If $a^{2} + b^{2} + c^{2} = x^{2} + y^{2} + z^{2} = 1$ then $a x + b y + c z$ is

$> 1$

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$< 1$

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$> 0$

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$< 0$

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Solution

The correct option is B
$< 1$

Explanation for the correct option:
To find the relation:
We know A.M. $>$ G.M.
$\Rightarrow \frac{(a^{2} + x^{2})}{2} > \sqrt{a^{2} x^{2}} \Rightarrow (a^{2} + x^{2}) > 2 a x$
Similarly, $(b^{2} + y^{2}) > 2 b y$ and $(c^{2} + z^{2}) > 2 c z$
Thus,
$(a^{2} + x^{2} + b^{2} + y^{2} + c^{2} + z^{2}) > 2 a x + 2 b y + 2 c z \Rightarrow (a^{2} + b^{2} + c^{2} + x^{2} + y^{2} + z^{2}) > 2 (a x + b y + c z) \Rightarrow 1 + 1 > 2 (a x + b y + c z) \Rightarrow a x + b y + c z < 1$
Hence, the correct option is B.

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