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Unformatted text preview: VECTOR ALGEBRA/ TRIG. IDENTITIES
Vector (K) : A mathematical object which has both a magnitude and a direction. Scalar (S): Has magnitude only I. If you multiply a vector Z by a scalar S you get a vector = S—K such that K] H K and has magnitude SV. This property allows us to express'any vector as a product of a scalar (magnitude)
and a unit vector (magnitude 1, direction only). Hence, we have written: 4=Aae as a vector of magnitude A in the +36 direction. Indeed, a vector along any direction a? can be
written as: ‘ V = Va? 11. Addition of Vectors. Given vectors VI and V2 we want to determine the Resultant Vector
5. : V! + V2 There are three methods for doing this: 1 i ) Geometry Choose a scale to represent V] and V2 , and draw a parallelogram. . The long diagonal gives
you a = V. + V2 You can get magnitude of R by using a scale, and of course measure (*9 R with a protractor. Also,
B. : VI _ V2 is determined by the short diagonal. Repeatedapplication of this construct will allow you to add
many vectors. " §=K+n+m+n+ﬁ+n
 as the vector which connects the “tail of VI
to the head of V6 . Further, it immediately follows that if all the vectors are parallel to one another
§=K3+n3+né—m&n =m+a+a—m+o3 (ii) Alg ebra/ Trig.
We want to calculate R, so as shown , drop a J. from C tom extended. Clearly, I
22 = 517169  
V2 L
59 = Cos® I
V2 ’J
4......
using Pythagoras’ Theorem 0 V; ,4 1)
R2 = 01)2 + CD2 I
=(Vl +V2Cos®)2 + (VZLS’z'nGD)2
=V] + V2C0s2® + 2V1V2C0s® + stmze
That is R = 4V3 + V22 + 2V1V2C0S® [2] Also
tan®R avg: V2S1n® . A
OD V1 + V2 0‘05 9
So indeed we have determined both the magnitude [Eq2] and direction [Eq3] of the vector E=(VI+V2) Again, if we have more than 2 vectors we can use Eq. [2] and [3] repeatedly to arrive at
§=K+Z+K+m were.» [3] (iii) The Method of Components This is the most elegant procedure for adding (or subtracting) many vectors.
We begin by deﬁning that the component of a vector K along any direction a? is a _Sc_alar
quantity. Vd = Vcos(Z,c?) That is, Vd = [magnitude of V]x [Cosine of angle between K and d ] Let us put our vector Z V N. B. If light were
in the xy coordinate falling straight down system and we see . Vx would be the
immediately that: ‘ ' “shadow” of V along x. V,c = VCos®
Vy = V cos(90 — (9) = VSin® and clearly V = «{sz + V; V tan (9 = —
. Vx This tells us that a vector can be speciﬁed either by writing magnitude (V) and direction (3) or
by writing the magnitudes of its components.
So now if we have many vectors: V1 = V1x£+Vlyj>
V2 = V2x£+ szf’ V; = Vix'ﬁiViyj) R:ZV}=EV;‘C£+ZVW}3 9[4]
=Rx£+Ryj2 9[4']
and hence R = Jsz + Ry2 [5]
R
tan®R =4 [6]
R x
where (9,. is the angle between R and )2. TRIG IDENTITIES Take two unit vectors and V2
making angles .91 and 82 with
the axis of x as shown. B. : 121 + V2
From Eq(1) R = ‘/1+1+2Cos(®2 —@,) [7]
Also 2 = Cos®lfc + Sin®1j2
V2 = Cos®2£+Sin®2f2
so Rx = (00.9031 + Cos®2) Ry = (5M9l + 5171(92) R = 1/(Cos®1 + 603692)2 + (Sin®l + 3mm)2 = 1/C0s2®, + (10326)2 + 200591692 +Sm2®, +Sin2®2 +ZSin®lSin®2
= 4/1 +1 + 2[Cos®,Cos®2 + Sin®,sm®z] [8] Compare Eqs [7] and [8] and you get the trig identity:
Cos(®1 ~®2) = C'osGDleSGD2 + Sin®lSin®2 9 Il Next, let (9, = ($493) x (mg—(93 —®2) = same)3 +®2) = a)“; — ®3)Cos®2+Sin(% — @3)Sz'n®2
Which gives another identity
Sin(®3 + @2) = Sin®3Cos®2 + Cos®38in®2 —> 12 if in 11 you put 34: ~32 and remember that
Sin(—®) = ~Sin® you get
Cos(®, +694) 2 Cos®lC0s®4 — Sin®1Sin®4) —> 1’3 and similarly
Sz'n(®3 —®5) = Sin®3Cos®5 —;S‘z'n®5Cos®3 —9 I4 ...
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 Spring '10
 Shawhan
 Physics

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