Mathematics - Vector Question with Solution | TestHub

Given,

$\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, with $10\left|a_i\right|<1,i=1,2,3$, 

Let us assume that $\left|a_1\right|\le \left|a_2\right|\le \left|a_3\right|$

Now we know that,

${\left|\overrightarrow{a}\right|}^2={\left|{\overrightarrow{a}}_1\right|}^2+{\left|{\overrightarrow{a}}_2\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2$

And ${\left|{\overrightarrow{a}}_3\right|}^2\le {\left|{\overrightarrow{a}}_1\right|}^2+{\left|{\overrightarrow{a}}_2\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2$

So, combining both above equations we get,

$\Rightarrow {\left|\overrightarrow{a}\right|}^2\ge {\left|{\overrightarrow{a}}_3\right|}^2$

$\Rightarrow \left|\overrightarrow{a}\right|\ge \left|{\overrightarrow{a}}_3\right|=\text{max}\left\{\left|{\overrightarrow{a}}_1\right|,\left|{\overrightarrow{a}}_2\right|\left|{\overrightarrow{a}}_3\right|\right\}$

Hence, $\left(A\right)$ is true

Now again solving,

${\left|\overrightarrow{a}\right|}^2={\left|{\overrightarrow{a}}_1\right|}^2+{\left|{\overrightarrow{a}}_2\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2$

And ${\left|{\overrightarrow{a}}_1\right|}^2+{\left|{\overrightarrow{a}}_2\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2\le {\left|{\overrightarrow{a}}_3\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2$

$\Rightarrow {\left|{\overrightarrow{a}}_1\right|}^2+{\left|{\overrightarrow{a}}_2\right|}^2+{\left|{\overrightarrow{a}}_3\right|}^2\le 3{\left|{\overrightarrow{a}}_3\right|}^2$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2\le 3{\left|{\overrightarrow{a}}_3\right|}^2$

$\Rightarrow \left|\overrightarrow{a}\right|\le 3\left|{\overrightarrow{a}}_3\right|=\text{max}\left\{\left|{\overrightarrow{a}}_1\right|,\left|{\overrightarrow{a}}_2\right|\left|{\overrightarrow{a}}_3\right|\right\}$

Hence, $\left(B\right)$ is also true.