"Show" that the direction cosines of a vector satisfies... "Show" that the direction cosines of a vector satisfies
$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$
I am stumped on these things:


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*"SHOW" that the direction cosines corresponds to a given vector to satisfy the relation above. ----> How do you "show" this? What does this mean? Does this mean to use the direction cosines of a vector?

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*I'm sure this is a proof but I don't know what the end result would look like or better, what I am expected to learn from this proof.



I am not looking for a mere answer but really an in-depth explanation of the problem.
Greatly appreciated. :)

Given two vectors $A$, $B$, I'll use $\angle AB$ for the angle between them. So, the direction cosines of a vector $V=(x,y,z)$ are 
$$\cos\alpha=\cos\angle (x,y,z)(1,0,0)=\frac{(x,y,z)\cdot(1,0,0)}{\|V\|\cdot\|(1,0,0)\|}$$
$$\cos\beta=\cos\angle (x,y,z)(0,1,0)=\frac{(x,y,z)\cdot(0,1,0)}{\|V\|\cdot\|(0,1,0)\|}$$
$$\cos\gamma=\cos\angle (x,y,z)(0,0,1)=\frac{(x,y,z)\cdot(0,0,1)}{\|V\|\cdot\|(0,0,1)\|}$$
but this is
$$\cos\alpha=\frac{x}{\sqrt{x^2+y^2+z^2}}$$
$$\cos\beta=\frac{y}{\sqrt{x^2+y^2+z^2}}$$
$$\cos\gamma=\frac{z}{\sqrt{x^2+y^2+z^2}}$$
so that
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1$$
Note it is easier to work with the normalized vectors since they are "the same" in terms of angles (you could have used $(x,0,0)$ or $(10,0,0)$, or any parallel vector), but these make the calculations much easier.
