Question
If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0},|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5,|\vec{\text{c}}|=7,$then the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is:
  1. $\frac{\pi}{6}$
  2. $\frac{2\pi}{3}$
  3. $\frac{5\pi}{3}$
  4. $\frac{\pi}{3}$

Answer

  1. $\frac{\pi}{3}$

Solution:

Given, $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5$ and $|\vec{\text{c}}|=7\dots(1)$

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$

Given that

$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$

$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$

$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=|-\vec{\text{c}}|^2$

$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2$

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2-|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=7^2-3^3-5^2$ [using (1)]

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=15$

$\Rightarrow2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=15$

$\Rightarrow2(3)(5)\cos\theta=15$ [using (1)]

$\Rightarrow\cos\theta=\frac{1}{2}$

$\therefore\theta=\frac{\pi}{3}$

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