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

A× (b×c)=(a.c) b (a.b) c (a×b)×c = (c.b) a+(a.c) b we have been given, (a×(b×c)).(( a×b) ×c)=0 ⇒ [(a.c) b (a. b )c] .[(a.c) b (c.b) a] =0 ⇒ (a. c)^2 |b|^2 ...

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Byju's Answer

Standard XIII

Mathematics

Dot Product of Two Vectors

If a× b×c is ...

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,
$(\to a \times (\to b \times \to c)) . ((\to a \times \to b) \times \to c) = 0 \Rightarrow [(\to a . \to c) \to b - (\to a . \to b) \to c] . [(\to a . \to c) \to b - (\to c . \to b) \to a] = 0 \Rightarrow (\to a . \to c)^{2} | \to b |^{2} - 2 (\to a . \to c) (\to b . \to c) (\to a . \to b) + (\to a . \to b) (\to b . \to c) (\to c . \to a) = 0 \Rightarrow (\to a . \to c)^{2} | \to b |^{2} = (\to a . \to c) (\to a . \to b) (\to b . \to c) \Rightarrow (\to a . \to c) ((\to a . \to c) (\to b . \to b) - (\to a . \to b) (\to b . \to c)) = 0 \Rightarrow \to a . \to c = 0 or (\to a . \to c) | \to b |^{2} = (\to a . v b) (\to b . \to c)$

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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,