Plane Waves – GATE Problems (Part – I)

1. A plane electromagnetic wave traveling along the + z – direction, has its

electric field given by 𝐸𝑥 = 2 cos(𝜔𝑡) 𝑎𝑛𝑑 𝐸𝑦 = 2 cos(𝜔 + 900) the

wave is

(a) linearly polarized

(b) right circularly polarized

(c) left circularly polarized

(d) elliptically polarized

[GATE 1994: 1 Mark]

Soln. 𝑬𝒙(𝒕) = 𝟐 𝐜𝐨𝐬𝝎𝒕

𝑬𝒚(𝒕) = 𝟐 𝐜𝐨𝐬(𝝎𝒕 + 𝟗𝟎𝟎) = −𝟐 𝐬𝐢𝐧𝝎𝒕

𝑬𝒙𝟐(𝒕) + 𝑬𝒚

𝟐(𝒕) = 𝟐𝟐

It represents a circle in the 𝑬𝒙 − 𝑬𝒚 plane with radius 2 as shown in

figure. Hence the wave is circularly polarized.

-2

-2 -1-1

1

2

2

1

𝐸𝑦(𝑡)

𝐸𝑥(𝑡)

𝜔𝑡 → 0 𝑡𝑜 2𝜋

𝝎𝒕 = 𝟎 𝝅 𝟐⁄ 𝝅 𝟑𝝅

𝟐 𝟐𝝅

𝑬𝒙 = 𝟐 𝟎 − 𝟐 𝟎 𝟐

𝑬𝒚 = 𝟎 − 𝟐 𝟎 𝟐 𝟎

When the fingers of the left hand follows the clock wise direction

(direction of rotation of the E vector), the thumb is pointing in the

given direction of propagation (+ z direction). The wave is left

circularly polarized.

Option (c)

2. The intrinsic impedance of a lossy dielectric medium is given by

(a) 𝑗𝜔𝜇

𝜎

(b) 𝑗𝜔∈

𝜇

(c) √𝑗𝜔𝜇

(𝜎+𝑗𝜔𝜖)

(d) √𝜇

∈

[GATE 1995: 1 Mark]

Soln. Conductivity = 𝝈 𝒎𝒉𝒐𝒔/𝒎

Permittivity = ∈ 𝒇𝒂𝒓𝒂𝒅 / 𝒎

Permeability = 𝝁 𝒉𝒆𝒏𝒓𝒚 / 𝒎

E / H for a lossy dielectric medium = 𝜼 = √𝒋𝝎𝝁

𝝈+𝒋𝝎𝝐

Option (c)

3. Copper behaves as a

(a) Conductor always.

(b) Conductor or dielectric depending on the applied electric field strength

(c) Conductor or dielectric depending on the frequency

(d) Conductor or dielectric depending on the electric current density

[GATE 1995: 1 Mark]

Soln. For a conductor 𝝈 ≫ 𝝎𝝐

For copper with 𝝈 = 𝟓. 𝟖 × 𝟏𝟎𝟕 𝒎𝒉𝒐𝒔/𝒎

∈ = ∈𝟎=𝟏

𝟑𝟔𝝅×𝟏𝟎𝟗 𝒇𝒂𝒓𝒂𝒅/𝒎

at relatively large frequency

𝒇 = 𝟑 × 𝟏𝟎𝟏𝟓 𝑯𝒛

𝝈

𝝎𝝐=

𝟓. 𝟖 × 𝟏𝟎𝟕 × 𝟑𝟔𝝅 × 𝟏𝟎𝟗

𝟐𝝅 × 𝟑 × 𝟏𝟎𝟏𝟓= 𝟑𝟒𝟖

Copper is good conductor for the frequencies used in practice

Option (a)

4. The intrinsic impedance of copper at high frequency is

(a) Purely resistive

(b) Purely inductive

(c) Complex with a capacitive component

(d) Complex with a inductive component

[GATE 1998: 1 Mark]

