If θ.∅ = π/2 then show that [cos2θ cosθ sinθ /cosθ sinθ sin2θ] [cos2∅ cos∅sin∅/cos∅ sin∅sin2∅] =0
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Answer:
Given, AB=0
2×2
⇒[
cos
2
θ
cosθsinθ
cosθsinθ
sin
2
θ
][
cos
2
ϕ
cosϕsinϕ
cosϕsinϕ
sin
2
ϕ
]=[
0
0
0
0
]
⇒[
cos
2
θcos
2
ϕ+cosθcosϕsinθsinϕ
cosθsinθcos
2
ϕ+sin
2
θcosϕsinϕ
cos
2
θcosϕsinϕ+sin
2
ϕcosθsinθ
cosθsinθcosϕsinϕ+sin
2
θsin
2
ϕ
]=[
0
0
0
0
]
⇒[
cosθcosϕcos(θ−ϕ)
cosϕsinθcos(θ−ϕ)
cosθsinϕcos(θ−ϕ)
sinϕsinθcos(θ−ϕ)
]=[
0
0
0
0
]
⇒cos(θ−ϕ)=0
⇒cos(θ−ϕ)=cos
2
π
⇒θ−ϕ=nπ+
2
π
⇒θ−ϕ=(2n+1)
2
π
,n∈Z
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