For any vector → a = a _ 1 ^ i + a _ 2 ^ j + a _ 3 ^ k , with 10| a...

Given,

\overset{\to}{a} = {a}_{1}\overset{^}{i} + {a}_{2}\overset{^}{j} + {a}_{3}\overset{^}{k}, with $10\left|{a}_{i}\right| < 1, i = 1, 2, 3$,

Let us assume that \left|{a}_{1}\right| \le \left|{a}_{2}\right| \le \left|{a}_{3}\right|$</p><p>Now we know that,</p><p>${\left|\overset{\to}{a}\right|}^{2} = {\left|\overset{\to}{a}1\right|}^{2} + {\left|\overset{\to}{a}2\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2}$</p><p>And${\left|\overset{\to}{a}3\right|}^{2} \le {\left|\overset{\to}{a}1\right|}^{2} + {\left|\overset{\to}{a}2\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2}$</p><p>So, combining both above equations we get,</p><p>$⇒ {\left|\overset{\to}{a}\right|}^{2} \ge {\left|\overset{\to}{a}3\right|}^{2}$</p><p>$⇒ \left|\overset{\to}{a}\right| \ge \left|\overset{\to}{a}3\right| = \text{max}\left\{\left|\overset{\to}{a}1\right|, \left|\overset{\to}{a}2\right|\left|\overset{\to}{a}3\right|\right\}$</p><p>Hence, \left(A\right)$ is true

Now again solving,

${\left|\overset{\to}{a}\right|}^{2} = {\left|\overset{\to}{a}1\right|}^{2} + {\left|\overset{\to}{a}2\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2}$

And ${\left|\overset{\to}{a}1\right|}^{2} + {\left|\overset{\to}{a}2\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2} \le {\left|\overset{\to}{a}3\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2}$

$⇒ {\left|\overset{\to}{a}1\right|}^{2} + {\left|\overset{\to}{a}2\right|}^{2} + {\left|\overset{\to}{a}3\right|}^{2} \le 3{\left|\overset{\to}{a}3\right|}^{2}$

$⇒ {\left|\overset{\to}{a}\right|}^{2} \le 3{\left|\overset{\to}{a}3\right|}^{2}$

$⇒ \left|\overset{\to}{a}\right| \le 3\left|\overset{\to}{a}3\right| = \text{max}\left\{\left|\overset{\to}{a}1\right|, \left|\overset{\to}{a}2\right|\left|\overset{\to}{a}3\right|\right\}$

Hence, \left(B\right) is also true.