Question

If $\left| \overrightarrow{a} \right|=1,\,\,\left| \overrightarrow{b} \right|=4,\,\,a.b=2$ and $\overrightarrow{c}=(2\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b}$ then the angle between b and c is?
A. $\dfrac{\pi }{6}$
B. $\dfrac{5\pi }{6}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{2\pi }{3}$

Answer 1

Hint : In this particular problem we have to find the angle between b and c for that first of all we are finding the angle between a and b using dot product that is $\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\theta )$ and then substitute the $\theta$ in the formula to get $\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\theta )$ then by simplifying this we have found the angle between b and c.

Complete step-by-step answer :
In this particular problem it is given that $\left| \overrightarrow{a} \right|=1,\,\,\left| \overrightarrow{b} \right|=4,\,\,a.b=2$ and $\overrightarrow{c}=(2\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b}$
So, first of all we have to find the angle between a as well as b that is formula is given below.
$\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\theta )$
So, we have to substitute the value of $\left| \overrightarrow{a} \right|=1,\,\,\left| \overrightarrow{b} \right|=4,\,\,a.b=2$ in the above formula then we get:
$2=1\times 4\times \cos (\theta )$
After substituting the values simplify further and solve we get the value of $\cos (\theta )$
$\dfrac{2}{4}=\cos (\theta )$
After simplifying further we get
$\therefore \dfrac{1}{2}=\cos (\theta )$
Further solving and we also know that ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{60}^{\circ }}$
$\therefore \theta ={{60}^{\circ }}$ (Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ )
Now, to find the value of $\overrightarrow{a}\times \overrightarrow{b}$ we apply the same value of $\theta$ which we get in the above values.
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\theta )$
So, we have to substitute the value of $\left| \overrightarrow{a} \right|=1,\,\,\left| \overrightarrow{b} \right|=4,\,\,a.b=2$ and $\theta ={{60}^{\circ }}$ in the above formula then we get:
$\overrightarrow{a}\times \overrightarrow{b}=1\times 4\times \sin ({{60}^{\circ }})$
As we know that $\sin ({{60}^{\circ }})=\dfrac{\sqrt{3}}{2}$ substitute the this above equation
$\overrightarrow{a}\times \overrightarrow{b}=4\times \dfrac{\sqrt{3}}{2}$
After simplifying further we get:
$\overrightarrow{a}\times \overrightarrow{b}=2\sqrt{3}$
Now we have to find the value of $\overrightarrow{c}$ for that we need to square it and by further simplification we get:
$\overrightarrow{c}=(2\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b}$
${{\left| \overrightarrow{c} \right|}^{2}}={{\left[ (2\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b} \right] }^{2}}$
By applying the property ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ on above equation we get:
${{\left| \overrightarrow{c} \right|}^{2}}=4{{\left| \overrightarrow{a}\times \overrightarrow{b} \right|}^{2}}+9{{\left| \overrightarrow{b} \right|}^{2}}-2\times \left( 3\overrightarrow{b}\left( 2\overrightarrow{a}\times \overrightarrow{b} \right) \right)$
Further solving we get:
${{\left| \overrightarrow{c} \right|}^{2}}=4{{\left| \overrightarrow{a}\times \overrightarrow{b} \right|}^{2}}+9{{\left| \overrightarrow{b} \right|}^{2}}-0$
( $\overrightarrow{b},\,\,2\overrightarrow{a}\times \overrightarrow{b}$ Are perpendicular to each other)
${{\left| \overrightarrow{c} \right|}^{2}}=4{{\left( 2\sqrt{3} \right)}^{2}}+9{{\left( 4 \right)}^{2}}$
After solving this above equation we get:
${{\left| \overrightarrow{c} \right|}^{2}}=4\times 12+9\times 16=48+144=192$
We get the value of $\left| \overrightarrow{c} \right|$
$\left| \overrightarrow{c} \right|=8\sqrt{3}$
Now, we have to find the value of $\overrightarrow{c}.\overrightarrow{b}$ for that multiply $\overrightarrow{b}$ on both sides on this equation $\overrightarrow{c}=(2\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b}$
After substituting this we get:
$\overrightarrow{c}.\overrightarrow{b}=\left( 2\overrightarrow{a}\times \overrightarrow{b} \right).\overrightarrow{b}-3\overrightarrow{b}.\overrightarrow{b}$
$\left( \overrightarrow{b}.\left( 2\overrightarrow{a}\times \overrightarrow{b} \right)\,\,and\,\,\cos ({{90}^{\circ }})=0 \right)$
$\overrightarrow{c}.\overrightarrow{b}=-3{{\left| \overrightarrow{b} \right|}^{2}}$
Now, finally we have to find the angle between b and c formula is given below:
$\left| \overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos (\alpha )=\overrightarrow{c}.\overrightarrow{b}$
Substitute the value of $\overrightarrow{c}.\overrightarrow{b}=-3{{\left| \overrightarrow{b} \right|}^{2}}$ in above equation (1) we get:
$\left| \overrightarrow{c} \right|\left| \overrightarrow{b} \right|\cos (\alpha )=-3{{\left| \overrightarrow{b} \right|}^{2}}$
After simplifying this we get:
$\left| \overrightarrow{c} \right|\cos (\alpha )=-3\left| \overrightarrow{b} \right|$
Further solving we get:
$\cos (\alpha )=\dfrac{-3\left| \overrightarrow{b} \right|}{\left| \overrightarrow{c} \right|}$
After substituting the value of $\left| \overrightarrow{c} \right|=8\sqrt{3}$ as well as $\,\left| \overrightarrow{b} \right|=4$ on above equation we get:
$\cos (\alpha )=\dfrac{-3\times 4}{8\sqrt{3}}=\dfrac{-\sqrt{3}}{2}$
After simplifying this we get
$\alpha =\dfrac{5\pi }{6}$ (Because we know that $\cos \left( \dfrac{5\pi }{6} \right)=\dfrac{-\sqrt{3}}{2}$ )
So, the correct answer is “Option B”.

Note : In this type of problems always remember which formula is used and ready the question, what is asked and what is given based on the we have to apply the formula correctly. Concept of vectors should be strong. Angle between $\overrightarrow{b}$ and $\left( 2\overrightarrow{a}\times \overrightarrow{b} \right)$ is $\dfrac{\pi }{2}$ because, both are perpendicular to each other but $\cos \left( \dfrac{\pi }{2} \right)=0$ that’s why $\overrightarrow{b}\left( 2\overrightarrow{a}\times \overrightarrow{b} \right)=0$ . Here, we get the value of $\cos (\alpha )$ is negative. We must be aware of the sign in each quadrant that is 1st quadrant is all positive 2nd quadrant is only sine and cosecant are positive and remaining are negative. In the 3rd quadrant only tangent and cotangent functions are positive and remaining are negative and in the 4th quadrant only cosine and secant functions are positive. Here, in this problem $\cos (\alpha )$ is negative, which means it lies on the 2nd quadrant. To find the angle that is $\alpha$ can be calculated as $\left( \pi -\dfrac{\pi }{6} \right)$ that is $\dfrac{5\pi }{6}$ because, as we know that ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}$ therefore, we have to subtract $\dfrac{\pi }{6}$ from $\pi$ hence, we get the value of $\alpha$ .