Let a-b-c be the three vectors having magnitudes 1-1-2 respectively- If a-a-a-b-0 then the angle between a and c is

The correct options are A π6 D 5π6 a×(a×c)+b=0 ∴b= a× nbsp; (a×c) = (a×c)×(a) ⇒ nbsp; nbsp;a×c+a ⇒|b|=|a×c| |a| ...

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Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

Let a,b,c b...

Question

Let $\to a, \to b, \to c$ be the three vectors having magnitudes $1, 1, 2$ respectively. If $\to a \times (\to a \times \to a) + \to b = 0$ then the angle between $\to a$ and $\to c$ is

\frac{π}{6}

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\frac{π}{3}

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\frac{2 π}{3}

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\frac{5 π}{6}

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Solution

The correct options are
A $\frac{π}{6}$
D $\frac{5 π}{6}$
$\to a \times (\to a \times \to c) + \to b = 0$
$∴ \to b = - \to a \times (\to a \times \to c)$
$= (\to a \times \to c) \times \to (a)$
$\Rightarrow \to a \times \to c + \to a$
$\Rightarrow ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to a \times \to c ∣ ∣ ∣ ∣ \to a ∣ ∣$
$\Rightarrow 1 = 1 \times 2 sin θ \times 1 = 2 sin θ$
$∴ sin θ = \frac{1}{2}, θ = \frac{π}{6}, \frac{5 π}{6}$

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Let →a,→b,→c be the three vectors having magnitudes 1,1,2 respectively. If →a×(→a×→a)+→b=0 then the angle between →a and →c is

The correct options are A π6 D 5π6→a×(→a×→c)+→b=0∴→b=−→a×(→a×→c)=(→a×→c)×→(a)⇒→a×→c+→a⇒∣∣∣→b∣∣∣=∣∣→a×→c∣∣∣∣→a∣∣⇒1=1×2sinθ×1=2sinθ∴sinθ=12,θ=π6,5π6

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