- $\frac{\pi}{6}$
- $\frac{2\pi}{3}$
- $\frac{5\pi}{3}$
- $\frac{\pi}{3}$
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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Statement $2$ : A function $f : R \to R$ is discontinuous at $x_0$ if and only if, $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right)$ exists and $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right) \ne f\left( {{x_0}} \right)$
E(X2)
E(X2) + (E(X))2
E(X2) - (E(X))2
$\sqrt{\text{E}(\text{X}^2)-(\text{E}(\text{X}))^2}$