Question regarding 3D vectors and their angles

Question

I recently came across this in my Vectors textbook.

I don't quite follow the logic that ends with $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. I understand that unit vectors have magnitude 1, and that they are achieved by dividing a vector $\overrightarrow v$ by its magnitude $\vert \vec v \vert$ which produces the unit vector $\hat v$. Is $a$ considered a vector here, such that dividing it by $\vert \vec u \vert$ produces a unit vector?

As I am just starting to learn about vectors, a rudimentary explanation of why $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ would be greatly appreciated.

Answer 1

If $\hat u=u_x\hat i + u_y\hat j + u_z\hat k$ is a unit vector then its magnitude equals one as does the square of its magnitude.

Therefore, $u_x^2 +u_y^2 + u_z^2=1$.

But $$u_x=\cos\alpha$$ $$u_y=\cos\beta$$ $$u_z=\cos\gamma$$ so plugging those into the above gives the result.

Answer 2

No, $a$ is one component of the vector $\vec{u}$. A unit vector is obtained by dividing $\vec{u}$, one component at a time, by $|\vec{u}|$. So $\cos^2\alpha+\cdots=\frac{a^2}{u^2}+\cdots=\frac{a^2+\cdots}{u^2}=\frac{u^2}{u^2}=1$.