Q74 / 160 · TS EAMCET 2009 · Mathematics
If $m_1, m_2, m_3$ and $m_4$ are respectively the magnitudes of the vectors
$\overrightarrow{\mathbf{a}}_1=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{a}}_2=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$,
$\overrightarrow{\mathbf{a}}_3=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \quad$ and $\quad \overrightarrow{\mathbf{a}}_4=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$,
then the correct order of $m_1, m_2, m_3$ and $m_4$ is
- Correct answerA$m_3 < m_1 < m_4 < m_2$
- B$m_3 < m_1 < m_2 < m_4$
- C$m_3 < m_4 < m_1 < m_2$
- D$m_3 < m_4 < m_2 < m_1$
Solution
Given,
$\begin{aligned}
& m_1=\left|\overrightarrow{\mathbf{a}}_1\right|=\sqrt{2^2+(-1)^2+(1)^2}=\sqrt{6} \\
& m_2=\left|\overrightarrow{\mathbf{a}}_2\right|=\sqrt{3^2+(-4)^2+(-4)^2}=\sqrt{41} \\
& m_3=\left|\overrightarrow{\mathbf{a}}_3\right|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}
\end{aligned}$
and
$m_4=\left|\overrightarrow{\mathbf{a}}_4\right|=\sqrt{(-1)^2+(3)^2+(1)^2}=\sqrt{11}$
$\therefore \quad m_3 < m_1 < m_4 < m_2$
Hence, the correct answer is option: 'A'.