If -b- are three unit vectors such that b and c are non-parallel and - b- - 1 2 b- then the angle between - and - is

The correct option is C 60^0 Since #160; ( a⃗×( b⃗×c⃗ #160;) #160;) = 1 2 b⃗ ⇒( a⃗ .c⃗ #160;) b⃗ ( a⃗ .b⃗ #160;) c⃗ = 1 2 b⃗ ⇒ ...

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Byju's Answer

Standard XII

Physics

Conduction Law

If â,b̂,ĉ a...

Question

If $^a,^b,^c$ are three unit vectors such that $\to b$ and $\to c$ are non-parallel and $^a \times (^b \times^c) = (\frac{1}{2})^b$ , then the angle between $^a$ and $^c$ is

30^{0}

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45^{0}

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60^{0}

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90^{0}

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Solution

The correct option is C $60^{0}$
Since $(\to a \times (\to b \times \to c)) = \frac{1}{2} \to b$
$\Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c = \frac{1}{2} \to b$
$\Rightarrow ((\to a . \to c) - \frac{1}{2}) \to b - (\to a . \to b) \to c = 0$
$\Rightarrow ((\to a . \to c) - \frac{1}{2}) = 0$ and $(\to a . \to b) = 0$ (as $\to b, \to c$ are non-collinear vectors)
$\Rightarrow cos θ = \frac{1}{2}$ where $θ$ is the angle between $\to a$ and $\to c$
Thus $θ = \frac{π}{3}$

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If ^a,^b,^c are three unit vectors such that →b and →c are non-parallel and ^a×(^b×^c)=(12)^b, then the angle between ^a and ^c is

The correct option is C 600Since (→a×(→b×→c))=12→b⇒(→a.→c)→b−(→a.→b)→c=12→b⇒((→a.→c)−12)→b−(→a.→b)→c=0⇒((→a.→c)−12)=0 and (→a.→b)=0 (as →b,→c are non-collinear vectors)⇒cosθ=12 where θ is the angle between →a and →cThus θ=π3

If $^a,^b,^c$ are three unit vectors such that $\to b$ and $\to c$ are non-parallel and $^a \times (^b \times^c) = (\frac{1}{2})^b$ , then the angle between $^a$ and $^c$ is