Question

If \[{m_1}\], \[{m_2}\], \[{m_3}\], \[{m_4}\] are respectively the magnitude of the vectors \[{{\bf{a}}_{\bf{1}}} = \overline {2{\bf{i}}} - \overline {\bf{j}} + \overline {\bf{k}} \], \[{{\bf{a}}_2} = \overline {3{\bf{i}}} - \overline {4{\bf{j}}} - \overline {4{\bf{k}}} \], \[{{\bf{a}}_3} = - \overline {\bf{i}} + \overline {\bf{j}} - \overline {\bf{k}} \], \[{{\bf{a}}_4} = - \overline {\bf{i}} + \overline {3{\bf{j}}} + \overline {\bf{k}} \], choose the lowest magnitude
(a) \[{m_1}\]
(b) \[{m_2}\]
(c) \[{m_3}\]
(d) \[{m_4}\]

Answer 1

Hint:
Here, we have to find which magnitude is the lowest. We will calculate the magnitudes of the given vectors and then compare them to get the answer.
Formula Used: The magnitude of a vector \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \] is given by \[\left| {\bf{v}} \right| = \sqrt {{x^2} + {y^2} + {z^2}} \].

Complete step by step solution:
We will first find the magnitude of the vector \[{{\bf{a}}_{\bf{1}}} = \overline {2{\bf{i}}} - \overline {\bf{j}} + \overline {\bf{k}} \].
Comparing \[{{\bf{a}}_{\bf{1}}} = \overline {2{\bf{i}}} - \overline {\bf{j}} + \overline {\bf{k}} \] and \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \], we get
\[x = 2\], \[y = - 1\] and \[z = 1\]
Substituting \[x = 2\], \[y = - 1\] and \[z = 1\] in the formula for magnitude of a vector, we get
\[\left| {{{\bf{a}}_{\bf{1}}}} \right| = \sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {1^2}} \]
Simplifying the expression, we get the magnitude \[{m_1}\] as
\[\begin{array}{l} \Rightarrow {m_1} = \sqrt {4 + 1 + 1} \\ \Rightarrow {m_1} = \sqrt 6 \end{array}\]
Next, we will find the magnitude of the vector \[{{\bf{a}}_2} = \overline {3{\bf{i}}} - \overline {4{\bf{j}}} - \overline {4{\bf{k}}} \].
Comparing \[{{\bf{a}}_2} = \overline {3{\bf{i}}} - \overline {4{\bf{j}}} - \overline {4{\bf{k}}} \] and \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \], we get
\[x = 3\], \[y = - 4\] and \[z = - 4\]
Substituting \[x = 3\],
\[y = - 4\] and \[z = - 4\] in the formula for magnitude of a vector, we get
\[\left| {{{\bf{a}}_2}} \right| = \sqrt {{3^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} \]
Simplifying the expression, we get the magnitude \[{m_2}\] as
\[\begin{array}{l} \Rightarrow {m_2} = \sqrt {9 + 16 + 16} \\ \Rightarrow {m_2} = \sqrt {41} \end{array}\]
Now, we will find the magnitude of the vector \[{{\bf{a}}_3} = - \overline {\bf{i}} + \overline {\bf{j}} - \overline {\bf{k}} \].
Comparing \[{{\bf{a}}_3} = - \overline {\bf{i}} + \overline {\bf{j}} - \overline {\bf{k}} \] and \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \], we get
\[x = - 1\], \[y = 1\] and \[z = - 1\]
Substituting \[x = - 1\], \[y = 1\] and \[z = - 1\] in the formula for magnitude of a vector, we get
\[\left| {{{\bf{a}}_3}} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} \]
Simplifying the expression, we get the magnitude \[{m_3}\] as
\[\begin{array}{l} \Rightarrow {m_3} = \sqrt {1 + 1 + 1} \\ \Rightarrow {m_3} = \sqrt 3 \end{array}\]
Last, we will find the magnitude of the vector \[{{\bf{a}}_4} = - \overline {\bf{i}} + \overline {3{\bf{j}}} + \overline {\bf{k}} \].
Comparing \[{{\bf{a}}_4} = - \overline {\bf{i}} + \overline {3{\bf{j}}} + \overline {\bf{k}} \] and \[{\bf{v}} = \overline {x{\bf{i}}} + \overline {y{\bf{j}}} + \overline {z{\bf{k}}} \], we get
\[x = - 1\], \[y = 3\] and \[z = 1\]
Substituting \[x = - 1\], \[y = 3\] and \[z = 1\] in the formula for magnitude of a vector, we get
\[\left| {{{\bf{a}}_4}} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {3^2} + {1^2}} \]
Simplifying the expression, we get the magnitude \[{m_4}\] as
\[\begin{array}{l} \Rightarrow {m_4} = \sqrt {1 + 9 + 1} \\ \Rightarrow {m_4} = \sqrt {11} \end{array}\]
Now, we will compare the magnitudes.
We know that \[\sqrt 3 < \sqrt 6 < \sqrt {11} < \sqrt {41} \].
Therefore, we get
\[{m_3} < {m_1} < {m_4} < {m_2}\]

Hence, the lowest magnitude is \[{m_3}\]. The correct option is option (C).

Note:
We know that a vector has both direction as well as magnitude. Magnitude is termed as the length of the vector. We can derive the formula of magnitude from the distance formula. It is important for us to remember the formula for magnitude correctly. We can make a mistake by using the formula \[\left| {\bf{v}} \right| = \sqrt {x + y + z} \], which is incorrect. Here we will calculate the magnitude of each vector separately and then compare them.