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# If m1, m2, m3, m4 are the magnitudes of the vectors ${\vec a_1} = \,2\vec i - \vec j + \vec k,\,{\vec a_2} = \,3\vec i - 4\vec j - 4\vec k,\,{\vec a_3} = - \vec i + \vec j - \vec k,\,{\vec a_4} = - \vec i + 3\vec j + \vec k$ then the correct order of m1, m2, m3, m4 is:A) m3 < m1 < m4 < m2 B) m3 < m1 < m2 < m4 C) m3 < m4 < m1 < m2 D) m3 < m4 < m2 < m1  Hint – In order to solve this problem use the formula of finding the magnitude of a given vector. After doing this you will get the right answer.

As we know that if the vector is $\vec p = a\vec i + b\vec j + c\vec k$ the its magnitude will be $|\vec p| = \sqrt {{a^2} + {b^2} + {c^2}}$.
Therefore the magnitude of the vector ${\vec a_1} = \,2\vec i - \vec j + \vec k$ is $|{\vec a_1}| = \sqrt {{2^2} + {{( - 1)}^2} + {{(1)}^2}} = \sqrt 6 = {m_1}$
And the magnitude of the vector ${\vec a_2} = \,3\vec i - 4\vec j - 4\vec k$ is $|{\vec a_2}| = \sqrt {{3^2} + {{( - 4)}^2} + {{( - 4)}^2}} = \sqrt {41} = {m_2}$
The magnitude of the vector ${\vec a_3} = - \vec i + \vec j - \vec k$ is $|{\vec a_3}| = \sqrt {{{( - 1)}^2} + {{(1)}^2} + {{( - 1)}^2}} = \sqrt 3 = {m_3}$
The magnitude of the vector ${a_4} = - \vec i + 3\vec j + \vec k$ = $|{\vec a_4}| = \sqrt {{{( - 1)}^2} + {{(3)}^2} + {{(1)}^2}} = \sqrt {11} = {m_4}$
We can clearly see that m3 < m1 < m4 < m2.
So, the correct option is A.

Note - Whenever you face such type of problems of finding magnitude of vectors you have to use the formula for finding magnitudes of vectors. For example the vector is $\vec p = a\vec i + b\vec j + c\vec k$ then its magnitude will be $|\vec p| = \sqrt {{a^2} + {b^2} + {c^2}}$. Proceeding like this you will get the right answer.
View Notes  Types of Vector  Coplanarity of Vectors  Vectors  Coplanar Vectors  Multiplication of Vector with Scalar  Dot Product of Two Vectors        