Question

# If a and b are unit vectors inclined at an angle $\alpha ,\alpha \in \left[ {0,\pi } \right]$ to each other and $\left| {a + b} \right| < 1$ . Then $\alpha$ belongs to (a) $\left( {\dfrac{\pi }{3},\dfrac{{2\pi }}{3}} \right)$ (b) $\left( {\dfrac{{2\pi }}{3},\pi } \right)$ (c) $\left( {0,\dfrac{\pi }{3}} \right)$ (d) $\left( {\dfrac{\pi }{4},\dfrac{{3\pi }}{4}} \right)$

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 $\alpha$.

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.