Answer:
Step-by-step explanation:
From the information given:
The instantaneous expression of the electric field in the wave is:
![E(r,t)= (i_x √(3) -i_z) 2 \ sin (2 \pi*10^8t + 2 \pi x/3+2 \pi z /√(3) + 30 ^0) , \ V/m](https://img.qammunity.org/2022/formulas/engineering/college/riqswldq55q2h0i9bc7192l6m7b7v25yuc.png)
To determine the unit vector in line with the wave electric field, we take the first term in E(r,t) for
as:
![I_E^\to = i_x √(3)-i_z \\ \\ I_E^\to = (i_x √(3)-i_z)/(√(3 +1)) \\ \\ \mathbf{ I_E = (i_x√(3) -i_z)/(2)}](https://img.qammunity.org/2022/formulas/engineering/college/ys1y7f10jbkb7x9d6tx6zkpic3vps5vdwl.png)
The amplitude is denoted by the numerical value after the first term, which is:
![\mathbf{E_o = 2}](https://img.qammunity.org/2022/formulas/engineering/college/rm4cekwpb1cxib1q82whzm2bm69i7xulqw.png)
The wavelength can be determined by using the expression:
![\beta =(2 \pi)/(\lambda )](https://img.qammunity.org/2022/formulas/engineering/college/jhlforf7l4kyjan72ccg7mssozmt5kjmjn.png)
from the given instantaneous expression:
![\beta = (2 \pi)/(3)x+(2 \pi)/(√(3))z](https://img.qammunity.org/2022/formulas/engineering/college/2kb41mugewru59jdudo1bxxrl4x2rqxzvm.png)
![\beta = \sqrt{(2 \pi)/((3)^2)+((2 \pi)/((√(3))^2)}](https://img.qammunity.org/2022/formulas/engineering/college/pnhs9287llazaa1pyskoy9g0gbpd5bm30k.png)
![\beta = \sqrt{(2 \pi)/(9)+\frac{2 \pi}{{3}}}](https://img.qammunity.org/2022/formulas/engineering/college/ls5y781fpdvd3nui8mynq5vfktbvjapbec.png)
Factorizing 2π
![\beta =2 \pi \sqrt{(1)/(9)+\frac{1}{{3}}}](https://img.qammunity.org/2022/formulas/engineering/college/z0kqu3oikdp0mz85bezx99chegdlelbsn3.png)
![\beta =2 \pi \sqrt{(9+3)/(9*3)}}](https://img.qammunity.org/2022/formulas/engineering/college/r1wyekdm0j2bu4nsvblbwuybi4zc282g8l.png)
![\beta =2 \pi \sqrt{(12)/(27)}}](https://img.qammunity.org/2022/formulas/engineering/college/ycuyxvmc0ut6r06cy1unabwxoun80lefr7.png)
![\beta =2 \pi \sqrt{(4*3)/(9*3)}}](https://img.qammunity.org/2022/formulas/engineering/college/t6ngafz84gm5ke9hpocr1y4kj5pmywy03t.png)
![\beta =2 \pi \sqrt{(4)/(9)}}](https://img.qammunity.org/2022/formulas/engineering/college/m0af00xygu30eltxviz4baeuxnjr74ypok.png)
![\beta =2 \pi* {(2)/(3)}}](https://img.qammunity.org/2022/formulas/engineering/college/awp7z4ntl79z0h9wg2u2ayoxw61yvraag0.png)
recall from the expression using in calculating wavelength:
![\beta =(2 \pi)/(\lambda )](https://img.qammunity.org/2022/formulas/engineering/college/jhlforf7l4kyjan72ccg7mssozmt5kjmjn.png)
∴
equating both together, we have:
![(2 \pi)/(\lambda )= 2 \pi* {(2)/(3)}}](https://img.qammunity.org/2022/formulas/engineering/college/ulv9xt941mjniqc4xyvve1mq4rny5s5kl7.png)
![\lambda = (3)/(2)](https://img.qammunity.org/2022/formulas/engineering/college/ol4wvyr6ubbelse6tzfpb360w9w0pqauk2.png)
λ = 1.5 m
In line with the wave direction; unit vector
can be computed as follows:
![i_k = - [ \beta_1x +\beta_2z]/\beta](https://img.qammunity.org/2022/formulas/engineering/college/pw8epaguw2eqaoc9iebp8b1l0l5xc62f6o.png)
where;
![\beta_1 = (2 \pi )/(3) \ ; \ \beta_2 = (2 \pi )/(√(3)) \ ; \ \beta = (2 \pi * 2)/(3) ;](https://img.qammunity.org/2022/formulas/engineering/college/bvt1gwzjmalrmlnd53yi4491qppv0uyr1b.png)
∴
![i_k = - \Big[(2 \pi)/(3)x + (2 \pi)/(√(3)) z\Big]* (1)/((2 \pi *2)/(3))](https://img.qammunity.org/2022/formulas/engineering/college/8cvtsrd8fr6f6tbzfm3e44d01q3xmy3zyr.png)
![i_k = - \Big[(x)/(2) + \sqrt\frac{{3}}{4}} z\Big]](https://img.qammunity.org/2022/formulas/engineering/college/yfiu998xay96sirmo4hbnuja3eks39q0nj.png)
![i_k = - \Big[(1)/(2)x + \sqrt{(3)/(4) }z\Big]](https://img.qammunity.org/2022/formulas/engineering/college/p80ohliv3kbvs2fkgiffo62b8jgk1dcrij.png)
![\mathbf{i_k = - \Big[0.5x +0.86 z\Big]}](https://img.qammunity.org/2022/formulas/engineering/college/ri33c9r1id1v97lqyi5wt8vcgnw54mesee.png)