Answer:
[tex]|R|=4.373km[/tex]
Explanation:
Given data
For first plate let it be p₁
[tex]p_{z1}=1162 m\\ p_{x1}=10.1km\\\alpha _{1}=34.2^{o}[/tex]
For second plate let it be p₂
[tex]p_{z2}=4162 m\\p_{x2}=9.5km\\\alpha _{2}=21.5^{o}[/tex]
To find
Distance R between them
Solution
To find distance between two plates first we need to find p₁ and p₂
Finding p₁
According to vector algebra
[tex]p_{1}=p_{x1}i+p_{y1}j+p_{z1}k\\ as\\tan\alpha =tan(34.2^{o} )=(p_{y1}/p_{x1})\\p_{y1}=10.1tan(34.2^{o} )\\p_{y1}=6.864km[/tex]
So we get
[tex]p_{1}=-10.1i-6.864j+1.162k[/tex]
Now to find p₂
[tex]p_{2}=p_{x2}i+p_{y2}j+p_{z2}k\\ as\\tan\alpha =tan(21.5^{o} )=(p_{y2}/p_{x2})\\p_{y2}=9.5tan(21.5^{o} )\\p_{y2}=3.74km[/tex]
So we get
[tex]p_{2}=-9.5i-3.74j+4.162k[/tex]
Now for distance R
According to vector algebra the position vector R between p₁ and p₂
[tex]R=p_{1}-p_{2}\\ R=(p_{x1}-p_{x2})i+(p_{y1}-p_{y2})j+(p_{z1}-p_{z2})k\\R=(-10.1-(-9.5))i+(-6.864-(-3.74))j+(1.162-4.162)k\\R=-0.6i-3.124j-3k\\|R|=\sqrt{(0.6)^{2}+(3.124)^{2}+(3)^{2} }\\ |R|=4.373km[/tex]