Soln. The intrinsic impedance 𝜼 = √𝒋𝝎𝝁

𝝈+𝒋𝝎𝝐

For a good conductor 𝝈 ≫ 𝝎 ∈

𝜼 = √𝒋𝝎𝝁

𝝈

= √𝝎𝝁

𝝈 𝒆𝒋𝟒𝟓𝟎

= 𝜼𝑹 + 𝒋𝜼𝒙

𝜼𝑹 = 𝜼𝒙 = √𝝎𝝁

𝟐𝝈

η is complex with inductive component

Option (d)

5. The wavelength of wave with propagation constant (0.1 𝜋 + 𝑗 0.2𝜋)𝑚−1

is

(a) 2

√0.05𝑚

(b) 10 m

(c) 20 m

(d) 30 m

[GATE 1998: 1 Mark]

Soln. Propagation constant = 𝜶 + 𝒋𝜷

= (𝟎. 𝟏𝝅 + 𝒋 𝟎. 𝟐𝝅)𝒎−𝟏

𝛂 = 𝐚𝐭𝐭𝐞𝐧𝐮𝐚𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 = 𝟎. 𝟏 𝛑

𝛃 = 𝐩𝐡𝐚𝐬𝐞 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 = 𝟎. 𝟐 𝛑

wavelength 𝝀 =𝟐𝝅

𝜷=

𝟐𝝅

𝟎.𝟐𝝅= 𝟏𝟎𝒎

Option (b)

6. The depth of penetration of a wave in a lossy dielectric increases with

increasing

(a) Conductivity

(b) Permeability

(c) Wavelength

(d) Permittivity

[GATE 1998: 1 Mark]

Soln. Depth of penetration 𝜹 =𝟏

𝜶

𝛂 = 𝐚𝐭𝐭𝐞𝐧𝐮𝐚𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

𝜸 = 𝐏𝐫𝐨𝐩𝐚𝐠𝐚𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

= 𝜶 + 𝒋𝜷

= √𝒋𝝎𝝁 (𝝈 + 𝒋𝝎𝝐)

𝜶 = 𝝎 √𝝁𝝐

𝟐[√𝟏 +

𝝈𝟐

𝝎𝟐𝝐𝟐− 𝟏]

𝜷 = 𝝎 √𝝁𝝐

𝟐[√𝟏 +

𝝈𝟐

𝝎𝟐𝝐𝟐+ 𝟏]

For a lossy dielectric with 𝝈 ≠ 𝟎

α increases with increasing µ and σ good conductor with 𝝈

𝝎𝝐≫ 𝟏

𝜶 = 𝜷 = 𝝎√𝝁𝝐

𝟐

𝝈

𝝎𝝐

= √𝝎𝟐𝝁𝝈

𝟐𝝎

= √𝝎𝝁𝝈

𝟐

𝜶 ∝ 𝒇 𝒐𝒓 𝜹 ∝ 𝟏

𝒇 𝒐𝒓 𝜹 ∝ 𝝀

Option (c)

7. The polarization of a wave with electric field vector

�⃗� = 𝐸0𝑒𝑗(𝜔𝑡−𝛽𝑧)(𝑎 𝑥 + 𝑎 𝑦)

(a) Linear

(b) Elliptical

(c) Left hand circular

(d) Right hand circular

[GATE 1998: 1 Mark]

Soln. �⃗⃗� = 𝑬𝟎𝒆𝒋(𝝎𝒕−𝜷𝒛)(�⃗⃗� 𝒙 + �⃗⃗� 𝒚)

It is a wave propagating in Z direction with electric field components

in x and y direction

at 𝒛 = 𝟎

𝑬𝒙 = 𝑬𝟎 𝐜𝐨𝐬𝝎𝒕

𝑬𝒚 = 𝑬𝟎 𝐜𝐨𝐬𝝎𝒕

𝑬𝒚 = 𝑬𝒙 𝐚𝐭 𝐚𝐧𝐲 𝐭𝐢𝐦𝐞 𝐭

𝜔𝑡 = 𝜋/2, 3𝜋/2

𝐸𝑦

𝐸𝑥

−𝐸0

𝐸0

−𝐸0

𝐴 𝜔𝑡 = 0, 2𝜋

0

𝜔𝑡 = 𝜋

𝐵

𝐸0

45

1

As 𝝎𝒕 varies from 0 to π the tip of �⃗⃗� vector moves along the straight

line AB from A to B and as 𝝎𝒕 varices from π to 2π the tip of �⃗⃗� vector moves back from B to A and the cycle repeats. The

polarization of the wave is linear

Option (a)

8. A TEAM wave is incident normally upon a perfect conductor. The E and

H fields at the boundary will be respectively

(a) Minimum and minimum

(b) Maximum and maximum

(c) Minimum and maximum

(d) Maximum and minimum

[GATE 2000: 1 Mark]

Soln. IN the case of a plane wave incident normally upon the surface of a

perfect conductor, the wave is entirely reflected, neither E nor H can

exist within a perfect conductor.

