If m_1, m_2, m_3 and m_4 are respectively the | vedclass

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(A) The magnitude of a vector $\overrightarrow{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\overrightarrow{a}| = \sqrt{x^2 + y^2 + z^2}$.Calcul

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$m_3 < m_1 < m_4 < m_2$

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(A) The magnitude of a vector $\overrightarrow{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\overrightarrow{a}| = \sqrt{x^2 + y^2 + z^2}$.Calculating the magnitudes:$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.45$$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40$$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73$$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \approx 3.32$Comparing the values: $\sqrt{3} < \sqrt{6} < \sqrt{11} < \sqrt{41}$,which implies $m_3 < m_1 < m_4 < m_2$.

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