If $m_{1}, m_{2}, m_{3}$ and $m_{4}$ are respectively the magnitudes of the vectors
$a_{1} = 2 i - j + k, a_{2} = 3 i - 4 j - 4 k,$
$a_{3} = i + j - k$ and $a_{4} = - i + 3 j + k,$
then the correct order of $m_{1}, m_{2}, m_{3}$ and $m_{4}$ is

m_{3} < m_{1} < m_{4} < m_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m_{3} < m_{1} < m_{2} < m_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m_{3} < m_{4} < m_{4} < m_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m_{3} < m_{4} < m_{2} < m_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $m_{3} < m_{1} < m_{4} < m_{2}$
Given that:
${\to a}_{1} = 2 i - j + k, {\to a}_{2} = 3 i - 4 j - 4 k$
${\to a}_{3} = i + j - k$ and ${\to a}_{4} = - i + 3 j + k$
Then,
$m_{1} = \sqrt{2^{2} + (- 1)^{2} + 1^{2}} = \sqrt{6}$
$m_{2} = \sqrt{3^{2} + (- 4)^{2} + (- 4)^{2}} = \sqrt{41}$
$m_{3} = \sqrt{1^{2} + 1^{2} + (- 1)^{2}} = \sqrt{3}$ and
$m_{4} = \sqrt{(- 1)^{2} + 3^{2} + 1^{2}} = \sqrt{} 11$
Correct order of magnitudes will be $m_{3} < m_{1} < m_{4} < m_{2} .$
Hence, A is the correct option.

Suggest Corrections

Similar questions

m_{1}, m_{2}, m_{3}, m_{4}

_{1} = ¯ ¯¯¯ ¯ 2 i - ¯ j + ¯ ¯ ¯ k,_{2} = ¯ ¯¯¯ ¯ 3 i - ¯ ¯¯¯ ¯ 4 j - ¯ ¯¯¯¯ ¯ 4 k,_{3} = - ¯ i + ¯ j - ¯ ¯ ¯ k,_{4} = - ¯ i + 3 ¯ j + ¯ ¯ ¯ k

m_{1}, m_{2}, m_{3}, m_{4}

m_{1}, m_{2}, m_{3}, m_{4},

a_{1} = ¯ ¯¯¯ ¯ 2 i - ¯ j + ¯ ¯ ¯ k, a_{2} = ¯ ¯¯¯ ¯ 3 i - ¯ ¯¯¯ ¯ 4 j - ¯ ¯¯¯¯ ¯ 4 k, a_{3} = ¯ ¯¯¯¯ ¯ - i + ¯ j - ¯ ¯ ¯ k, a_{4} = ¯ ¯¯¯¯ ¯ - i + ¯ ¯¯¯ ¯ 3 j + ¯ ¯ ¯ k

M 1, M 2, M 3

m_{1}

m_{2}

m_{3}

m_{4}

1 + 4 i, 3 + i, 1 - i a n d 2 - 3 i

m_{4}

(m_{1} + m_{2}) > (m_{3} + m_{4})

Join BYJU'S Learning Program