According to the boundary condition

𝑬𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒊𝒂𝒍 = 𝑬𝒊𝒏𝒄 + 𝑬𝒓𝒆𝒇

= 0

𝑬𝒊𝒏𝒄 is reflected with phase reversal

E is minimum equal to zero at the surface of a perfect conductor

The magnetic field H must be reflected without reversal of phase. It

both E and H are reversed with phase reversal there would be no

reversal of direction of propagation. The phase of the reflected

magnetic field strength 𝑯𝒓 is the same.

𝑯𝒕𝒂𝒏 = 𝑯𝒊 + 𝑯𝒓 = 𝟐𝑯𝒊 = 𝑱𝑺

Where 𝑱 𝑺 is the linear current density in A/m on the surface.

H is maximum equal to twice the incident value at the boundary

Per

fect

Co

nd

uct

or

𝐸𝑇

𝐻𝑇

Option (c)

9. If a plane electromagnetic wave satisfies the equation 𝜕2𝐸𝑥

𝜕𝑧2 = 𝑐2 𝜕2𝐸𝑥

𝜕𝑡2 ,

the wave propagates in the

(a) x – direction

(b) z – direction

(c) y – direction

(d) x z plane at an angle of 450 between the x and z directions

[GATE 2001: 1 Mark]

Soln. The wave equations for free space (In a perfect dielectric containing

no charges)

𝛁𝟐𝑬 = 𝝁𝝐 �̈�

𝛁𝟐𝑯 = 𝝁𝝐 �̈�

The wave equation reduces to a simple form where E and H are

considered to be independent of two dimensions (x and y)

𝝏𝟐𝑬

𝝏𝒛𝟐= 𝝁𝝐

𝝏𝟐𝑬

𝝏𝒕𝟐

For uniform plane propagating in the Z direction, E may have

components 𝑬𝒙 and 𝑬𝒚

𝝏𝟐𝑬𝒙

𝝏𝒛𝟐= 𝝁𝝐

𝝏𝟐𝑬𝒙

𝝏𝒕𝟐

Where 𝒗𝟎 =𝟏

√𝝁𝝐 is the velocity of propagation

Option (b)

10. The depth of penetration of electromagnetic wave in a medium having

conductivity 𝜎 at a frequency 1 KHz is 25 cm. The depth of penetration at

a frequency of 4 KHz will be

(a) 6.25 cm

(b) 12.50 cm

(c) 50.00 cm

(d) 100. 00 cm

[GATE 2003: 1 Mark]

Soln. Depth of penetration 𝜹𝟏 = 𝟐𝟓𝒄𝒎 𝒂𝒕 𝒇𝟏 = 𝟏𝑲𝑯𝒛 conductivity 𝝈

For a medium to be good conductor 𝝈

𝝎𝝐≫ 𝟏

𝜹 =𝟏

∝ 𝐰𝐡𝐞𝐫𝐞 ∝ = 𝐚𝐭𝐭𝐞𝐧𝐮𝐚𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

𝜸 = 𝜶 + 𝒋𝜷 = √𝒋𝝎𝝁 + (𝝈 + 𝒋𝝎𝝐)

For a lossy dielectric, considered as a good conductor 𝝈

𝝎𝝐≫ 𝟏

𝜶 = 𝜷 = √𝝎𝝁𝝈

𝟐= √𝝅𝒇𝝁𝝈

𝜹 =𝟏

𝜶=

𝟏

√𝝅𝒇𝝁𝝈

𝒐𝒓 𝜹𝜶𝟏

√𝒇 , 𝜹𝟐 𝐛𝐞 𝐭𝐡𝐞 𝐝𝐞𝐩𝐭𝐡 𝐨𝐟 𝐩𝐞𝐧𝐞𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐚𝐭 𝒇𝟐 = 𝟒𝑲𝑯𝒛

