Vector subtraction proof

Asked Jan 19 '20 at 13:35

Active Jan 19 '20 at 14:02

Viewed 37 times

Prove for any vectors $\mathbf{a}$ and $\mathbf{b}$

(i) $|\mathbf{a}—\mathbf{b}|\ge|\mathbf{a}|—|\mathbf{b}|$

(ii) $|\mathbf{a} — \mathbf{b}|\ge |\mathbf{b}| — |\mathbf{a}|$

(iii) Deduce that $|\mathbf{a} — \mathbf{b}|\ge |\ |\mathbf{a}| — |\mathbf{b}|\ |$

I have drawn parallelograms for (i) and (ii) so I know what $|\mathbf{a} — \mathbf{b}|$ looks like but I cannot prove them.

I cannot even fathom out what (iii) means.

edited Jan 19 '20 at 14:02

almagest

asked Jan 19 '20 at 13:35

Steblo

$(ii)$ seems to have an extra set of magnitudes, $a-|b|$ doesn’t make sense. – Michael Burr Jan 19 '20 at 13:39
You should look up the reverse triangle inequality. There should be plenty of proofs. And the right side of part (iii) means the absolute value of $|a|-|b|$, since that is a real number. And as Michael Burr says, the left side of (ii) should be $|a-b|$. – Milten Jan 19 '20 at 13:40
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Does this answer your question? Reverse Triangle Inequality Proof – Milten Jan 19 '20 at 13:42
Thanks. I’ve edited out the mistake in (ii) – Steblo Jan 19 '20 at 13:55
As a hint, if you don't wanna just read the proof, try proving $|a-b|+|b| \ge |a|$ for part (i), using the triangle inequality. For part (iii), remember that $|x| \le y \iff x\le y \wedge -x\le y$. – Milten Jan 19 '20 at 14:18
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I cannot believe that the earlier question attracted 89 votes and the answer 113. – almagest Jan 19 '20 at 14:20

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