Given $a^{2} + b^{2} + c^{2} = 25; x^{2} + y^{2} + z^{2} = 36 and a x + b y + c z = 30$ for a,b,c being real. How will you prove $\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{5}{6}$ ?

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Solution

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Explanation:
Suppose you have three points in $R^{3}$

$p_{1} = (a, b, c)$
$p_{2} = (x, y, z)$
$p_{0} = (0, 0, 0)$

The respective lengths of sides $p_{1} - p_{0}$ and $p_{2} - p_{0}$
are

$∥ p_{1} - p_{0} ∥ = \sqrt{a^{2} + b^{2} + c^{2}}$

$∥ p_{2} - p_{0} ∥ = \sqrt{x^{2} + y^{2} + z^{2}}$

and their scalar product

$(p_{1} - p_{0}) . (p_{2} - p_{0}) = a x + b y + c z$

but $(p_{1} - p_{0}) . (p_{2} - p_{0}) = ∥ p_{1} - p_{0} ∥ ∥ p_{2} - p_{0} ∥ cos (^p_{1} p_{0} p_{2})$

so

$cos (^p_{1} p_{0} p_{2}) = \frac{a x + b y + c z}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{x^{2} + y^{2} + z^{2}}} = \frac{30}{5 \times 6} = 1$

Then the sides $p_{1} - p_{0}$ and $p_{2} - p - 0$ are aligned so
$(p_{1} - p_{0}) = λ (p_{2} - p_{0})$ and of course

$λ = \frac{5}{6}$

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