JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 8)
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) : $$|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$$
Only (B) is true
Only (A) is true
Neither (A) nor (B) is true
Both (A) and (B) are true
Explanation
We have,
$$ \begin{aligned} & 10\left|a_i\right|<1, i=1,2,3 \\\\ & \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\ & |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\ & \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. } \end{aligned} $$
Hence, (A) is true.
$$ \begin{array}{rlrl} & |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\ & =\sqrt{3}\left|a_1\right| \\\\ \therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\ \therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \end{array} $$
Hence, (B) is also true.
$$ \begin{aligned} & 10\left|a_i\right|<1, i=1,2,3 \\\\ & \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\ & |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\ & \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. } \end{aligned} $$
Hence, (A) is true.
$$ \begin{array}{rlrl} & |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\ & =\sqrt{3}\left|a_1\right| \\\\ \therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\ \therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \end{array} $$
Hence, (B) is also true.
Comments (0)
