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

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(A) Given vectors are $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\h

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$m_3 < m_1 < m_4 < m_2$

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(A) Given vectors are $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$, and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$.Calculating the magnitudes $m_1, m_2, m_3, m_4$:$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9+16+16} = \sqrt{41}$$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1+1+1} = \sqrt{3}$$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1+9+1} = \sqrt{11}$Comparing the values: $\sqrt{3} Therefore, $m_3 < m_1 < m_4 < m_2$.

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