Question

If $\left| a \right|=\left| b \right|=\left| a+b \right|=1$, then the value of $\left| a-b \right|$ should be equal to
a) 1
b) $\sqrt{2}$
c) $\sqrt{3}$
d) None of these

Answer 1

Hint: In this question, we are given the modulus of a, b and a+b and we have to find the modulus of a-b. Therefore, we should try to use the formula for modulus of a sum in terms of the modulus of the individual objects and then obtain sufficient information to calculate the modulus of a-b.

Complete step-by-step answer:

We know that the magnitude of the sum of two objects is given by
$\left| a+b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+2\left| a \right|\left| b \right|\cos (\theta )}.............(1.1)$
Where $\cos (\theta )$ is the angle between a and b.
Here, it is given that $\left| a \right|=\left| b \right|=\left| a+b \right|=1$. Using this in equation (1.1), we get
$\begin{aligned} & 1=\sqrt{1+1+2\times 1\times 1\cos (\theta )} \\\ & \Rightarrow 1=2(1+\cos (\theta ))\Rightarrow \cos (\theta )=\dfrac{1}{2}-1=\dfrac{-1}{2}..............(1.2) \\\ \end{aligned}$
Also, we know that the magnitude of the difference of two objects is given by
$\left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\left| b \right|\cos (\theta )}.............(1.3)$
Thus, by using the values given in the question $\left| a \right|=\left| b \right|=\left| a+b \right|=1$, and using the value of $\cos (\theta )$ from equation(1.3), we obtain
$\begin{aligned} & \left| a-b \right|=\sqrt{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}-2\left| a \right|\left| b \right|\cos (\theta )} \\\ & =\sqrt{1+1-2\times 1\times 1\cos (\theta )}=\sqrt{2-2\times \left( \dfrac{-1}{2} \right)} \\\ & =\sqrt{2+1}=\sqrt{3} \\\ \end{aligned}$
Thus, we obtain the answer to the given question as
$\left| a-b \right|=\sqrt{3}$
Which matches option (c) of the question. Hence, option (c) is the correct answer to this question.

Note: We note that in this case, $\left| a-b \right|$ is greater than the value of $\left| a+b \right|$. One might wonder why the magnitude was less when a and b are added rather than when they were subtracted. This is because the obtained value of $\cos (\theta )$ was negative and hence a and b were towards opposite directions. Therefore., the magnitude was more when they were subtracted rather than when they were added.