors to find the

#a$ angle between the edgealong the )-axis #line '+$ andthe diagonal of the cube #line'$.

#b$ angle between the diagonalof a face #line '$ and line '.

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PHY10 CO1 - Vectors

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