For any vector a=a_1i+a_2j+a_3k, with 10|a_i|(A): max|a_1|,|a_2|,|a_3| |a| (B) :

Question

For any vector $\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$, with $10|{a}_{i}|(A): $max{|{a}\_{1}|,|{a}\_{2}|,|{a}\_{3}|}\le |\vec{a}|$

(B) : $

Accepted Answer

We have,

$\begin{gathered} 10|{a}_{i}|<1,i=1,2,3 \ |{a}_{1}|\ge |{a}_{2}|\ge |{a}_{3}| \ |\vec{a}|=\sqrt{{a}_{2}^{1}+{a}_{2}^{2}+{a}_{2}^{3}}\ge \sqrt{{a}_{2}^{1}} \ 	herefore |\vec{a}|\ge |{a}_{1}|max{|{a}_{1}|,|{a}_{2}|,|{a}_{3}|} \end{gathered}$

Hence, (A) is true.

$\begin{gathered} |\vec{a}|=\sqrt{{a}_{2}^{1}+{a}_{2}^{2}+{a}_{2}^{3}}\le \sqrt{{a}_{2}^{1}+{a}_{2}^{1}+{a}_{2}^{1}} \ \ =\sqrt{3}|{a}_{1}| \ \ 	herefore |\vec{a}|=\sqrt{3}|{a}_{1}|<3|{a}_{1}| \ \ 	herefore |\vec{a}|<3max{|{a}_{1}|,|{a}_{2}|,|{a}_{3}|} \end{gathered}$

Hence, (B) is also true.

Solutions · 1