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**Hint:**For this question we have to know that the $\left| {a + b} \right| = \sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} + 2\left| a \right|\left| b \right|\cos \alpha } $ on squaring on both side we get ${\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right|\cos \alpha $ and use of $\left| a \right| = 1$ , $\left| b \right| = 1$ and $\left| {a + b} \right| < 1$ we will find the range of the $\alpha $.

**Complete step-by-step answer:**

It is given that in the question that $\left| {a + b} \right| < 1$ and if we square on both side we get ;

${\left| {a + b} \right|^2} < 1$

Now we know that from the vectors property $\left| {a + b} \right| = \sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} + 2\left| a \right|\left| b \right|\cos \alpha } $

Where $a$ and $b$ are the unit vectors and $\alpha $ is the angle between them .

Now that the for the unit vectors $\left| a \right| = 1$ , $\left| b \right| = 1$

On Squaring the above equation became : ${\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right|\cos \alpha $

Put the values $\left| a \right| = 1$ $\left| b \right| = 1$

${\left| {a + b} \right|^2} = {1^2} + {1^2} + 2\cos \alpha $

From above it is proved that

${\left| {a + b} \right|^2} < 1$

$\left| {a + b} \right| < 1$

hence ${\left| {a + b} \right|^2} = {1^2} + {1^2} + 2\cos \alpha $

$2 + 2\cos \alpha < 1$

$2\cos \alpha < - 1$

$\cos \alpha < - \dfrac{1}{2}$

Hence the

$\alpha \in \left[ {\dfrac{{2\pi }}{3},\pi } \right]$

It is given that the

\[\alpha \in \left[ {0,\pi } \right]\]

Now we know that the cosine is negative in the second quadrant and by taking the intersection from the given value we will find out that $\alpha \in \left[ {\dfrac{{2\pi }}{3},\pi } \right]$

**So, the correct answer is “Option B”.**

**Note:**Always be careful on solving the trigonometric inequalities . If you have any problem in solving it then try to make the graph of the given trigonometric function . So it will become easier to solve .

The dot product tells you what amount of one vector goes in the direction of another .

Dot product of the perpendicular vector is always zero .

Cross product always gives the vector which is perpendicular to the both vectors and its direction would be determined by Right hand rule.

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