If a-b-c- is perpendicular to a-b- -c- then- we may have-

The correct option is D a⃗ . c⃗ = 0 a⃗× (b⃗×c⃗)=(a⃗.c⃗) b⃗ (a⃗.b⃗) c (a⃗×b⃗)×c⃗ = (c⃗.b⃗) a⃗+(a⃗.c⃗) b⃗ we have been given, (a⃗×(b⃗×c⃗)).(( a⃗×b⃗) ×c⃗)=0 ⇒ ...

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Byju's Answer

Standard IX

Mathematics

Dot Product of Two Vectors

If a⃗×b⃗×c⃗ i...

Question

If $\to a \times (\to b \times \to c)$ is perpendicular to $(\to a \times \to b) \times \to c$ then, we may have,

$\to b . \to c = 0$

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$\to a . \to b = 0$

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$\to a . \to c = 0$

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None of these

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Solution

The correct option is C
$\to a . \to c = 0$

$\to a \times (\to b \times \to c) = (\to a . \to c) \to b - (\to a . \to b) c (\to a \times \to b) \times \to c = - (\to c . \to b) \to a + (\to a . \to c) \to b$
we have been given,
$(\to a \times (\to b \times \to c)) . ((\to a \times \to b) \times \to c) = 0 \Rightarrow [(\to a . \to c) \to b - (\to a . \to b) \to c] . [(\to a . \to c) \to b - (\to c . \to b) \to a] = 0 \Rightarrow (\to a . \to c)^{2} | \to b |^{2} - 2 (\to a . \to c) (\to b . \to c) (\to a . \to b) + (\to a . \to b) (\to b . \to c) (\to c . \to a) = 0 \Rightarrow (\to a . \to c)^{2} | \to b |^{2} = (\to a . \to c) (\to a . \to b) (\to b . \to c) \Rightarrow (\to a . \to c) ((\to a . \to c) (\to b . \to b) - (\to a . \to b) (\to b . \to c)) = 0 \Rightarrow \to a . \to c = 0 o r (\to a . \to c) | \to b |^{2} = (\to a . \to b) (\to b . \to c)$

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If →a×(→b×→c) is perpendicular to (→a×→b)×→c then, we may have,

If $\to a \times (\to b \times \to c)$ is perpendicular to $(\to a \times \to b) \times \to c$ then, we may have,