Q56 / 120 · VITEEE 2009 · Mathematics
If $\mathrm{m}_{1}, \mathrm{~m}_{2}, \mathrm{~m}_{3}$ and $\mathrm{m}_{4}$ are respectively the magnitudes of the vectors
$\vec{a}_{1}=2 \hat{i}-\hat{j}+\hat{k}, \vec{a}_{2}=3 \hat{i}-4 \hat{j}-4 \hat{k}$
$\vec{a}_{3}=\hat{i}+\hat{j}-\hat{k}, \text { and } \vec{a}_{4}=-\hat{i}+3 \hat{j}+\hat{k}$
then the correct order of $\mathrm{m}_{1}, \mathrm{~m}_{2}, \mathrm{~m}_{3}$ and $\mathrm{m}_{4}$ is
- Correct answerA$\mathrm{m}_{3} < \mathrm{m}_{1} < \mathrm{m}_{4} < \mathrm{m}_{2}$
- B$\mathrm{m}_{3} < \mathrm{m}_{1} < \mathrm{m}_{2} < \mathrm{m}_{4}$
- C$\mathrm{m}_{3} < \mathrm{m}_{4} < \mathrm{m}_{1} < \mathrm{m}_{2}$
- D$\mathrm{m}_{3} < \mathrm{m}_{4} < \mathrm{m}_{2} < \mathrm{m}_{1}$
Solution
$m_{1}=\left|\vec{a}_{1}\right|=\sqrt{2^{2}+(-1)^{2}+(1)^{2}}=\sqrt{6}$
$$
\begin{array}{l}
m_{2}=\left|\vec{a}_{2}\right|=\sqrt{3^{2}+(-4)^{2}+(-4)^{2}}=\sqrt{41} \\
m_{3}=\left|\vec{a}_{3}\right|=\sqrt{1^{2}+1^{2}+(-1)^{2}}=\sqrt{3}
\end{array}
$$
and $m_{4}=\left|\vec{a}_{4}\right|=\sqrt{(-1)^{2}+(3)^{2}+(1)^{2}}=\sqrt{11}$
$$
\therefore m_{3} < m_{1} < m_{4} < m_{2}
$$
Hence, the correct answer is option: 'A'.