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Hint: Let us assume that two unit vectors are $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$, and the sum of both the unit vectors is also a unit vector say $\overset{\to }{\mathop{c}}\,$.
By using the vector sum formula: $\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$, get the magnitude of resultant of addition of vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$.
It is given that the magnitude of the sum of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$is also 1. So, put the value of resultant equal to 1 and get the value of $\cos \text{ }\!\!\theta\!\!\text{ }$.
Now substitute the value of $\cos \text{ }\!\!\theta\!\!\text{ }$ in the vector difference formula: $\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$ and get the magnitude of difference of vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$.
Complete step by step answer:
Let us assume that two unit vectors are $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$, and the sum of both the unit vectors is also a unit vector say $\overset{\to }{\mathop{c}}\,$.
Since, it is given that $\overset{\to }{\mathop{a}}\,$ , $\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{c}}\,$ are unit vectors. Therefore, we can say that:
$\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=1......(1)$
Now, by using vector addition formula, we get magnitude of resultant of addition of two vectors as:
$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$
So, we can write that:
$\left| \overset{\to }{\mathop{c}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }......(2)$
Since, we have $\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=1$, from equation (1), put the value in equation (2):
$\begin{align}
& \Rightarrow 1=\sqrt{1+1+2\cos \text{ }\!\!\theta\!\!\text{ }} \\
& \Rightarrow 1=2+2\cos \text{ }\!\!\theta\!\!\text{ } \\
& \Rightarrow 1=2\left( \cos \text{ }\!\!\theta\!\!\text{ }+1 \right) \\
& \Rightarrow \dfrac{1}{2}=\left( \cos \text{ }\!\!\theta\!\!\text{ }+1 \right) \\
& \Rightarrow \cos \text{ }\!\!\theta\!\!\text{ }=-\dfrac{1}{2}......(3) \\
\end{align}$
Now, we need to find the magnitude of difference of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$. Let us assume that the difference of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$is a vector $\left| \overset{\to }{\mathop{d}}\, \right|$.
By using the vector difference formula, we get the magnitude of difference of two vectors as:
$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$
So, we can write:
$\left| \overset{\to }{\mathop{d}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }......(4)$
From equation (1) and (3) we have:
$\begin{align}
& \left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=1 \\
& \cos \text{ }\!\!\theta\!\!\text{ }=-\dfrac{1}{2} \\
\end{align}$
Put the values in equation (4), we have:
$\begin{align}
& \left| \overset{\to }{\mathop{d}}\, \right|=\sqrt{1+1-2\left( -\dfrac{1}{2} \right)} \\
& =\sqrt{3}
\end{align}$
Hence the magnitude of difference of two unit vectors is $\sqrt{3}$.
So, the correct answer is “Option C”.
Note: Never assume that the vector addition or subtraction is similar to algebraic addition or subtraction. Vectors do not obey the algebraic identities. So, remember the formulae for vector addition and subtraction.
By using the vector sum formula: $\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$, get the magnitude of resultant of addition of vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$.
It is given that the magnitude of the sum of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$is also 1. So, put the value of resultant equal to 1 and get the value of $\cos \text{ }\!\!\theta\!\!\text{ }$.
Now substitute the value of $\cos \text{ }\!\!\theta\!\!\text{ }$ in the vector difference formula: $\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$ and get the magnitude of difference of vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$.
Complete step by step answer:
Let us assume that two unit vectors are $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$, and the sum of both the unit vectors is also a unit vector say $\overset{\to }{\mathop{c}}\,$.
Since, it is given that $\overset{\to }{\mathop{a}}\,$ , $\overset{\to }{\mathop{b}}\,$and $\overset{\to }{\mathop{c}}\,$ are unit vectors. Therefore, we can say that:
$\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=1......(1)$
Now, by using vector addition formula, we get magnitude of resultant of addition of two vectors as:
$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$
So, we can write that:
$\left| \overset{\to }{\mathop{c}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}+2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }......(2)$
Since, we have $\left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=\left| \overset{\to }{\mathop{c}}\, \right|=1$, from equation (1), put the value in equation (2):
$\begin{align}
& \Rightarrow 1=\sqrt{1+1+2\cos \text{ }\!\!\theta\!\!\text{ }} \\
& \Rightarrow 1=2+2\cos \text{ }\!\!\theta\!\!\text{ } \\
& \Rightarrow 1=2\left( \cos \text{ }\!\!\theta\!\!\text{ }+1 \right) \\
& \Rightarrow \dfrac{1}{2}=\left( \cos \text{ }\!\!\theta\!\!\text{ }+1 \right) \\
& \Rightarrow \cos \text{ }\!\!\theta\!\!\text{ }=-\dfrac{1}{2}......(3) \\
\end{align}$
Now, we need to find the magnitude of difference of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$. Let us assume that the difference of unit vectors $\overset{\to }{\mathop{a}}\,$ and $\overset{\to }{\mathop{b}}\,$is a vector $\left| \overset{\to }{\mathop{d}}\, \right|$.
By using the vector difference formula, we get the magnitude of difference of two vectors as:
$\left| \overset{\to }{\mathop{r}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }$
So, we can write:
$\left| \overset{\to }{\mathop{d}}\, \right|=\sqrt{{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}}+{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}-2\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|\cos \theta }......(4)$
From equation (1) and (3) we have:
$\begin{align}
& \left| \overset{\to }{\mathop{a}}\, \right|=\left| \overset{\to }{\mathop{b}}\, \right|=1 \\
& \cos \text{ }\!\!\theta\!\!\text{ }=-\dfrac{1}{2} \\
\end{align}$
Put the values in equation (4), we have:
$\begin{align}
& \left| \overset{\to }{\mathop{d}}\, \right|=\sqrt{1+1-2\left( -\dfrac{1}{2} \right)} \\
& =\sqrt{3}
\end{align}$
Hence the magnitude of difference of two unit vectors is $\sqrt{3}$.
So, the correct answer is “Option C”.
Note: Never assume that the vector addition or subtraction is similar to algebraic addition or subtraction. Vectors do not obey the algebraic identities. So, remember the formulae for vector addition and subtraction.
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