𝜹𝟏

𝜹𝟐= √

𝒇𝟐

𝒇𝟏

𝜹𝟏

𝜹𝟐= √

𝟒

𝟏= 𝟐

𝜹𝟐 =𝜹𝟏

𝟐= 𝟏𝟐. 𝟓𝒄𝒎

Option (b)

11. The magnetic field intensity vector of a plane wave is given by

�⃗⃗� (𝑥, 𝑦, 𝑧, 𝑡) = 10 sin(50000𝑡 + 0.004𝑥 + 30)�̂�𝑦 𝑤ℎ𝑒𝑟𝑒 �̂�𝑦 denotes the

unit vector in y direction. The wave is propagating with a phase velocity

(a) 5 × 104𝑚/𝑠

(b) −3 × 108𝑚/𝑠

(c) −1.25 × 107𝑚/𝑠

(d) 3 × 108𝑚/𝑠

[GATE 2005: 1 Mark]

Soln. 𝑯(𝒙, 𝒚, 𝒛, 𝒕) = 𝟏𝟎 𝐬𝐢𝐧[𝟓𝟎𝟎𝟎𝟎𝒕 + 𝟎. 𝟎𝟎𝟒𝒙 + 𝟑𝟎]𝒂𝒚⃗⃗ ⃗⃗ = 𝑯𝒚 𝒂𝒚⃗⃗ ⃗⃗

𝑯𝒚 = 𝟏𝟎𝐬𝐢𝐧[𝝎𝒕 + 𝜷𝒙 + 𝟑𝟎]

𝝎 = 𝟓𝟎, 𝟎𝟎𝟎 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 / 𝒔𝒆𝒄

𝜷 = −𝟎. 𝟎𝟎𝟒 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 / 𝒎

Phase velocity 𝑽𝒑 =𝝎

𝜷=

𝟓𝟎×𝟏𝟎𝟑

−𝟒×𝟏𝟎−𝟑

= −𝟏𝟐. 𝟓 × 𝟏𝟎𝟔𝒎/𝒔𝒆𝒄

= −𝟏. 𝟐𝟓 × 𝟏𝟎𝟕𝒎/𝒔𝒆𝒄

Represents a wave traveling in the negative x direction

Option (c)

12. The electric field of an electromagnetic wave propagating in the positive

z – direction is given by

𝐸 = �̂�𝑥 sin(𝜔𝑡 − 𝛽𝑧) + �̂�𝑦 sin(𝜔𝑡 − 𝛽𝑧 + 𝜋/2)

The wave is

(a) linearly polarized in the z – direction

(b) elliptically polarized

(c) left – hand circularly polarized

(d) right – hand circularly polarized

[GATE 2006: 1 Mark]

Soln. 𝑬 = �⃗⃗� 𝒙 𝐬𝐢𝐧(𝝎𝒕 − 𝜷𝒛) + 𝒂𝒚 𝐬𝐢𝐧 (𝝎𝒕 − 𝜷𝒛 +𝝅

𝟐)

= �⃗⃗� 𝒙 𝐬𝐢𝐧(𝝎𝒕 − 𝜷𝒛) + 𝒂𝒚 𝐜𝐨𝐬(𝝎𝒕 − 𝜷𝒛)

𝑬𝒙(𝒛, 𝒕) = 𝐬𝐢𝐧(𝝎𝒕 − 𝜷𝒛)

𝑬𝒚(𝒛, 𝒕) = 𝐜𝐨𝐬(𝝎𝒕 − 𝜷𝒛)

𝑬𝒙𝟐 + 𝑬𝒚

𝟐 = 𝟏

It represents a circle in the 𝑬𝒙 − 𝑬𝒚 plane with radius 1

The tip of the E vector is tracing the circle in the clock wise direction

over a cycle from 𝝎𝒕 = 𝟎 𝒕𝒐 𝟐𝝅.

