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Prove that a...

Question

Prove that
$\to a \times (\to b \times \to c) \neq (\to a \times \to b) \times \to c$ if
$\to a = 2^i + 3^j - 5^k, \to b = -^i +^j + \sqrt{2}^k, \to c = 4^i - 2^j + \sqrt{3}^k$

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Solution

Given that
$\to a = 2^i + 3^j - 5^k$
$\to b = -^i +^j + \sqrt{2}^k$
$\to c = 4^i - 2^j + \sqrt{3}^k$
$\to a . \to b = (2^i + 3^j - 5^k) . (-^i +^j + \sqrt{2}^k)$
$= - 2 + 3 - 5 \sqrt{2} = 1 - 5 \sqrt{2}$
$\to b . \to c = (-^i -^j + \sqrt{2}^k) . (4^i - 2^j + \sqrt{3}^k)$
$= - 4 - 2 + \sqrt{6} = - 6 + \sqrt{6}$
$\to a . \to c = (2^i + 3^j - 5^k) . (4^i - 2^j + \sqrt{3}^k)$
$= 8 - 6 - 5 \sqrt{3} = 2 - 5 \sqrt{3}$
Now $\to a \times (\to b \times \to c) = (\to a . \to c) \to b - (\to a . \to b) \to c$
$\Rightarrow \to a \times (\to b \times \to c) = (2 - 5 \sqrt{3}) (-^i +^j + \sqrt{2}^k) - (1 - 5 \sqrt{2}) (4^i - 2^j + \sqrt{3}^k)$
$\Rightarrow \to a \times (\to b \times \to c) = (- 2 + 5 \sqrt{3} - 4 + 20 \sqrt{2})^i + (2 - 5 \sqrt{3} + 2 - 10 \sqrt{2})^j + (2 \sqrt{2} - 5 \sqrt{6} - \sqrt{3} + 5 \sqrt{6})^k$
$\Rightarrow \to a \times (\to b \times \to c) = (- 6 + 20 \sqrt{2} + 5 \sqrt{3})^i + (4 - 10 \sqrt{2} - 5 \sqrt{3})^j + (2 \sqrt{2} - \sqrt{3})^k . . . . (1)$
Again $\Rightarrow (\to a \times \to b) \times \to c = (\to a . \to c) \to b - (\to b . \to c) \to a$
$\Rightarrow (\to a \times \to b) \times \to c = (2 - 5 \sqrt{3}) (-^i +^j + \sqrt{2}^k) - (- 6 + \sqrt{6}) (2^i + 3^j - 5^k)$
$\Rightarrow (\to a \times \to b) \times \to c = (- 3 + 5 \sqrt{3} + 12 - 2 \sqrt{6})^i + (2 - 5 \sqrt{3} + 18 - 3 \sqrt{6})^j + (2 \sqrt{2} - 5 \sqrt{6} + 30 - 5 \sqrt{6})^k$
$\Rightarrow (\to a \times \to b) \times \to c = (10 + 5 \sqrt{3} - 2 \sqrt{6})^i + (20 - 5^3 - 3 \sqrt{6})^j + (30 + 2 \sqrt{2} - 10 \sqrt{6})^k . . . . (2)$
From equation (1) and (2)
$\to a \times (\to b \times \to c) \neq (\to a \times \to b) \times \to c$

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\to a = 3^i - 2^j +^k,

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