B-c-a-d-c-a-b-d- -a-b-c-d-

The correct option is A 0 Solving =b⃗.(c⃗×a⃗×b⃗)+c⃗.(a⃗×b⃗×d⃗)+a⃗.(b⃗×ĉ×d⃗) =b⃗.((c⃗.d⃗)a⃗ (c⃗.a⃗)d⃗)+c⃗.((a⃗.d⃗)b⃗ (a⃗.d⃗))+a⃗.((b⃗.d⃗)c⃗ (b⃗.c⃗ ...

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Byju's Answer

Standard XII

Mathematics

Modulus Function

b⃗×c⃗.a⃗×d⃗+c...

Question

$(\to b \times \to c) . (\to a \times \to d) + (\to c \times \to a) . (\to b \times \to d)$ $+ (\to a \times \to b) . (\to c \times \to d) =$

0

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Solution

The correct option is A $0$
Solving
$= \to b . (\to c \times \to a \times \to b) + \to c . (\to a \times \to b \times \to d) + \to a . (\to b \times^c \times \to d)$
$= \to b . ((\to c . \to d) \to a - (\to c . \to a) \to d) + \to c . ((\to a . \to d) \to b - (\to a . \to d)) + \to a . ((\to b . \to d) \to c - (\to b . \to c) \to d)$
$= (\to a \cdot \to b) (\to c \cdot \to d) - (\to c \cdot \to a) \to b \cdot \to d + (\to c \cdot \to b) (\to a \cdot \to d) - (\to a \cdot \to b) \to c \cdot \to d + (\to a \cdot \to c) (\to b \cdot \to d) - (\to b \cdot \to c) \to d \to a$
$= 0$

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(→b×→c).(→a×→d)+(→c×→a).(→b×→d)+(→a×→b).(→c×→d)=

$(\to b \times \to c) . (\to a \times \to d) + (\to c \times \to a) . (\to b \times \to d)$ $+ (\to a \times \to b) . (\to c \times \to d) =$