If $m_{1}, m_{2}, m_{3}, m_{4}$ are respectively the magnitudes of the vectors $a_{1} = 2 i - j + k, a_{2} = 3 i - 4 j - 4 k, a_{3} = - i + j - k, a_{4} = - i + 3 j + k$ , then the correct order of $m_{1}, m_{2}, m_{3}, m_{4}$ is;

$m_{3} < m_{1} < m_{4} < m_{2}$

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$m_{3} < m_{1} < m_{2} < m_{4}$

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$m_{3} < m_{4} < m_{1} < m_{2}$

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$m_{3} < m_{4} < m_{2} < m_{1}$

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Solution

The correct option is A
$m_{3} < m_{1} < m_{4} < m_{2}$

Explanation for the correct answer:
Given, $a_{1} = 2 i - j + k, a_{2} = 3 i - 4 j - 4 k, a_{3} = - i + j - k, a_{4} = - i + 3 j + k$
We know that the magnitude of vector is given by $\sqrt{x^{2} + y^{2} + z^{2}}$
So,
$\Rightarrow m_{1} = \sqrt{2^{2} + {(- 1)}^{2} + {(1)}^{2}} = \sqrt{6} \Rightarrow m_{2} = \sqrt{3^{2} + {(- 4)}^{2} + {(- 4)}^{2}} = \sqrt{41} \Rightarrow m_{3} = \sqrt{{(- 1)}^{2} + 1^{2} + {(- 1)}^{2}} = \sqrt{3} \Rightarrow m_{4} = \sqrt{{(- 1)}^{2} + 3^{2} + 1^{2}} = \sqrt{11}$
So, $m_{3} < m_{1} < m_{4} < m_{2}$
Therefore, option (A) is the correct answer.

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