Question

If ${m_1}$,${m_2}$,${m_3}$,${m_4}$, are respectively the magnitudes of the vectors ${\overline a _1} = 2\overline i - \overline j + \overline k ,$ ${\overline a _2} = 3\overline i - 4\overline j - 4\overline k ,{\overline a _3} = - \overline i + \overline j - \overline k ,$ ${\overline a _4} = - \overline i + 3\overline j + \overline k$, then the correct order of ${m_1}$,${m_2}$,${m_3}$,${m_4}$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 formula for to calculate the determinant of vector is:
${\overline A _1} = {a_1}\overline i - {a_2}\overline j + {a_3}\overline k$
$\left| {{{\overline A }_1}} \right| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{a_2}} \right)}^2} + {{\left( {{a_3}} \right)}^2}}$
After finding the determinant, compare all the values and arrange them in descending order.

• terms are said to be in descending order if every term on the right hand side is less than the previous term. Terms in descending order have a sign of less than (<) between them.

Complete step-by-step answer :
We calculate all the determinants separately.
${\overline a _1} = 2\overline i - \overline j + \overline k$
Calculate the determinant by using this formula $\left| {{{\overline A }_1}} \right| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{a_2}} \right)}^2} + {{\left( {{a_3}} \right)}^2}}$
Substitute the values of coefficients of i, j and k in the formula
${m_1} = \left| {{{\overline a }_1}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}}$
$= \sqrt {4 + 1 + 1}$
$= \sqrt 9$

${\overline a _2} = 3\overline i - 4\overline j - 4\overline k$
Calculate the determinant by using this formula $\left| {{{\overline A }_1}} \right| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{a_2}} \right)}^2} + {{\left( {{a_3}} \right)}^2}}$
Substitute the values of coefficients of i, j and k in the formula
${m_2} = \left| {{{\overline a }_2}} \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}}$
$= \sqrt {9 + 16 + 16}$
$= \sqrt {41}$

${\overline a _3} = - \overline i + \overline j - \overline k$
Calculate the determinant by using this formula $\left| {{{\overline A }_1}} \right| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{a_2}} \right)}^2} + {{\left( {{a_3}} \right)}^2}}$
Substitute the values of coefficients of i, j and k in the formula
${m_3} = \left| {{{\overline a }_3}} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}}$
$= \sqrt {1 + 1 + 1}$
$= \sqrt 3$

${\overline a _4} = - \overline i + 3\overline j + \overline k$
Calculate the determinant by using this formula $\left| {{{\overline A }_1}} \right| = \sqrt {{{\left( {{a_1}} \right)}^2} + {{\left( {{a_2}} \right)}^2} + {{\left( {{a_3}} \right)}^2}}$
Substitute the values of coefficients of i, j and k in the formula
${m_4} = \left| {{{\overline a }_4}} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 1 \right)}^2}}$
$= \sqrt {1 + 9 + 1}$
$= \sqrt {11}$
The value of ${m_1}$,${m_2}$,${m_3}$,${m_4}$are $\sqrt 6 ,\sqrt {41} ,\sqrt 3 ,\sqrt 9$.

$\therefore$ ${m_3} < {m_1} < m{}_4 < {m_2}$. So, Option A is correct.

Note:
All the value of ${m_1}$,${m_2}$,${m_3}$,${m_4}$are in root mode or in decimal form otherwise so student can easily place in correct order otherwise get error in answer.
Students are likely to make wrong descending order of terms as they might get confused with the under root terms, they can arrange such terms by assuming that there is no under root as all the values contain under root so value of under root of lesser value will be less than under root of greater value.