Question

If $M_{1}, M_{2}, M_{3}$ and $M_{4}$, are respectively the magnitudes of the vectors $a_{1}=2 \hat{ i }-\hat{ j }+\hat{ k }$
$a _{2}=-3 \hat{ i }-4 \hat{ j }-4 \hat{ k }, a _{3}=-\hat{ i }+\hat{ j }-\hat{ k }$
$a _{4}=-\hat{ i }+3 \hat{ j }+\hat{ k }$, 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)

Given that, $a _{1}=2 \hat{ i }-\hat{ j }+\hat{ k }$
$a _{2} =3 \hat{ i }-4 \hat{ j }-4 \hat{ k } $
$a _{3} =-\hat{ i }+\hat{ j }-\hat{ k } $
$a _{4} =-\hat{ i }+3 \hat{ j }+\hat{ k } $
$M =| a |$
$M_{1} =\left| a _{1}\right|=|2 \hat{ i }-\hat{ j }+\hat{ k }| $
$=\sqrt{(2)^{2}+(-1)^{2}+(1)^{2}} $
$=\sqrt{4+1+1}=\sqrt{6} $
$=|3 \hat{ i }-4 \hat{ j }-4 \hat{ k }|$
$=\sqrt{(3)^{2}+(-4)^{2}+(-4)^{2}} $
$=\sqrt{9+16+16}=\sqrt{41} $
$ M_{3} =|- i +\hat{ j }-\hat{ k }| $
$=\sqrt{(-1)^{2}+(1)^{2}+(-1)^{2}} $
$=\sqrt{1+1+1}=\sqrt{3}$
$ M_{4} =|-\hat{ i }+3 \hat{ j }+\hat{ k }| $
$=\sqrt{(-1)^{2}+(3)^{2}+(-1)^{2}} $
$=\sqrt{1+9+1}=\sqrt{11}$
Hence, the correct order of magnitudes
$M_{3} < M_{1} < M_{4} < M_{2}$