Question

If a and b are unit vectors and $\theta$ is the angle between them, then a-b will be a unit vector if $\theta$ =
(A). $\dfrac{\pi }{4}$
(B). $\dfrac{\pi }{3}$
(C). $\dfrac{\pi }{6}$
(D). $\dfrac{\pi }{2}$

Answer 1

Hint: In this question, we are given that a and b are unit vectors, therefore their magnitudes should be equal to 1. Now, if a-b is also a unit vector, it means that the magnitude of a-b should also be equal to one. We should, therefore, try to understand the method of finding out the magnitude of a vector through dot product and then the information is given in the question to obtain three equations which we can solve to get the value of $\theta$.

Complete step-by-step solution -
We are given that a and b are unit vectors, therefore their magnitudes should be equal to one. Therefore,
$\left| a \right|=1................(1.1)$
$\left| b \right|=1....................(1.2)$
We know that the dot product of two vectors x and y having an angle $\theta$ between them is given by
$x.y=\left| x \right|\left| y \right|\cos \theta ..........................(1.3)$
Also, we know that the magnitude of a vector is given by
$\left| x \right|=\sqrt{x.x}............(1.4)$
Therefore, using equation (1.3) and (1.4), the magnitude of a-b should be given by
$\begin{aligned} & \left| a-b \right|=\sqrt{\left( a-b \right).\left( a-b \right)}=\sqrt{a.a-a.b-b.a+b.b} \\\ & =\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\left| b \right|\cos \theta } \\\ \end{aligned}$
Using $\left| a \right|=1,\text{ }\left| b \right|=1$ from equation (1.1) and (1.2), we can write this as
$\begin{aligned} & \left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\left| b \right|\cos \theta }=\sqrt{1+1-2\times 1\times 1\times \cos \theta } \\\ & \Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta } \\\ \end{aligned}$
However, as it is also given that $\left| a-b \right|$ is a unit vector, $\left| a-b \right|=1$. Therefore, using this value in the above equation, we get
$\begin{aligned} & 1=\sqrt{2-2\cos \theta } \\\ & \Rightarrow 2\left( 1-\cos \theta \right)={{1}^{2}}=1 \\\ & \Rightarrow 1-\cos \theta =\dfrac{1}{2}\Rightarrow \cos \theta =1-\dfrac{1}{2}=\dfrac{1}{2}........(1.5) \\\ \end{aligned}$
However, we know that $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$. Therefore, using this value in equation (1.5), we get
$\cos \theta =\cos \left( \dfrac{\pi }{3} \right)\Rightarrow \theta =\dfrac{\pi }{3}...........(1.6)$
Which matches option (b) in the question. Thus, option (b) is the correct answer to this question.

Note: We should note that in equation (1.6), we equated $\theta$ to $\dfrac{\pi }{3}$ because the cosine of both these angles was the same. However, in the general case, as the cosine function is even function and has a periodicity of $2\pi$, the general value of $\theta$ should be $2n\pi \pm \dfrac{\pi }{3}$. However, increasing an angle by $2\pi$ returns the vectors to their original position and as we are just asked about the angle and not its orientation, a positive or negative angle represents the same thing. Thus, equating the angle $\theta$ to $\dfrac{\pi }{3}$ in equation (1.6) is justified.