Fundamentals of Applied Electromagnetics (7th Edition)
Fundamentals of Applied Electromagnetics (7th Edition)
7th Edition
ISBN: 9780133356816
Author: Fawwaz T. Ulaby, Umberto Ravaioli
Publisher: PEARSON
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Question
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Chapter 3, Problem 1P
To determine

The unit vector in the direction of A.

Expert Solution & Answer
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Answer to Problem 1P

The unit vector in the direction of A is (x^0.32+z^0.95)_.

Explanation of Solution

Given data:

The first point is (1,1,3).

The second point is (2,1,0).

Calculation:

Let the first point is A1 and the second point is A2.

The position vector of point (1,1,3) is given as,

OP=x^x+y^y+z^z (1)

Here,

O is the origin,

OP is the position vector at point P.

Substitute 1 for x, 1 for y, 3 for z and OA1 for OP in equation (1).

OA1=x^(1)+y^(1)+z^(3)=x^y^3z^

Substitute 2 for x, 1 for y, 0 for z and OA2 for OP in equation (1).

OA2=x^(2)+y^(1)+z^(0)=x^2y^

The distance vector between the points A1 and A2 is given as,

A=OA2OA1

Here,

A is the distance vector between the points A1 and A2.

Substitute x^y^3z^ for OA1 and x^2y^ for OA2 in the above expression.

A=(x^2y^)(x^y^3z^)

A=(x^+3z^) (2)

The general form of a vector is given as,

A=(x^Ax+y^Ay+z^Az) (3)

Compare equation (2) and (3), the values are written as,

Ax=1

Ay=0

Az=3

The magnitude of the vector A is calculated as,

A=(Ax)2+(Ay)2+(Az)2

Here,

A is the magnitude of the vector A.

Substitute 1 for Ax, 0 for Ay and 3 for Az in the above expression.

A=(1)2+(0)2+(3)2=1+9=10

The unit vector in the direction of A is given as,

a^=AA

Here,

a^ is the unit vector in the direction of A.

Substitute (x^+3z^) for A and 10 for A in the above expression.

a^=(x^+3z^)10=(x^0.32+z^0.95)

Conclusion:

Therefore, the unit vector in the direction of A is (x^0.32+z^0.95)_.

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Chapter 3 Solutions

Fundamentals of Applied Electromagnetics (7th Edition)

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