# Question regarding 3D vectors and their angles

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.

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.

• Thank you for your help! I'm confused about how you got the left side: $u_x^2 + u_y^2 + u_z^2 = 1$. If this side is supposed to be the square of $u_x + u_y + u_z$, don't you have to expand the trinomial (like FOIL for binomials)? Mar 27, 2020 at 18:02
• @Kman3 No. The expression $(u_x+u_y+u_z)$ is not the same as $u_x\hat i+ u_y\hat j +u_z\hat k$. We can't square the latter using FOIL because terms like $\hat i\hat j$ are undefined. Mar 27, 2020 at 18:11
• So when you stated that "its magnitude equals one as does the square of its magnitude", I assume you mean that $\vert \hat u \vert = \vert \hat u \vert^2 = 1$. So with that knowledge, how do I get to your conclusion, that $u_x^2 + u_y^2 + u_z^2 = 1$? Thanks Mar 27, 2020 at 18:23
• $|\hat x|=\sqrt{x_1^2+x_2^2+x_3^2}$. Mar 27, 2020 at 18:24
• Ah! I see. Just for reference, is that a formula with a certain name, or is it a logical conclusion due to properties of vectors? Mar 27, 2020 at 18:26

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 $$\cos^2\alpha+\cdots=\frac{a^2}{u^2}+\cdots=\frac{a^2+\cdots}{u^2}=\frac{u^2}{u^2}=1$$.

• Hi! I appreciate you helping me. Can you clarify why $a^2 + b^2 + c^2 = u^2$? Mar 27, 2020 at 18:03
• @Kman3 This is the Pythagorean theorem in $3$ dimensions.
– J.G.
Mar 27, 2020 at 18:26
• Thanks very much! Mar 27, 2020 at 18:31