The fingers of the left hand follows the clockwise direction, thumb is

pointing in the direction of propagation (+ z)

-1

-1 1

2

𝐸𝑦(𝑡)

𝐸𝑥(𝑡)

𝜔𝑡 =𝜋

2

𝜔𝑡 =3𝜋

2

𝜔𝑡 = 𝜋

𝜔𝑡 = 2𝜋

Option (c)

13. The electric field of a uniform plane electromagnetic wave in free space,

along the positive X direction is given by �⃗� = 10(�̂�𝑦 + 𝑗�̂�𝑧)𝑒−𝑗25𝑥. The

frequency and polarization of the wave respectively are

(a) 1.2 GHz and left circular

(b) 4 Hz and left circular

(c) 1.2 GHz and right circular

(d) 4 Hz and right circular

[GATE 2012: 1 Mark]

Soln. 𝑬 = 𝟏𝟎(�⃗⃗� 𝒚 + 𝒋�⃗⃗� 𝒛)𝒆−𝒋𝟐𝟓𝒙 in free space

�⃗⃗� = (𝑬𝒚 𝒂𝒚⃗⃗ ⃗⃗ + 𝑬𝒛 𝒂𝒛⃗⃗⃗⃗ )𝒆−𝒋𝜷𝒙

The wave is propagating in free space in x direction with components

in y and z direction.

𝜷 = 𝟐𝟓, 𝒗𝒑 =𝝎

𝜷= 𝟑 × 𝟏𝟎𝟖 𝒎/𝒔

𝝎 = 𝟑 × 𝟏𝟎𝟖 × 𝟐𝟓

= 𝟕𝟓 × 𝟏𝟎𝟖 𝒓𝒂𝒅 / 𝒔𝒆𝒄

𝒇 =𝟕𝟓×𝟏𝟎𝟖

𝟐𝝅=

𝟕.𝟓×𝟏𝟎𝟗

𝟐𝝅≈ 𝟏. 𝟐 𝑮𝑯𝒛

The field in circular polarization is found to be

𝑬𝒔 = 𝑬𝒐 = (𝒂𝒚 ∓ 𝒋𝒂𝒛)𝒆−𝒋𝜷𝒙

Propagating in +ve X direction where plus sign is used for left

circular polarization and minus for right circular polarization.

Option (a)

14. A plane wave propagating in air with �⃗� = (8�̂�𝑥 + 6�̂�𝑦 +

5�̂�𝑧)𝑒𝑗(𝜔𝑡+3𝑥−4𝑦)𝑉/𝑀 is incident on a perfectly conducting slab

positioned at 𝑥 ≤ 0. The �⃗� field of the reflected wave is

(a) (−8�̂�𝑥 − 6�̂�𝑦 − 5�̂�𝑧) 𝑒𝑗(𝜔𝑡+3𝑥+4𝑦)𝑉/𝑀

(b) (−8�̂�𝑥 + 6�̂�𝑦 − 5�̂�𝑧) 𝑒𝑗(𝜔𝑡+3𝑥+4𝑦)𝑉/𝑀

(c) (−8�̂�𝑥 − 6�̂�𝑦 − 5�̂�𝑧) 𝑒𝑗(𝜔𝑡−3𝑥−4𝑦)𝑉/𝑀

(d) (−8�̂�𝑥 + 6�̂�𝑦 − 5�̂�𝑧) 𝑒𝑗(𝜔𝑡−3𝑥−4𝑦)𝑉/𝑀

[GATE 2012: 1 Mark]

Soln. 𝑬 = (𝟖𝒂𝒙 + 𝟔𝒂𝒚 + 𝟓𝒂𝒛)𝒆𝒋(𝝎𝒕+𝟑𝒙−𝟒𝒚) 𝒗/𝒎

𝜎 = ∞

𝐸𝑖

𝐸𝑟

𝑥 = 0

Electric field inside a perfect conductor is zero

𝑬𝒕𝒓𝒂𝒏𝒔𝒎𝒊𝒕𝒕𝒆𝒅 = 𝟎

𝑬𝒊 + 𝑬𝒓 = 𝟎

𝑬𝒓 = −𝑬𝒊 = −𝟖�̂�𝒙 − 𝟔�̂�𝒚 − 𝟖�̂�𝒛

The x – component of Eincident which is normal to slab gets reflected

with 1800 phase change

Option (c)

