For any vector a=a_1 i+a_2 j+a_3 k, with 10|a_i|<1,i=1,2,3, consider the following statements: (A) : max |a_1|,|a_2|,|a_3|≤ | a| (B) : | a|≤ 3 max…

Given,

$\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$a=a1​i^+a2​j^​+a3​k^, with $10|{a}_{i}|<1,i=1,2,3$10∣ai​∣<1,i=1,2,3,

Let us assume that $|{a}_{1}|\leq |{a}_{2}|\leq |{a}_{3}|$∣a1​∣≤∣a2​∣≤∣a3​∣

Now we know that,

$${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$$∣a∣2=∣a1​∣2+∣a2​∣2+∣a3​∣2

And ${|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$∣a3​∣2≤∣a1​∣2+∣a2​∣2+∣a3​∣2

So, combining both above equations we get,

$$\Rightarrow {|\vec{a}|}^{2}\geq {|{\vec{a}}_{3}|}^{2}$$⇒∣a∣2≥∣a3​∣2

$$\Rightarrow |\vec{a}|\geq |{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$$⇒∣a∣≥∣a3​∣=max∣a1​∣,∣a2​∣∣a3​∣

Hence, $(A)$(A) is true

Now again solving,

$${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$$∣a∣2=∣a1​∣2+∣a2​∣2+∣a3​∣2

And ${|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}$∣a1​∣2+∣a2​∣2+∣a3​∣2≤∣a3​∣2+∣a3​∣2+∣a3​∣2

$$\Rightarrow {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$$⇒∣a1​∣2+∣a2​∣2+∣a3​∣2≤3∣a3​∣2

$$\Rightarrow {|\vec{a}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$$⇒∣a∣2≤3∣a3​∣2

$$\Rightarrow |\vec{a}|\leq 3|{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$$⇒∣a∣≤3∣a3​∣=max∣a1​∣,∣a2​∣∣a3​∣

Hence, $(B)$(B) is also true.