If $|b| \ge 1$ and $x=-|a|b$, then which one of the following is necessarily true? [closed]

Question

If $|b| \ge 1$ and $x=-|a|b$, then which one of the following is necessarily true?

(1)$a-xb \lt 0$

(2)$a-xb \ge 0$

(3)$a-xb \gt 0$

(4)$a-xb \le 0$

$|b|\ge 1$, means if $b$ is positive then, $b \ge 1$, else $b \le -1$

What do you know? What did you try? Can you really answer none of these items? (Subsidiary question: when will you start conforming to the recommandations about the proper ways of asking questions on this site? And start answering comments, by the way?)
The new tag is ludicrous. Where do you see some linear algebra in the question? By the way, what do you view as belonging to linear algebra? If you have no clue, why do you see proper to edit tags?
Pretty sure a homework tag should be up there somewhere, that's for sure...

lab bhattacharjee · Accepted Answer · 2012-08-25 08:52:28Z

$a-xb=a+|a|b\cdot b=a+|a|b^2$

If $a≤0,|a|=-a=>a-xb=(-a)(b^2-1)$ which is clearly $≥0$ as $b^2≥1$ as $|b|≥1$ assuming $b$ is real.

If $a>0, |a|=a=>a-xb=a(b^2+1)$ which is >0

So. $a-xb≥0$

Alternatively, $a-xb=a+|a|b\cdot b=a+|a|b^2≥a+|a|$ as $b^2≥1$ as $|b|≥1$ assuming $b$ is real.

which is 0 if $a≤0$,

and if $a>0$, $a-xb≥2a>0$

So. $a-xb≥0$

closed as too localized by Did, LVK, Steve D, William, Quixotic Sep 9 '12 at 15:26