Questions & Answers

Question

Answers

(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) $\left( {\dfrac{\pi }{4},\dfrac{{3\pi }}{4}} \right)$

Answer
Verified

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

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.