Q65 / 150 · MHT CET 2007 · Mathematics
If $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}},|\overrightarrow{\mathbf{a}}|=3,|\overrightarrow{\mathbf{b}}|=5,|\overrightarrow{\mathbf{c}}|=7$, then
the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is
- A$\pi / 6$
- B$2 \pi / 3$
- C$5 \pi / 3$
- Correct answerD$\pi / 3$
Solution
Given, $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$
$$
\begin{array}{ll}
\therefore \quad & \overrightarrow{\mathbf{c}}=-(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \\
\Rightarrow & |\overrightarrow{\mathbf{c}}|^{2}=\overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{c}}=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow \quad|\overrightarrow{\mathbf{c}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{c}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta \\
\Rightarrow \quad 49=9+25+2 \times 3 \times 5 \cos \theta \\
\Rightarrow \quad \quad 15=30 \cos \theta \quad \Rightarrow \quad \cos \theta=\frac{1}{2} \\
\Rightarrow \quad \theta=\frac{\pi}{3}
\end{array}
$$
Hence, the correct answer is option: 'D'.