If M_1, M_2, M_3 and M_4 are respectively the | vedclass

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(A) The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.Calculating the magnitudes:$

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$M_3 < M_1 < M_4 < M_2$

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(A) The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.Calculating the magnitudes:$M_1 = |\vec{a}_1| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.45$$M_2 = |\vec{a}_2| = \sqrt{(-3)^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40$$M_3 = |\vec{a}_3| = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73$$M_4 = |\vec{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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