Given $\left | _A \mathbf r _B \right | = \left | _C \mathbf r _B \right | = 10$, prove that $0 \le \left | _A \mathbf r _C \right | \le 20$

Question

The question:

If $\left | _A \mathbf r _B \right | = \left | _C \mathbf r _B \right | = 10$, prove that $0 \le \left | _A \mathbf r _C \right | \le 20$.

Note that

$_A \mathbf r _B:$ the position of $A$ relative to $B$, which is $\overline{OA}-\overline{OB}$.

$\left | \mathbf r \right |$: magnitude of vector $\mathbf r$.

I was able to complete the question, but using a diagram and some logic. My attempt:

$\left | _A \mathbf r _C \right |$ is at a minimum when $A=C \implies \left | _A \mathbf r _C \right |=0$. Maximum is when $B$ is on $_A \mathbf r _C \implies \left | _A \mathbf r _C \right | = 10+10 =20$. Hence $0 \le \left | _A \mathbf r _C \right | \le 20$.

Is there any way to do this question with pure algebra? I suppose the triangle inequality will appear somewhere but I have no idea how to link the information the question gives you with an inequality.

Answer 1

This can be done purely algebraically. It is an instance of the triangle inequality, which can be derived from the Cauchy–Schwarz inequality, which itself has a purely algebraic proof. The proof of Cauchy–Schwarz and the derivation of the triangle inequality are both on the linked Wikipedia page.

Answer 2

Well here you go: $$_C \mathbf r _B = -1\cdot _B \mathbf r _C \Rightarrow \lvert _C \mathbf r _B\rvert = \lvert -1\rvert \cdot \lvert _B \mathbf r _C\rvert= \lvert _B \mathbf r _C\rvert$$ and due to triangle inequality you have: $$\lvert _A \mathbf r _C\rvert \leq \lvert _A \mathbf r _B\rvert + \lvert _B \mathbf r _C\rvert = \lvert _A \mathbf r _B\rvert + \lvert _C \mathbf r _B\rvert = 10+10 = 20$$