Q73 / 160 · TS EAMCET 2016 · Mathematics
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors satisfying the relation $\mathbf{a}+\mathbf{b}+\sqrt{3} \mathbf{c}=0$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is
- A$\frac{\pi}{6}$
- B$\frac{\pi}{4}$
- Correct answerC$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
Solution
Given, $\mathbf{a}+\mathbf{b}+\sqrt{3} \mathbf{c}=0$
$\begin{array}{rlrl}<br />\Rightarrow & a+b & =-\sqrt{3} \mathbf{c} \\<br />\Rightarrow & (a+b) \cdot(a+b) & =(-\sqrt{3} \mathbf{c})(-\sqrt{3} \mathbf{c}) \\<br />\Rightarrow & |a|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=3|\mathbf{c}|^2<br />\end{array}$
Since, $|\mathbf{a}|,|\mathbf{b}|$ and $|\mathbf{c}|$ are unit vectors.
$\begin{array}{rrr}<br />\therefore & (1)^2+2(1)(1) \cos \theta+1=3(1) \\<br />\Rightarrow & 1+2 \cos \theta+1=3 \\<br />\Rightarrow & 2 \cos \theta=3-2<br />\end{array}$
$\begin{array}{rlrl}<br />\Rightarrow & 2 \cos \theta & =1 \\<br />\Rightarrow & & \cos \theta & =\frac{1}{2}=\cos \frac{\pi}{3} \\<br />\Rightarrow & \theta & =\frac{\pi}{3}<br />\end{array}$
$\begin{array}{rlrl}<br />\Rightarrow & a+b & =-\sqrt{3} \mathbf{c} \\<br />\Rightarrow & (a+b) \cdot(a+b) & =(-\sqrt{3} \mathbf{c})(-\sqrt{3} \mathbf{c}) \\<br />\Rightarrow & |a|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=3|\mathbf{c}|^2<br />\end{array}$
Since, $|\mathbf{a}|,|\mathbf{b}|$ and $|\mathbf{c}|$ are unit vectors.
$\begin{array}{rrr}<br />\therefore & (1)^2+2(1)(1) \cos \theta+1=3(1) \\<br />\Rightarrow & 1+2 \cos \theta+1=3 \\<br />\Rightarrow & 2 \cos \theta=3-2<br />\end{array}$
$\begin{array}{rlrl}<br />\Rightarrow & 2 \cos \theta & =1 \\<br />\Rightarrow & & \cos \theta & =\frac{1}{2}=\cos \frac{\pi}{3} \\<br />\Rightarrow & \theta & =\frac{\pi}{3}<br />\end{array}$
Hence, the correct answer is option: 'C'.