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Prove that fo...

Question

Prove that for any three vectors a, b, c [b + c c + a a + b] = 2 [abc].

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Solution

For any vector $∴ \to a, \to b$ and $\to c$ , prove that
$[\to a + \to b \to b + \to c \to c + \to a] = 2 [\to a \to b \to c]$
sowing LHS :-
$[\to a + \to b \to b + \to c \to c + \to a] = (\to a + \to b) . [(\to b + \to c) \times (\to c + \to a)]$
$[\to a \to b \to c = \to a . (\to b \times \to c]$
$= (\to a + \to b) . [(\to b \times \to c) + (\to b \times \to a) + (\to c \times \to c) + (\to c \times \to a)]$

$∵ [\to c \times \to c = | \to c | | \to c | sin O^n = o$
$= (\to a + \to b) . [(\to b \times \to c) + (\to b \times \to a) + o + (\to c \times \to a)]$
$= \to a . (\to b \times \to c) + \to b . (\to b \times \to c) + \to a . (\to b \times \to a) + \to b . (\to b \times \to a) + \to a . (\to c \times \to a) + \to b . (\to c \times \to a)$
$= [\to a, \to b, \to c] + [\to b \to b \to c] + [\to a \to b \to a] + [\to a \to c \to a] + [\to b \to c \to a]$

$∵ [\to b \to b \to c] = [\to c \to b \to b]$
$= \to c . (\to b \times \to b)$
As $(\to b \times \to b) = \to o$
$= \to c . o$
$= \to o$

$∵$ Using property
$[\to b \to b \to c] = 0$
$[\to a \to b \to a] = 0$
$[\to b \to b \to a] = 0$
$[\to a \to c \to a] = 0$

$= [\to a \to b \to c] + o + o + o + o + [\to b \to c \to a]$
$= [\to a \to b \to c] + [\to b \to c \to a]$
$= [\to a \to b \to c] + [\to a \to b \to c]$
$= 2 [\to a \to b \to c] = R . H . S$

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