Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 1, Problem 20P

a)

To determine

The vectors P and Q are unit vectors and this vectors make an angles θ1 and θ2 with the x axis.

a)

Expert Solution
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Explanation of Solution

Given:

P=Pxax+Pyay        (I)

Q=Qxax+Qyay        (II)

Calculation:

Consider the two vectors P and Q.

  P=Pxax+Pyay        (I)

  Q=Qxax+Qyay        (II)

Represent the vector P as shown in figure (1).

Elements Of Electromagnetics, Chapter 1, Problem 20P , additional homework tip  1

Refer the figure (1) and write the expression for the sin and cosine angle.

  sinθ1=Opposite sideHypotenusesinθ1=Py|P|Py=|P|sinθ1

Similarly,

  cosθ1=AdjacentHypotenusecosθ1=Px|P|Px=|P|cosθ1

Calculate the magnitude of P vector using the figure (1).

  |P|=Px2+Py2

Substitute Px=|P|cosθ1 and Py=|P|sinθ1 in equation (I).

  P=(|P|cosθ1)ax+(|P|sinθ1)ay

Calculate the unit vector along P.

  ap=Pxax+PyayPx2+Py2ap=(|P|cosθ1)ax+(|P|sinθ1)ay|P|ap=|P|(cosθ1ax+sinθ1ay)|P|ap=(cosθ1ax+sinθ1ay)

Thus, ap=(cosθ1ax+sinθ1ay) hence proved.

Similarly represent the vector Q as shown in figure (1).

Elements Of Electromagnetics, Chapter 1, Problem 20P , additional homework tip  2

Refer the figure (2) and write the expression for the sin and cosine angle.

  sinθ2=Opposite sideHypotenusesinθ2=Qy|Q|Qy=|Q|sinθ2

Similarly,

  cosθ2=AdjacentHypotenusecosθ2=Qx|Q|Qx=|Q|cosθ2

Calculate the magnitude of P vector using the figure (2).

  |Q|=Qx2+Qy2

Substitute Qx=|Q|cosθ1 and Qy=|Q|sinθ2 in equation (I).

  Q=(|Q|cosθ2)ax+(|Q|sinθ2)ay

Calculate the unit vector along P.

  aQ=Qxax+QyayQx2+Qy2aQ=(|Q|cosθ2)ax+(|Q|sinθ2)ay|Q|aQ=|Q|((cosθ2)ax+(sinθ2)ay)|Q|aQ=(cosθ2)ax+(sinθ2)ay

Thus, aQ=(cosθ2ax+sinθ2ay) hence proved.

b)

To determine

The formula for cos(θ2θ1) and cos(θ2+θ1).

b)

Expert Solution
Check Mark

Explanation of Solution

Consider P and Q is making an angle of θ1 and θ2 respectively with the x axis as shown in figure (3).

Elements Of Electromagnetics, Chapter 1, Problem 20P , additional homework tip  3

Calculate the dot product of P and Q.

  PQ=PQcosθPQ

Here, the smaller angle between P and Q is θPQ.

  P·Q|PQ|=cos(θ2θ1)P·Q|PQ|=cos(θ2θ1)(Px,Py)(Qx,Qy)|P||Q|=cos(θ2θ1)PzQx+PyQy|PQ|=cos(θ2θ1)PsQx+PyQx|P||Q|=cos(θ2θ1)Px|P|Qx|Q|+Py|P|Qy|Q|=cos(θ2θ1)

Substitute the respective value in the above equation and obtain.

  cos(θ2θ1)=Px|P|Qx|Q|+Py|P|Qy|Q|cos(θ2θ1)=cosθ1cosθ2+sinθ1sinθ2

Thus, the formula for cos(θ2θ1) is cosθ1cosθ2+sinθ1sinθ2.

Consider the vectors P and Q are making angle θ1 and θ2 with the x axis as shown in figure (4).

Elements Of Electromagnetics, Chapter 1, Problem 20P , additional homework tip  4

Calculate the dot product of P and Q.

  PQ=PQcosθPQ

Here, the smaller angle between P and Q is θPQ=θ1+θ2.

  P·Q=PQcosθPQP·Q|PQ|=cos(θ1+θ2)P·Q|PQ| =cos(θ1+θ2)

  (Px,Py)(Qx,Qy)|P||Q|=cos(θ1+θ2)PxQxPyQy|PQ|=cos(θ1+θ2)PxQxPyQy|PQ|=cos(θ1+θ2)

  PxQx|PQ|PyQy|PQ|=cos(θ1+θ2)

Substituting sin and cosine value in the above equation and simplified as shown below.

  cos(θ1+θ2)=Px|P|Qx|Q|Py|P|Qy|Q|cos(θ1+θ2)=cosθ1cosθ2sinθ1sinθ2

Thus, the formula for cos(θ1+θ2) is cosθ1cosθ2sinθ1sinθ2.

c)

To determine

The expression for 12|PQ| if P is making an angle θ with Q.

c)

Expert Solution
Check Mark

Explanation of Solution

Consider the two vectors P and Q making an angle θ1 and θ2 with the x axis.

Calculate the PQ.

  PQ=(cosθ1cosθ2)ax+(sinθ1sinθ2)ay

Calculate the magnitude of |PQ|.

  |PQ|=(cosθ1cosθ2)2+(sinθ1sinθ2)2=cos2θ1+cos2θ22cosθ1cosθ2+sin2θ1+sin2θ22sinθ1sinθ2=cos2θ1+sin2θ1+cos2θ2+sin2θ22cosθ1cosθ22sinθ1sinθ2=22(cosθ1cosθ2+sinθsinθ2)

Calculate the 12|PQ|.

  12|PQ|==122(1cos(θ2θ1))=(1cos(θ2θ1))2=sin2(θ2θ1)2=|sin(θ2θ1)2|

Thus, 12|PQ| is equal to in terms of θ is |sin(θ2θ1)2|.

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