If a-b-c- are unit vectors such that a-b-c-a-b- then -c- a- b-

The correct option is A 12sin 2α We have [ĉâb̂] =ĉ (â×b̂) =ĉ | â×b̂ |cos α #160; = |ĉ #160; | |â #160; | |b̂ #160; |sinα cosα ...

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Standard XII

Mathematics

Domain

If a⃗,b⃗,c⃗...

Question

If $\to a, \to b, \to c$ are unit vectors such that $(\to a, \to b) = α = (\to c, \to a \times \to b),$ then $[\to c$ $\to a$ $\to b] =$

1

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\frac{1}{2} sin 2 α

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sin 2 α

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\frac{1}{4} sin 2 α

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Solution

The correct option is A $\frac{1}{2} sin 2 α$
We have
$[^c^a^b]$
$=^c - (^a \times^b)$
$=^c ∣ ∣^a \times^b ∣ ∣ cos α$
$= |^c | |^a | ∣ ∣^b ∣ ∣ sin α cos α$
$= \frac{1}{2} sin 2 α$

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Similar questions

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If →a,→b,→c are unit vectors such that (→a,→b)=α=(→c,→a×→b), then [→c →a →b]=

The correct option is A 12sin2αWe have[^c^a^b]=^c−(^a×^b)=^c∣∣^a×^b∣∣cos α=|^c||^a|∣∣^b∣∣sinα cosα=12sin2α

If $\to a, \to b, \to c$ are unit vectors such that $(\to a, \to b) = α = (\to c, \to a \times \to b),$ then $[\to c$ $\to a$ $\to b] =$

The correct option is A $\frac{1}{2} sin 2 α$
We have
$[^c^a^b]$
$=^c - (^a \times^b)$
$=^c ∣ ∣^a \times^b ∣ ∣ cos α$
$= |^c | |^a | ∣ ∣^b ∣ ∣ sin α cos α$
$= \frac{1}{2} sin 2 α$