vector_algebra_trig

# vector_algebra_trig - VECTOR ALGEBRA/ TRIG. IDENTITIES...

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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 x-y 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'ﬁ-i-Viyj) 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 + 51-71(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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## vector_algebra_trig - VECTOR ALGEBRA/ TRIG. IDENTITIES...

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