1. ## Ratio help

Hi, I need a hint to start this problem:

If $(a^2 +b^2+c^2)(x^2+y^2+z^2) = (ax+by+cz)^2$

prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

Thanks.

2. Originally Posted by MagicS06
Hi, I need a hint to start this problem:

If $(a^2 +b^2+c^2)(x^2+y^2+z^2) = (ax+by+cz)^2$

prove that $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

Thanks.
That's not a ratio... but since you asked for a hint...

Expand all the brackets and collect like terms, you should get...

$a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 = 2abxy + 2acxz + 2bcyz$

$b^2x^2 - 2abxy + a^2y^2 + a^2z^2 -2acxz + c^2x^2 + b^2z^2 - 2bcyz + c^2y^2 = 0$

Recall that $(p-q)^2 = p^2 - 2pq + q^2$? Using this we get...

$(bx - ay)^2 + (az - cx)^2 + (bz - cy)^2 = 0$

Now, assuming a, b, c, x, y and z are all real... we have 3 square terms adding to become 0. Since a square term is never non-negative... all the terms must be 0.

So we have $(bx - ay)^2 = 0$ which gives $bx = ay$ or $\frac{x}{a} = \frac{y}{b}$.

Similarly, $(az - cx)^2 = 0$ gives $\frac{z}{c}=\frac{x}{a}$,

and $(bz - cy)^2 = 0$ gives $\frac{z}{c} = \frac{y}{b}$.

Thus

$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$.