Question

How do you find the magnitude and direction angle of the vector \[v=3\widehat{i}-4\widehat{j}\]?

Answer 1

Hint: We need to understand the representation of a vector in the different forms in order to find the magnitude of the vector and the direction angle or the cosines of the vector from the given form which can be used here to solve this problem easily.

Complete Solution:
We are given a vector whose magnitude and direction angles have to be found. The magnitude of a vector is the quantity independent of the direction of the vector. It gives the idea on how large or small the vector is. The magnitude along with the direction can only explain the complete form of the vector.

The magnitude of a vector is often found by using the trivial method of finding the square root of the sum of the squares of the components of the vector along the different coordinates. i.e., for given vector \[\overrightarrow{A}=a\widehat{i}+b\widehat{j}+c\widehat{k}\], the magnitude of vector A is given as –
\[\begin{align}
  & \overrightarrow{A}=a\widehat{i}+b\widehat{j}+c\widehat{k} \\
 & \therefore \left| \overrightarrow{A} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\
\end{align}\]

Similarly, for the given vector, we can find the magnitude of the vector as –
\[\begin{align}
  & \overrightarrow{A}=3\widehat{i}-4\widehat{j} \\
 & \left| \overrightarrow{A} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\
 & \Rightarrow \left| \overrightarrow{A} \right|=\sqrt{{{3}^{2}}+{{(-4)}^{2}}} \\
 & \therefore \left| \overrightarrow{A} \right|=5 \\
\end{align}\]

Now, we can find the direction angles of the vector by find the tangent of the vector as –
\[\begin{align}
  & \tan \theta =\dfrac{b}{a} \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{-4}{3} \right) \\
 & \therefore \theta =-{{53.28}^{0}} \\
\end{align}\]

The direction angle of the vector is as found. So, we have found the magnitude and the direction angle of the given vector A as follows –
\[\left| \overrightarrow{A} \right|=5\]
And,
\[\theta =-{{53.28}^{0}}\]
This is the required solution for this problem.

Note:
When we find the magnitude of a vector using the method we have used, we do not consider the sign of the number. The magnitude should be the quantity and the sign shouldn’t give us a wrong conclusion as the direction angle is used to determine the direction.