Q. If $m_{1},m_{2},m_{3}\ and\ m_{4}$ are respectively the magnitudes of the vectors $\overrightarrow{\mathbf{a}_{1}} = 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}},\ \overrightarrow{\mathbf{a}_{2}} = 3\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}},\ \overrightarrow{\mathbf{a}_{3}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}\ and\ \overrightarrow{\mathbf{a}_{4}} = \widehat{- \dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}},$ then the correct order of $m_{1},m_{2},m_{3}\ and\ m_{4}$ is
A) $m_{3} < m_{1} < m_{4} < m_{2}$
B) $m_{3} < m_{1} < m_{2} < m_{4}$
C) $m_{3} < m_{4} < m_{1} < m_{2}$
D) $m_{3} < m_{4} < m_{2} < m_{1}$
Explanation
(a)
Given, $m_{1} = \left| \overrightarrow{\mathbf{a}_{1}} \right| = \sqrt{2^{2} + {( - 1)}^{2} + {(1)}^{2}} = \sqrt{6}$
$m_{2} = \left| \overrightarrow{\mathbf{a}_{2}} \right| = \sqrt{3^{2} + {( - 4)}^{2} + {( - 4)}^{2}} = \sqrt{41}$
$m_{3} = \left| \overrightarrow{\mathbf{a}_{3}} \right| = \sqrt{1^{2} + 1^{2} + {( - 1)}^{2} = \sqrt{3}}$
and$\ m_{4} = \left| \overrightarrow{\mathbf{a}_{4}} \right| = \sqrt{{( - 1)}^{2} + {(3)}^{2} + {(1)}^{2}} = \sqrt{11}$
$\therefore\ \ m_{3} < m_{1} < m_{4} < m_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.