Q74 / 160 · TS EAMCET 2006 · Mathematics
If $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$ and $|\overrightarrow{\mathbf{a}}|=3,|\overrightarrow{\mathbf{b}}|=4$ and $|\overrightarrow{\mathbf{c}}|=\sqrt{37}$, then the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is :
- A$\frac{\pi}{4}$
- B$\frac{\pi}{2}$
- C$\frac{\pi}{6}$
- Correct answerD$\frac{\pi}{3}$
Solution
$\because \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0$
$\Rightarrow \quad \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=-\overrightarrow{\mathbf{c}}$
$\Rightarrow \quad(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})^2=(-\overrightarrow{\mathbf{c}})^2$
$\Rightarrow \quad|\overrightarrow{\mathbf{a}}|^2+|\overrightarrow{\mathbf{b}}|^2+2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad 9+16+2 \cdot 3 \cdot 4 \cos \theta=37$
$\Rightarrow \quad 24 \cos \theta=37-25$
$\Rightarrow \quad \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$
Hence, the correct answer is option: 'D'.