Answer

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Hint: We can use the formula for modulus of vectors as $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$ to solve this question. Also the modulus of a unit vector is 1.

Complete step-by-step answer:

Before proceeding to the solution we should know that the magnitude is also called the modulus or the length of the vector. Magnitude is represented by the length of the directed line segment. A unit vector is a vector of length 1. To obtain a unit vector in the direction of any vector we divide by its modulus. A unit vector is a vector of unit length, sometimes also called a direction vector. The unit vector is defined by \[\widehat{v}\].

\[\widehat{v}=\dfrac{v}{\left| v \right|}\],

Given, $\left| a+b \right|=1$

Using the formula $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ we get:

Since, a and b are unit vectors therefore, $\left| a \right|=1,\left| b \right|=1$ and $\cos \theta $ is the angle between the vectors a and b.

So, we can substitute the values in the above equation and we will get,

$\therefore 1=\sqrt{{{1}^{2}}+{{1}^{2}}+2\times 1\times 1\times \cos \theta }$

$\Rightarrow 1=\sqrt{2+2\cos \theta }$

Squaring both sides we get,

$\Rightarrow 2+2\cos \theta =1$

$\Rightarrow 2\cos \theta =-1$

$\Rightarrow \cos \theta =-\dfrac{1}{2}......(i)$

Now, we have to find out the value of $\left| a-b \right|$ therefore, we can use the formula for $\left| a-b \right|$.

Since, we know the values, we can substitute them in the equation and we will get,

$\therefore \left| a-b \right|=\sqrt{1+1-2\times 1\times 1\times \cos \theta }$

$\Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta }$

Then, we can substitute the value of$\cos \theta $ from equation (i), and we will get,

$\Rightarrow \left| a-b \right|=\sqrt{2-2\times -\dfrac{1}{2}}$

$\therefore \left| a-b \right|=\sqrt{3}$

Hence, the answer is option (d).

Note: It is important to use the cos theta term in the modulus equation. The dot product of two vectors along with the angle between them by cos theta should be used. If not, it would not be easy to solve the question. Be very carefully about the value of modulus, modulus can never be negative. Always remember the formula of modulus like:

$\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$, and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$

Complete step-by-step answer:

Before proceeding to the solution we should know that the magnitude is also called the modulus or the length of the vector. Magnitude is represented by the length of the directed line segment. A unit vector is a vector of length 1. To obtain a unit vector in the direction of any vector we divide by its modulus. A unit vector is a vector of unit length, sometimes also called a direction vector. The unit vector is defined by \[\widehat{v}\].

\[\widehat{v}=\dfrac{v}{\left| v \right|}\],

Given, $\left| a+b \right|=1$

Using the formula $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ we get:

Since, a and b are unit vectors therefore, $\left| a \right|=1,\left| b \right|=1$ and $\cos \theta $ is the angle between the vectors a and b.

So, we can substitute the values in the above equation and we will get,

$\therefore 1=\sqrt{{{1}^{2}}+{{1}^{2}}+2\times 1\times 1\times \cos \theta }$

$\Rightarrow 1=\sqrt{2+2\cos \theta }$

Squaring both sides we get,

$\Rightarrow 2+2\cos \theta =1$

$\Rightarrow 2\cos \theta =-1$

$\Rightarrow \cos \theta =-\dfrac{1}{2}......(i)$

Now, we have to find out the value of $\left| a-b \right|$ therefore, we can use the formula for $\left| a-b \right|$.

Since, we know the values, we can substitute them in the equation and we will get,

$\therefore \left| a-b \right|=\sqrt{1+1-2\times 1\times 1\times \cos \theta }$

$\Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta }$

Then, we can substitute the value of$\cos \theta $ from equation (i), and we will get,

$\Rightarrow \left| a-b \right|=\sqrt{2-2\times -\dfrac{1}{2}}$

$\therefore \left| a-b \right|=\sqrt{3}$

Hence, the answer is option (d).

Note: It is important to use the cos theta term in the modulus equation. The dot product of two vectors along with the angle between them by cos theta should be used. If not, it would not be easy to solve the question. Be very carefully about the value of modulus, modulus can never be negative. Always remember the formula of modulus like:

$\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$, and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$

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