NIMCET 2021 Question Paper with solutions

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Question : 48 of 120

Marks: +1, -0

If

F(\θ)=[\table \cos \θ,-\sin \θ, 0 ;\sin \θ , \cos \θ , 0 ; 0 , 0 , 1 ]

then find

f(\θ) \.f(\α)

is equal to

$f(\θ\α)$
$f(\θ\/ \α)$
${f}(\θ+\α)$
$f(\θ-\α)$

Solution:

F(\θ)=[\table \cos \θ,-\sin \θ, 0 ;\sin \θ , \cos \θ , 0 ; 0 , 0 , 1 ]

F(\θ).F(α)=[\table \cos \θ,-\sin \θ, 0 ;\sin \θ , \cos \θ , 0 ; 0 , 0 , 1 ]

[\table \cos \α,-\sin \α, 0 ;\sin \α, \cos \α, 0 ; 0 , 0 , 1 ]

=[\table \cos \θ \.\cos \α-\sin \θ\sin \α, -\cos \θ\sin \α-\sin \θ \. \cos \α, 0; \sin \θ\.\cos \α+\cos \θ\.\sin \α, -\sin \θ\.\sin \α+\cos \θ \.\cos \α, 0 ; 0, 0, 1\ ]

=[\table \cos( \θ+\α),-\sin ( \θ+\α), 0 ;\sin( \θ+\α) , \cos ( \θ+\α) , 0 ; 0 , 0 , 1 ]

={F}(\θ+\α)

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