Electromagnetic Waves, Antennas, and RF: Vector Fields for Wave Mechanics - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM02, vector fields for wave mechanics. All exercises are stated first. Complete worked solutions follow in Part II.

The problems remain intentionally within the EM02 mathematical foundation. They practice vector components, magnitude, unit vectors, dot products, projections, wave vectors, three-dimensional plane-wave phase, cross products, and the geometric relation among electric field, magnetic field, and propagation direction. Gradient, divergence, curl, and Maxwell’s equations are deferred to EM03 and later articles.

The central formulas used throughout are

(1)

with magnitude

(2)

the dot product

(3)

the scalar projection onto a unit direction n,

(4)

the cross product magnitude

(5)

and the three-dimensional plane-wave phase

(6)

These are the same definitions and conventions used in EM02 [1, 2, 3, 4].

How to use this problem set

Attempt all exercises in Part I before reading Part II. For every vector calculation, keep three questions separate:

1. What are the scalar components?
2. What is the vector direction?
3. What physical or geometric meaning does the result have?

A correct numerical result with the wrong direction is not a correct vector answer.

Part I: Exercises

Exercise 1: components, magnitude, and unit vector

Let

(7)

Find:

1. the three Cartesian components;
2. the magnitude |A|;
3. the unit vector A;
4. a numerical check that |A| = 1.

Exercise 2: displacement vector between two points

Two points are

(8)

and

(9)

Find the displacement vector from P to Q, its magnitude, and the corresponding unit direction vector.

Exercise 3: use the dot product to test perpendicularity

Let

(10)

and

(11)

Compute A ⋅ B and determine whether the vectors are perpendicular.

Exercise 4: projection onto a sensor axis

An electric-field vector at one point is

(12)

A sensor responds only along the unit direction

(13)

Find:

1. the scalar component E_∥ = E ⋅n;
2. the vector projection E_∥;
3. the perpendicular remainder E_⊥;
4. the magnitude |E_⊥|.

Exercise 5: find the angle from a dot product

Let

(14)

and

(15)

Find the angle between the vectors in degrees.

Exercise 6: construct a wave vector from wavelength and direction

A plane wave has wavelength

(16)

and propagates in the unit direction

(17)

Find:

1. the scalar Wavenumber k;
2. the wave vector k in Cartesian components.

Exercise 7: evaluate a three-dimensional plane-wave phase

Let

(18)

(19)

(20)

Find:

1. the spatial phase k ⋅ r;
2. the total phase k ⋅ r − ωt.

Exercise 8: constant-phase planes

A plane wave has

(21)

At a fixed time, constant-phase surfaces satisfy

(22)

Answer the following:

1. Write the constant-phase equation in x and y.
2. Show that the points

   (23)

   and

   (24)

   lie on the same constant-phase plane.

3. Explain geometrically why k is perpendicular to those planes.

Exercise 9: compute a cross product in Cartesian form

Let

(25)

and

(26)

Find:

1. A × B;
2. |A × B|;
3. B × A.

Exercise 10: parallel and perpendicular cross-product limits

Without using the component formula, evaluate each expression and explain the geometry:

1. x ×x;
2. x ×y;
3. y ×x;
4. z ×x.

Exercise 11: field direction versus propagation direction

Consider

(27)

Identify:

1. the electric-field direction;
2. the wave vector k;
3. the propagation direction;
4. whether the field is transverse to the propagation direction.

Exercise 12: complete a transverse electromagnetic triad

Suppose an ideal plane-wave geometry has

(28)

and

(29)

Find the direction of H so that

(30)

points along +z.

Then verify the three perpendicularity conditions using dot products.

Exercise 13: read a wave vector from the phase

Consider

(31)

Find:

1. the wave vector k;
2. the scalar wavenumber k;
3. the wavelength λ;
4. the propagation unit vector k;
5. a dot-product check that the electric field is transverse to k.

