If a- b-c- - a- -b- b- - 4-2- -sin- b- - - 2 -1 c- and c- -c- a- -c- where b- and c- are non-collinear- then the value of scalars - and - are

The correct option is A #160;β =1,α =( 4n+1 ) π #160; 2 ;n∈ I Given: #⃗1⃗6⃗0⃗;⃗a⃗×( #160;#⃗1⃗6⃗0⃗;⃗b⃗× #160;#⃗1⃗6⃗0⃗;⃗c⃗ #160;) +( #160;#⃗1 ...

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Critical Point

If a⃗× b⃗×c...

Question

If $\to a \times (\to b \times \to c) + (\to a . \to b) \to b = (4 - 2 β - sin α) \to b + (β^{2} - 1) \to c$ and $(\to c . \to c) \to a = \to c$ , where $\to b$ and $\to c$ are non-collinear, then the value of scalars $α$ and $β$ , are

β = 1, α = (4 n + 1) \frac{π}{2}; n \in I

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β = 1, α = (4 n - 1) \frac{π}{2}; n \in I

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β = - 1, α = (4 n + 1) \frac{π}{2}; n \in I

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β = - 1, α = (4 n - 1) \frac{π}{2}; n \in I

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Solution

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Similar questions

\to b

\to c

\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b

= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c

(\to c \cdot \to c) \to a = \to c

\to a, \to b, \to c

\to b, \to c

(\to a, \to b) = α, (\to a, \to c) = β

\to a \times (\to b \times \to c) = \frac{\to b + \to c}{2}

c o s (α + β) =

α

β

x^{2} - 15 x + 1 = 0

{(\frac{1}{α} - 15)}^{- 2} + {(\frac{1}{β} - 15)}^{- 2}

\to a, \to b, \to c

\to b

\to c

\to a

\to b

α

\to a

\to c

β

\to a \times (\to b \times \to c) = \frac{1}{2} \to b

x cos α + y sin α = x cos β + y sin β = a, 2 n π < α, β < (4 n + 1) \frac{π}{2},

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