For a=a_1 i+a_2 j+a_3 k, consider: (A) |a_1|,|a_2|,|a_3|| a| (B) | a|√3|a_1|,|a_2|,|a_3|

Question

For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$, consider:
(A) $\max\{|a_1|,|a_2|,|a_3|\}\le|\vec a|$
(B) $|\vec a|\le\sqrt3\max\{|a_1|,|a_2|,|a_3|\}$

Accepted Answer

WLOG let $|a_3|=\max\{|a_1|,|a_2|,|a_3|\}$.

$$|\vec a|^2 = a_1^2+a_2^2+a_3^2 \ge a_3^2 \implies |\vec a|\ge|a_3| = \max\{|a_1|,|a_2|,|a_3|\}$$

So (A) holds. Also
$$|\vec a|^2 \le a_3^2+a_3^2+a_3^2 = 3a_3^2 \implies |\vec a|\le\sqrt3\,\max\{|a_1|,|a_2|,|a_3|\}$$

So (B) holds too. Both are true.

Solutions · 1