If a-b-c - a- b b- 4-2- -sin- b- - 2 -1 c and c- c a- c- while b and c are non-collinear- then

The correct option is B #160;α =π #160;/ 2 ,β =1 We have, a×( b× c ) +( a.b ) b=( 4 2β sinα #160; #160;) b+( β #160;^ 2 1 ) c #160; .. ...

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Standard XII

Mathematics

Applications of Cross Product

If a×b×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 $(\to c . \to c) \to a = \to c$ , while $\to b$ and $\to c$ are non-collinear, then

α = \frac{π}{2}, β = - 1

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α = \frac{π}{2}, β = 1

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α = \frac{π}{3}, β = - 1

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α = \frac{π}{3}, β = 1

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Solution

The correct option is B $α = \frac{π}{2}, β = 1$
We have, $a \times (b \times c) + (a . b) b = (4 - 2 β - sin α) b + (β^{2} - 1) c$ ...(1)
and $(c . c) a = c$ ...(2)
where $b$ and $c$ are non-collienear vectors and $α, β$ are scalars
From (2), $(c . c) a . c = c . c \Rightarrow a . c = 1$ ...(3)
From (1), we get
$(a . c) b - (a . b) c + (a . b) b = (4 - 2 β - sin α) b + (β^{2} - 1) c \Rightarrow {1 + (a . b)} b - (a . b) c = (4 - 2 β - sin α) b + (β^{2} - 1) c$
$\Rightarrow 1 + (a . b) = 4 - 2 β - sin α$ ...(4)
and $a . b = - (β^{2} - 1)$ ...(5)
$∴ sin α = 1 + {(1 - β)}^{2} \Rightarrow β = 1, sin α = 1$
$\Rightarrow α = \frac{π}{2} + 2 n π, n \in I$

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If →a×(→b×→c)+(→a.→b)→b=(4−2β−sinα)→b+(β2−1)→c and (→c.→c)→a=→c, while →b and →c are non-collinear, then

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 $(\to c . \to c) \to a = \to c$ , while $\to b$ and $\to c$ are non-collinear, then