Question

If $m_{1}, m_{2}, m_{3}$ and $m_{4}$ are respectively the magnitudes of the vectors $\vec{ a }_{1}=2 \hat{ i }-\hat{ j }+\hat{ k }, \,\,\,\vec{ a }_{2}=3 \hat{ i }-4 \hat{ j }-4 \hat{ k }$ $\vec{ a }_{3}=\hat{ i }+\hat{ j }-\hat{ k } \,\,\,$ and $\,\,\,\vec{ a }_{4}=-\hat{ i }+3 \hat{ j }+\hat{ k }$ 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. $m_3 < m_4 < m_2 < m_1$

Answer 1

Answer: (A)

$m_3 < m_1 < m_4 < m_2$

Answer 2

The correct answer is A:$m_3<m_1<m_4<m_2$
Given that: $m_1,m_2,m_3,m_4$ are the magnitudes of vectors;
$\vec{a_1}=\hat{2i}+\hat{j}+\hat{k}$
$\vec{a_2}=3\hat{i}-4\hat{j}-4\hat{k}$
$\vec{a_3}=-\hat{i}+\hat{j}-\hat{k}$
$\vec{a_4}=-\hat{i}+3\hat{j}+\hat{k}$
$\therefore m_{1}=\left|\vec{a_{1}}\right|=\sqrt{2^{2}+\left(-1\right)^{2}+\left(1\right)^{2}}=\sqrt{6}$
$m_{2}=\left|\vec{a_{2}}\right|=\sqrt{3^{2}+\left(-4\right)^{2}+\left(-4\right)^{2}}=\sqrt{41}$
$m_{3}=\left|\vec{a_{3}}\right|=\sqrt{1^{2}+1^{2}+\left(-1\right)^{2}}=\sqrt{3}$
and $m_{4}=\left|\vec{a_{4}}\right|=\sqrt{\left(-1\right)^{2}+\left(3\right)^{2}+\left(1\right)^{2}}=\sqrt{11}$
As question asked for the order of arrangement of $m_1,m_2,m_3,m_4$ so it will be;
$\therefore m_{3} < m_{1} < m_{4} < m_{2}$