Question

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=5,\left|\overrightarrow{c}\right|=7,$ then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

$\frac{\pi }{3}$

Let $\theta$ be the angle between $a$ and $b$

$\therefore a.b=\left|a\right|\left|b\right|\cos \theta$

$\therefore \cos \theta =\frac{a.b}{\left|a\right|\left|b\right|}=\frac{a.b}{\left(3\right)\left(5\right)}=\frac{a.b}{15}$ ...(1)

Now $a+b+c=0\Rightarrow a+b=-c$

$\Rightarrow |a+b|=|-c|=\left|c\right|\Rightarrow {|a+b|}^2={\left|c\right|}^2\Rightarrow (a+b).(a+b)={\left(7\right)}^2\Rightarrow {\left|a\right|}^2+2a.b+{\left|b\right|}^2=49\Rightarrow {\left(3\right)}^2+2a.b+{\left(5\right)}^2=49$

$\Rightarrow 2a.b=49-9-25=15\Rightarrow a.b=\frac{15}{2}$

$\cos \theta =\frac{15}{2\times 15}=\frac{1}{2}=\cos {60}^0\Rightarrow \theta ={60}^0$