Question

How do you find the unit vector in the direction of the given vector of $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ ?

Answer 1

Hint: We start solving the problem by making use of the result that the magnitude of the vector $ a\widehat{i}+b\widehat{j}+c\widehat{k} $ is $ \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} $ to find the magnitude of the given vector $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ . We then make use of the fact that the unit vector in the direction of the vector $ \overrightarrow{x}=a\widehat{i}+b\widehat{j}+c\widehat{k} $ is defined as $ \dfrac{\overrightarrow{x}}{\left| \overrightarrow{x} \right|} $ . We use this result to find the required vector of the given vector $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ .

Complete step by step answer:
According to the problem, we are asked to find the unit vector in the direction of the given vector of $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ .
Let us find the magnitude of the given vector $ \overrightarrow{\omega } $ .
We know that the magnitude of the vector $ a\widehat{i}+b\widehat{j}+c\widehat{k} $ is $ \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} $ . Let us use this result to find the magnitude of the given vector $ \overrightarrow{\omega } $ .
So, the magnitude of the given vector $ \overrightarrow{\omega } $ is $ \left| \overrightarrow{\omega } \right|=\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{1+4}=\sqrt{5} $ ---(1).
We know that the unit vector in the direction of the vector $ \overrightarrow{x}=a\widehat{i}+b\widehat{j}+c\widehat{k} $ is defined as $ \dfrac{\overrightarrow{x}}{\left| \overrightarrow{x} \right|} $ . Using this result, we get the unit vector in the direction of vector $ \overrightarrow{\omega } $ as $ \dfrac{\overrightarrow{\omega }}{\left| \overrightarrow{\omega } \right|} $ .
From equation (1), we get the unit vector as $ \dfrac{\overrightarrow{\omega }}{\left| \overrightarrow{\omega } \right|}=\dfrac{\widehat{i}-2\widehat{j}}{\sqrt{5}} $ .
$ \therefore $ We have found the unit vector in the direction of given vector $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ as $ \dfrac{\widehat{i}-2\widehat{j}}{\sqrt{5}} $ .

Note:
We can verify the obtained result by finding the magnitude of the obtained vector as the magnitude of the unit vectors is 1 and the angle of the obtained unit vector with the given vector should be 0 as both are in the same direction. We can also solve the problem by making use of the fact that the vector in the direction of vector $ \overrightarrow{x} $ is $ \lambda \overrightarrow{x} $ and then equating its magnitude to 1. Similarly, we can expect problems to find the unit vector opposite to the given vector $ \overrightarrow{\omega }=\widehat{i}-2\widehat{j} $ .