QJEEM23Ma4299: JEE Main 2023 Physics - Class 11 JEEM23Ma4299 Solution: For any vector , with , consider the following statements :…

Question

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For any vector $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$ , with $10\left|a_{i}\right|<1, i=1,2,3$ , consider the following statements :

(A): $\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|$

(B) : $|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$

Accepted Answer

We have,

\begin{aligned}&amp; 10\left|a_i\right|&lt;1, i=1,2,3 \\\\&amp; \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\&amp; |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\&amp; \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. }\end{aligned}

Hence, (A) is true.

\begin{array}{rlrl} &amp; |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\&amp; =\sqrt{3}\left|a_1\right| \\\\\therefore &amp; |\vec{a}|=\sqrt{3}\left|a_1\right|&lt;3\left|a_1\right| \\\\\therefore &amp; |\vec{a}|&lt;3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\}\end{array}

Hence, (B) is also true.

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