Exercise 14: synthesis - vector field, phase, sensor projection, and triad geometry

A plane-wave electric field is

(32)

At

(33)

and

(34)

answer the following:

1. Find k and k = |k|.
2. Find the wavelength λ.
3. Find the propagation unit vector k.
4. Evaluate the phase at (r₀,t₀).
5. Evaluate E(r₀,t₀) as a multiple of E₀.
6. A sensor responds along

   (35)

   Find the sensor-measured scalar field component as a multiple of E₀.

7. Determine a unit direction for H such that E × H points along k.

Part II: Complete Worked Solutions

Solution 1: components, magnitude, and unit vector

The vector is

(36)

Therefore the scalar components are

(37)

Its magnitude is

|A|	=	(38)
	=	(39)
	=	(40)
	= 7.	(41)

Hence

(42)

The unit vector is

A	=	(43)
	= x −y + z.	(44)

Thus

(45)

Finally,

|A|	=	(46)
	=	(47)
	= 1.	(48)

So the normalization check passes.

Solution 2: displacement vector between two points

The displacement from P to Q is

(49)

Therefore

d	= (5 − 1)x + (1 − (−2))y + (0 − 0)z	(50)
	= 4x + 3y m.	(51)

Hence

(52)

Its magnitude is

(53)

The unit direction is

(54)

Solution 3: use the dot product to test perpendicularity

Compute

A ⋅ B	= (2)(3) + (3)(−2) + (−1)(0)	(55)
	= 6 − 6	(56)
	= 0.	(57)

For nonzero vectors, zero dot product means the vectors are perpendicular. Therefore

(58)

Solution 4: projection onto a sensor axis

The scalar component along the sensor axis is

E∥	= E ⋅n	(59)
	= ⋅	(60)
	= +	(61)
	= V/m.	(62)

Thus

(63)

The vector projection is

E∥	= (E ⋅n)n	(64)
	= 5.20 V/m	(65)
	= 3.12x + 4.16y V/m.	(66)

Therefore

(67)

The perpendicular part is

E⊥	= E − E∥	(68)
	= (6 − 3.12)x + (2 − 4.16)y V/m	(69)
	= 2.88x − 2.16y V/m.	(70)

Hence

(71)

Its magnitude is

|E⊥|	=	(72)
	= 3.60 V/m.	(73)

Thus

(74)

Solution 5: find the angle from a dot product

First compute the dot product:

A ⋅ B	= (1)(2) + (2)(1) + (2)(2)	(75)
	= 8.	(76)

The magnitudes are

(77)

and

(78)

Therefore

(79)

Hence

(80)

Solution 6: construct a wave vector from wavelength and direction

The scalar wavenumber is

k	=	(81)
	=	(82)
	≈ 10.472 rad/m.	(83)

Therefore

(84)

The wave vector is

k	= kn	(85)
	= 10.472 rad/m.	(86)

Thus

(87)

Solution 7: evaluate a three-dimensional plane-wave phase

The spatial phase is

k ⋅ r	= (2)(1) + (3)(2) + (6)(−1)	(88)
	= 2 + 6 − 6	(89)
	= 2 rad.	(90)

Hence

(91)

The time-dependent contribution is

(92)

Therefore

(93)

Solution 8: constant-phase planes

With

(94)

the constant-phase condition is

(95)

Therefore

(96)

For

(97)

we obtain

(98)

For

(99)

we obtain

k ⋅ r2	= 3(−1) + 4(1.5)	(100)
	= −3 + 6	(101)
	= 3.	(102)

Thus both points have the same value of k ⋅ r and lie on the same constant-phase plane.

Geometrically, the equation

(103)

defines a plane whose normal vector is k. Any displacement Δr lying within the plane satisfies

(104)

so in-plane directions are perpendicular to k.

Solution 9: compute a cross product in Cartesian form

Use

(105)

and

(106)

The cross product is

A × B	= (AyBz − AzBy)x + (AzBx − AxBz)y + (AxBy − AyBx)z	(107)
	= (2 ⋅ 3 − 0)x + (0 − 1 ⋅ 3)y + (0 − 0)z	(108)
	= 6x − 3y.	(109)

Therefore

(110)

Its magnitude is

|A × B|	=	(111)
	=	(112)
	= 3.	(113)

Hence

(114)

Reversing the order changes the sign:

(115)

Solution 10: parallel and perpendicular cross-product limits

For parallel unit vectors,

(116)

This follows from sin 0 = 0.

