Question

If $ \overrightarrow{a} $ is a non-zero vector of magnitude $ a $ and \[\lambda \overrightarrow{a}\] is a unit vector, then find the value of $ \lambda . $

Answer 1

Hint: If we have a unit vector $ \widehat{b} $ then it’s value is given by the following expression $ \widehat{b}=\dfrac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|} $ .
where $ \left| \overrightarrow{b} \right| $ is the magnitude of $ \overrightarrow{b} $ ,
Using the above equation and information given in the question we can find out the value of $ \lambda $ .

Complete step-by-step answer:
As we know value of a unit vector $ \widehat{a} $ is given by,
 $ \widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}......(1) $
Where $ \left| \overrightarrow{a} \right| $ is the magnitude of $ \overrightarrow{a} $ .
For example, suppose
 $ \overrightarrow{a}=2\widehat{i}+2\overset\frown{j}+\widehat{k} $
 where $ \overset\frown{i},\overset\frown{j},\overset\frown{k} $ are the unit vectors in the direction $ x,y\,and\,z $ respectively
so, to find the magnitude of \[\overrightarrow{a}\] i.e. $ \left| \overrightarrow{a} \right| $ we need to add the individual squares of the coefficients of the unit vectors and take the square root of the whole expression so, $ \left| \overrightarrow{a} \right| $ would be equal to $ \sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}} $
hence, $ \left| \overrightarrow{a} \right|=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}=\sqrt{4+4+1}=\sqrt{9}=3 $ and hence,
 $ \widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}=\dfrac{2\widehat{i}+2\overset\frown{j}+\widehat{k}}{3} $
Now, moving back to the given question,
We are given that:
 $ \widehat{a}=\lambda \overrightarrow{a}......(2) $
Comparing and equating both the equations (1) and (2) we get,
 $ \lambda \widehat{a}=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|} $
Now, by dividing both sides by $ \overrightarrow{a} $ , we get
 $ \lambda =\dfrac{1}{\left| \overrightarrow{a} \right|} $
Hence value of $ \lambda $ is equal to $ \lambda =\dfrac{1}{\left| \overrightarrow{a} \right|} $ .

Note: To solve this problem you need to know the basics of vectors and unit vectors.
You should have a good grasp over what is a unit vector and what is it’s magnitude which is understood by its name (i.e. unit) that is 1. Always remember the following equation of unit vectors for future references:
 If we are given a vector $ \overrightarrow{b} $ , then suppose \[\overset\frown{b}\] is the unit vector $ \overrightarrow{b} $ and it’s value will be equal to $ \dfrac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|} $ , i.e. $ \widehat{b}=\dfrac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|} $ where $ \left| \overrightarrow{b} \right| $ is the magnitude of $ \overrightarrow{b} $ .