15. A two – port network has scattering parameters given by

[𝑆] = [𝑆11 𝑆12

𝑆21 𝑆22]. If the port – 2 of the two – port is short circuited, the

S11 parameter for the resultant one – port network is

(a) 𝑆11−𝑆11𝑆22+𝑆12𝑆21

1+𝑆22

(b) 𝑆11+𝑆11𝑆22−𝑆12𝑆21

1+𝑆22

(c) 𝑆11+𝑆11𝑆22+𝑆12𝑆21

1−𝑆22

(d) 𝑆11−𝑆11𝑆22+𝑆12𝑆21

1−𝑆22

[GATE 2014: 1 Mark]

Soln. In put reflection coefficient = 𝑻𝒊𝒏

𝑎2

𝑏2

𝑎1

𝑏1

𝑍𝐿 [S]

𝒁𝒍 = 𝟎

𝒃𝟏 = 𝑺𝟏𝟏 𝒂𝟏 + 𝑺𝟏𝟐 𝒂𝟐 − − − − − − − −(𝑰)

𝒃𝟐 = 𝑺𝟐𝟏 𝒂𝟏 + 𝑺𝟐𝟐 𝒂𝟐 − − − − − − − −(𝑰𝑰)

𝑻𝒊𝒏 =𝒃𝟏

𝒂𝟏= 𝑺𝟏𝟏 + 𝑺𝟏𝟐

𝒂𝟐

𝒂𝟏

From (II)

𝒃𝟐

𝒂𝟐= 𝑺𝟐𝟏

𝒂𝟏

𝒂𝟐+ 𝑺𝟐𝟐

𝒃𝟐

𝒂𝟐=

𝟏

𝑻𝑳 𝐰𝐡𝐞𝐫𝐞 𝑻𝑳 𝐢𝐬 𝐥𝐨𝐚𝐝 𝐫𝐞𝐟𝐥𝐞𝐜𝐭𝐢𝐨𝐧 𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭

𝑻𝑳 = −𝟏, 𝒁𝑳 = 𝟎

𝒐𝒓 𝟏

𝑻𝑳= 𝑺𝟐𝟏

𝒂𝟏

𝒂𝟐+ 𝑺𝟐𝟐

𝒐𝒓 𝑺𝟐𝟐 𝑻𝑳−𝟏

𝑻𝑳= −𝑺𝟐𝟏

𝒂𝟏

𝒂𝟐

𝒐𝒓 𝒂𝟏

𝒂𝟐=

𝟏−𝑺𝟐𝟐 𝑻𝑳

𝑺𝟐𝟏 𝑻𝑳

𝑻𝒊𝒏 = 𝑺𝟏𝟏 +𝑺𝟏𝟐 𝑺𝟐𝟏 𝑻𝑳

𝟏−𝑺𝟐𝟐 𝑻𝑳

𝑻𝑳 = −𝟏

𝒔𝒐, 𝑻𝒊𝒏 = 𝑺𝟏𝟏 −𝑺𝟏𝟐 𝑺𝟐𝟏

𝟏+𝑺𝟐𝟐

=𝑺𝟏𝟏+𝑺𝟏𝟏 𝑺𝟐𝟐−𝑺𝟏𝟐 𝑺𝟐𝟏

𝟏+𝑺𝟐𝟐

Option (b)

16. Which one of the following filed patterns represents a TEM wave

traveling in the positive x direction?

(a) 𝐸 = +8�̂�, 𝐻 = −4�̂�

(b) 𝐸 = −2�̂�, 𝐻 = −3�̂�

(c) 𝐸 = +2�̂�,𝐻 = +2�̂�

(d) 𝐸 = −3�̂�, 𝐻 = +4�̂�

[GATE 2014: 1 Mark]

Soln. The possible combinations are

�̂�𝒚 × �̂�𝒛 = �̂�𝒙

−�̂�𝒛 × �̂�𝒚 = �̂�𝒙

−�̂�𝒚 × −�̂�𝒛 = �̂�𝒙

𝑌

𝑋

𝑍

Option (b)