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

The Correct Option is A

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;
\(\therefore m_{3} < m_{1} < m_{4} < m_{2}\)