Question

If \[\vec b = 3\hat i + 4\hat j\] and \[\vec a = \hat i - \hat j\], the vector having the same magnitude as that of \[\vec b\] and parallel to \[\vec a\] is
A. \[\dfrac{5}{{\sqrt 2 }}\left( {\hat i - \hat j} \right)\]
B. \[\dfrac{5}{{\sqrt 2 }}\left( {\hat i + \hat j} \right)\]
C. \[5\left( {\hat i - \hat j} \right)\]
D. \[5\left( {\hat i + \hat j} \right)\]

Answer 1

Hint: Determine the magnitude of vector b. The unit vector of \[\hat a\] is the ratio of vector \[\vec a\] and magnitude of vector \[\vec a\]. The vector \[\vec c\] will be equal to vector magnitude of \[\vec b\] and parallel to \[\vec a\] if \[\vec c = \left| {\vec b} \right| \times {\text{unit vector of }}\hat a\].

Complete step by step answer:
We have given two vectors \[\vec b = 3\hat i + 4\hat j\] and \[\vec a = \hat i - \hat j\].Let us calculate the magnitude of vector \[\vec b\] as follows,
\[\left| {\vec b} \right| = \sqrt {{3^2} + {4^2}} \]
\[ \Rightarrow \left| {\vec b} \right| = \sqrt {25} \]
\[ \Rightarrow \left| {\vec b} \right| = 5\] …… (1)
The unit vector of \[\hat a\] is the ratio of vector \[\vec a\] and magnitude of vector \[\vec a\]. Therefore,
\[{\text{unit vector }}\hat a = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}\]
\[ \Rightarrow {\text{unit vector }}\hat a = \dfrac{{\hat i - \hat j}}{{\sqrt {{1^2} + {1^2}} }}\]
\[ \Rightarrow {\text{unit vector }}\hat a = \dfrac{{\hat i}}{{\sqrt 2 }} - \dfrac{{\hat j}}{{\sqrt 2 }}\] …… (2)

Let us calculate the vector \[\vec c\] of vector magnitude equal to \[\vec b\] and parallel to \[\vec a\] as,
\[\vec c = \left| {\vec b} \right| \times {\text{unit vector of }}\hat a\]
Using equation (1) and (2) in the above equation, we get,
\[\vec c = 5 \times \left( {\dfrac{{\hat i}}{{\sqrt 2 }} - \dfrac{{\hat j}}{{\sqrt 2 }}} \right)\]
\[ \Rightarrow \vec c = \dfrac{5}{{\sqrt 2 }}\hat i - \dfrac{5}{{\sqrt 2 }}\hat j\]
This vector has the same magnitude as that of \[\vec b\] and it will be parallel to \[\vec a\]. To verify this, let us do the following procedure.

Let us calculate the magnitude of vector \[\vec c\] as follows,
\[\left| {\vec c} \right| = \sqrt {{{\left( {\dfrac{5}{{\sqrt 2 }}} \right)}^2} + {{\left( { - \dfrac{5}{{\sqrt 2 }}} \right)}^2}} \]
\[ \Rightarrow \left| {\vec c} \right| = \sqrt {\left( {\dfrac{{25}}{2}} \right) + \left( {\dfrac{{25}}{2}} \right)} \]
\[ \Rightarrow \left| {\vec c} \right| = \sqrt {\dfrac{{50}}{2}} \]
\[ \Rightarrow \left| {\vec c} \right| = 5\]
Let us calculate the angle of \[\vec a\] as follows,
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{1}} \right)\]
\[ \Rightarrow \theta = - 45^\circ \]
Let us calculate the angle of \[\vec c\] as follows,
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - \dfrac{5}{{\sqrt 2 }}}}{{\dfrac{5}{{\sqrt 2 }}}}} \right)\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( { - 1} \right)\]
\[ \therefore \theta = - 45^\circ \]

Thus, the angle made by \[\vec a\] and angle made by \[\vec c\] is the same. Therefore, these two vectors must be the same.

Note: The unit vector has a magnitude equal to 1. You can verify it in the above solution. While determining the angle made by the vector, always take the ratio of the y-component of the vector to the x-component of the vector and take the tan-inverse of the answer.