Using the right-handed Cartesian basis,

(117)

Reversing the order reverses the sign:

(118)

Finally,

(119)

The perpendicular cases all have unit magnitude because both input vectors have unit magnitude and sin(π∕2) = 1.

Solution 11: field direction versus propagation direction

The field is

(120)

The vector prefactor x gives the electric-field direction. Thus

(121)

The spatial phase is 5z, so

(122)

Because the phase has the form kz − ωt, the pattern propagates toward increasing z:

(123)

Finally,

(124)

so the electric field is transverse to the propagation direction.

Solution 12: complete a transverse electromagnetic triad

We are given

(125)

and

(126)

We require

(127)

to point along +z.

Since

(128)

we choose

(129)

The three perpendicularity checks are

(130)

(131)

and

(132)

Thus the triad is mutually perpendicular and right handed.

Solution 13: read a wave vector from the phase

The phase is

(133)

Therefore

(134)

Its magnitude is

k	=	(135)
	= 5 rad/m.	(136)

Thus

(137)

The wavelength is

λ	=	(138)
	= m	(139)
	≈ 1.257 m.	(140)

Hence

(141)

The propagation unit vector is

(142)

The electric-field direction is y. Therefore

y ⋅k	= y ⋅	(143)
	= 0.	(144)

Thus the electric field is transverse to the propagation direction.

Solution 14: synthesis - vector field, phase, sensor projection, and triad geometry

The field is

(145)

The spatial phase gives

(146)

Its magnitude is

k	=	(147)
	= 10 rad/m.	(148)

Thus

(149)

The wavelength is

λ	= m	(150)
	≈ 0.6283 m.	(151)

Therefore

(152)

The propagation unit vector is

(153)

At

(154)

the spatial phase is

k ⋅ r0	= (6)(0.10) + (8)(0.20)	(155)
	= 0.6 + 1.6	(156)
	= 2.2 rad.	(157)

The temporal term is

(158)

Therefore the total phase is

(159)

The electric field is

(160)

Since

(161)

we obtain

(162)

The sensor axis is

(163)

The measured scalar component is

Emeas	= E ⋅n	(164)
	= 0.3624E0y ⋅	(165)
	= 0.3624E0	(166)
	≈ 0.2174E0.	(167)

Thus

(168)

Finally, the electric-field direction is y and the propagation direction is

(169)

A unit magnetic-field direction that completes the right-handed triad is

(170)

Therefore

H	= ×y	(171)
	= 0.6z − 0.8x.	(172)

Hence

(173)

A check gives

(174)

so the orientation is correct.

Common mistakes

• Mistake: treating a scalar component such as A_x as though it were itself a vector. The vector contribution is A_xx.
• Mistake: allowing a vector magnitude to be negative. Component signs encode direction; magnitude is nonnegative.
• Mistake: using the dot product when a perpendicular vector is required. The dot product returns a scalar; the cross product returns a vector.
• Mistake: forgetting to normalize a measurement direction before interpreting A⋅n as a scalar component.
• Mistake: confusing k with k. The scalar k is the magnitude; the vector k also contains propagation direction.
• Mistake: reading field direction from the phase. The vector prefactor determines field direction; the spatial phase determines propagation direction.
• Mistake: assuming A × B = B × A. Reversing the order reverses the sign.
• Mistake: applying the ideal plane-wave triad to arbitrary electromagnetic fields. EM02 and EM02E use it only as a preview of the uniform plane-wave geometry derived later.

What EM02E reinforces

The component form

(175)

makes vector direction and magnitude explicit.

The dot product provides alignment, projection, and perpendicularity tests:

(176)

The cross product provides a perpendicular direction with right-handed orientation:

(177)

For wave mechanics, the vector phase

(178)

contains the propagation geometry, while the vector field prefactor contains the field direction.

The electromagnetic geometry that later follows from Maxwell’s equations is previewed by

(179)

EM03 next introduces spatial derivatives of fields: gradient, divergence, and curl.

References