The magnitude of $5 \vec u + 4 \vec v + 2 \vec w$ where $\vec u,\vec v,\vec w$ are mutually perpendicular and of unit magnitude

Question

Of course the answer is $\sqrt{25+16+4} = \sqrt{45}$.

It is easy to see it when we consider the (extended) Pythagorean Theorem, or even more easily just taking $\vec u = \hat i, \ \vec v = \hat j$ and $\vec w = \hat k$.

But how do we show it without geometry and for the general case? For instance using dot/vector product?

Answer 1

By brute force:

$$\begin{align} \|5u+4v+2w\|^2 &= \langle 5u+4v+2w,5u+4v+2w\rangle \\ &= 25 \langle u,u\rangle + 16\langle v,v\rangle + 4\langle w,w\rangle + 2\left(20\langle u,v\rangle + 10\langle u,w\rangle + 8\langle v,w\rangle\right) \\ &= 25 \cdot 1 + 16 \cdot 1 + 4 \cdot 1 + 2(20 \cdot 0 + 10\cdot 0+ 8\cdot 0) \\ &= 25+16+4 \\ &= 45, \end{align}$$

so $\|5u+4v+2w\| = \sqrt{45}$. The point is to use the Pythagoeran theorem to reduce these boring computations.