Question

Find the vector of magnitude \[\left( {\dfrac{5}{2}} \right)\]units which is parallel to the vector \[3\widehat i + 4\widehat j\]

Answer 1

Hint: We use the formula of the resultant vector in the direction of the given vector using unit vectors. Calculate the value of magnitude of a given vector using the formula of magnitude and write the unit vector. Multiply the given magnitude to the unit vector obtained.
* Unit vector is given by dividing the vector by its magnitude. Let \[\overrightarrow m = a\hat i + b\hat j + c\hat k\]be a vector then its magnitude is given by \[\left| {\overrightarrow m } \right| = \sqrt {{a^2} + {b^2} + {c^2}} \]. And unit vector becomes \[\hat m = \dfrac{{a\hat i + b\hat j + c\hat k}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
* Two vectors are said to be parallel to each other if they are scalar multiples of each other.

Complete step by step solution:
We are given the vector \[3\widehat i + 4\widehat j\]
We find the unit vector associated with this vector by calculating its magnitude.
Let \[\overrightarrow p = 3\widehat i + 4\widehat j\]..................… (1)
Use the formula of magnitude of a vector
\[ \Rightarrow \left| {\overrightarrow p } \right| = \sqrt {{3^2} + {4^2}} \]
Square the terms under the square root
\[ \Rightarrow \left| {\overrightarrow p } \right| = \sqrt {9 + 16} \]
\[ \Rightarrow \left| {\overrightarrow p } \right| = \sqrt {25} \]
Since \[25 = {5^2}\]
\[ \Rightarrow \left| {\overrightarrow p } \right| = \sqrt {{5^2}} \]
Cancel square root by square power in RHS
\[ \Rightarrow \left| {\overrightarrow p } \right| = 5\]....................… (2)
Since we know unit vector is obtained by dividing the vector by its magnitude
\[ \Rightarrow \hat p = \dfrac{{\overrightarrow p }}{{\left| {\overrightarrow p } \right|}}\]
Substitute the values of vector and magnitude of vector from equations (1) and (2) in the formula
\[ \Rightarrow \hat p = \dfrac{{3\hat i + 4\hat j}}{5}\].............… (3)
We know that two vectors are parallel to each other if they are scalar multiples of each other. Here we have to find a vector of magnitude \[\left( {\dfrac{5}{2}} \right)\]units which is parallel to the vector \[3\widehat i + 4\widehat j\]
So, we multiply the unit vector in the direction with the scalar magnitude \[\left( {\dfrac{5}{2}} \right)\]
\[ \Rightarrow \]The vector of magnitude \[\left( {\dfrac{5}{2}} \right)\]units which is parallel to the vector \[3\widehat i + 4\widehat j = \dfrac{5}{2} \times \dfrac{{3\hat i + 4\hat j}}{5}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \]The vector of magnitude \[\left( {\dfrac{5}{2}} \right)\]units which is parallel to the vector \[3\widehat i + 4\widehat j = \dfrac{{3\hat i + 4\hat j}}{2}\]

\[\therefore \]The vector of magnitude \[\left( {\dfrac{5}{2}} \right)\] units which is parallel to the vector \[3\widehat i + 4\widehat j\] is \[\dfrac{{3\hat i + 4\hat j}}{2}\]

Note: Many students make mistake of calculating the parallel vector by just multiplying the magnitude given to the given vector as they think the direction will be given by the given vector and we just need to multiply the magnitude, which is wrong because on calculation of this vector we will see the magnitude will not be equal to \[\left( {\dfrac{5}{2}} \right)\] which contradicts our requirement.