Question

Unit vector \[\widehat{a}\] and \[\widehat{b}\] are inclined at an angle $2\theta $ and \[\left| \widehat{a}-\widehat{b} \right|<1\]. If $0\le \theta <\pi $, then $\theta $ may belong to

(a) $\left[ 0,\dfrac{\pi }{6} \right)$
(b) $\left( \dfrac{\pi }{6},\dfrac{5\pi }{6} \right)$
(c) \[\left[ \dfrac{\pi }{6},\dfrac{\pi }{2} \right]\]
(d) None of these.

Answer 1

Hint: This question is based on the unit vector’s dot product and magnitude concept. In this, we use trigonometric graphs also.
Formula of dot product –
\[\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta \] , where $\theta $is the angle between \[\overrightarrow{a}\]and \[\overrightarrow{b}\].

By using this formula, we get the value of $\left( \widehat{a}.\widehat{b} \right)$ and then by this formula
\[{{\left| \overrightarrow{a}+\overrightarrow{b} \right|}^{2}}={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+2\left| \overrightarrow{a}.\overrightarrow{b} \right|\]

We will solve inequality, where we will use the formula –
$\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
\[\cos 2\theta =2{{\cos }^{2}}-1\]

Complete step by step answer:
Here we use an inequality concept.
Let an inequality - $\left( x-\alpha \right)\left( x-\beta \right)>0$.
To solve, firstly we have to calculate critical points:
$\left( x-\alpha \right)\left( x-\beta \right)=0$
$\Rightarrow x=\alpha $ and $x=\beta $are critical points.

Let $\beta >\alpha $.
So in number line, we have:

For $x>\beta $, $\left( x-\beta \right)=positive$
$\left( x-\alpha \right)=positive$
So, $\left( x-\alpha \right)\left( x-\beta \right)=positive$

For $\alpha < x<\beta $, $\left( x-\beta \right)=negative$
$\left( x-\alpha \right)=positive$
So, $\left( x-\alpha \right)\left( x-\beta \right)=negative$

For $x<\alpha $, $\left( x-\beta \right)=negative$
$\left( x-\alpha \right)=negative$
$\left( x-\alpha \right)\left( x-\beta \right)=positive$

We have to take $\left( x-\alpha \right)\left( x-\beta \right)=positive$
So, for $x<\alpha $ and $x>\beta $
$\left( x-\alpha \right)\left( x-\beta \right)>0$

So, $x\in \left( -\infty ,\alpha \right)\cup \left( \beta ,\infty \right)$

Now, as given in the question \[\widehat{a}\] and \[\widehat{b}\]are unit vectors and inclined at an angle $2\theta $.
Then, \[\left| \overrightarrow{a} \right|=1\], \[\left| \overrightarrow{b} \right|=1\]and \[\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos 2\theta \]
\[\Rightarrow \overrightarrow{a}.\overrightarrow{b}=\cos 2\theta \]
Now, as given in the question:
\[\left| \widehat{a}-\widehat{b} \right|<1\]
As magnitude of vector is always positive, so we can square both side. So by squaring both sides,
\[{{\left| \widehat{a}-\widehat{b} \right|}^{2}}<1\]

By formula:
\[{{\left| \overrightarrow{a}+\overrightarrow{b} \right|}^{2}}={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+2\left| \overrightarrow{a}.\overrightarrow{b} \right|\]
\[\Rightarrow {{\left| \widehat{a} \right|}^{2}}+{{\left| -\widehat{b} \right|}^{2}}+2\left| \widehat{a}.\left( -\widehat{b} \right) \right|<1\]
\[\Rightarrow 1+1-2\left| \widehat{a}.\widehat{b} \right|<1\]
\[\Rightarrow 2-2\left| \widehat{a}.\widehat{b} \right|<1\]
$\because \left| \widehat{a}.\widehat{b} \right|=\cos 2\theta $
\[\Rightarrow 2-2\cos 2\theta <1\]
\[\Rightarrow 2\cos 2\theta >1\]
\[\Rightarrow \cos 2\theta >\dfrac{1}{2}\]

\[\because \cos 2\theta =2{{\cos }^{2}}-1\]
\[\Rightarrow 2{{\cos }^{2}}\theta -1>1\]
\[\Rightarrow 2{{\cos }^{2}}\theta >\dfrac{3}{2}\]
\[\Rightarrow {{\cos }^{2}}\theta >\dfrac{3}{4}\]
\[\Rightarrow {{\cos }^{2}}\theta -\dfrac{3}{4}>0\]
\[\Rightarrow {{\cos }^{2}}\theta -{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}>0\]

$\because \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
\[\Rightarrow \left( \cos \theta +\dfrac{\sqrt{3}}{2} \right)\left( \cos \theta -\dfrac{\sqrt{3}}{2} \right)>0\]

So, $\cos \in \left[ -1,\dfrac{\sqrt{3}}{2} \right)\cup \left( \dfrac{\sqrt{3}}{2},1 \right]$

Graph of $\cos \theta $

So, according to graph –
$\cos \in \left[ -1,\dfrac{\sqrt{3}}{2} \right)\cup \left( \dfrac{\sqrt{3}}{2},1 \right]$
$\Rightarrow \theta \in \left[ 0,\dfrac{\pi }{6} \right)\cup \left( \dfrac{5\pi }{6},\pi \right]$

Hence, in option (a), $\left[ 0,\dfrac{\pi }{6} \right)$ is correct.

Note:
In this question, students should take care of the inequality signs (< or >) since this can change your answer. So, during calculation especially check this sign is correct or not.