"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. :)
 A: If $$x^2+y^2+z^2=r^2,$$ then $$\frac{x^2}{r^2}+\frac{y^2}{r^2}+\frac{z^2}{r^2}=1,$$ or $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma=1.$$
A: 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.
A: Suppose the vector is $(x,y,z)$.  The direction cosine $\alpha$ is the cosine of the angle between this vector and the $x$-axis.  The vector $(x,0,0)$ points in the direction of the $x$-axis.  So you want the cosine of the angle between the two vectors $(x,y,z)$ and $(x,0,0)$.  The law of cosines tells us that
$$
\| (x,y,z) \|^2 + \|(x,0,0)\|^2 - 2\|(x,y,z)\|\|(x,0,0)\|\cos\alpha = \|(x,y,z)-(x,0,0)\|^2
$$
(since $(x,y,z)-(x,0,0)$ corresponds to the third side of the triangle).  Hence
$$
(x^2+y^2+z^2) + x^2 - 2\sqrt{x^2+y^2+z^2}\,\cdot |x|\cos\alpha = y^2+z^2.
$$
And then
$$
x^2 - \sqrt{x^2+y^2+z^2}\,\cdot |x|\cos\alpha = 0.
$$
Solve this for $\cos\alpha$.  Similarly find $\cos\beta$ and $\cos\gamma$.  Then add their squares.
(And remember that $x^2/|x| = |x|$.)
Later edit: The comment by "Blue" below gives a simplification: The points $(0,0,0)$, $(x,0,0)$, and $(x,y,z)$ are the vertices of a right triangle.  From $(0,0,0)$ to $(x,y,z)$ is the hypotenuse; from $(0,0,0)$ to $(x,0,0)$ is the side that is "adjacent" to the angle $\alpha$.  So
$$
\cos\alpha = \frac{\text{adjacent}}{\text{hypontenuse}} = \frac{|x|}{\sqrt{x^2+y^2+z^2}}.
$$
The other two cosines are found in the same way, with $|y|$ and $|z|$ in their numerators instead of $|x|.$
