MCQ
Angle between vectors $\bar{a}$ and $\bar{b}$, where $\bar{a}, \bar{b}$ and $\overline{ c }$ are unit vectors satisfying $\bar{a}+\bar{b}+\sqrt{3} \bar{c}=\overline{0}$, is
- A$\frac{\pi}{6}$
- B$\frac{\pi}{4}$
- ✓$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
✓
Answer
Correct option: C.
$\frac{\pi}{3}$
(C) Given,
$\overline{ a }+\overline{ b }+\sqrt{3} \overline{ c }=\overline{0}$
$\Rightarrow \overline{ a }+\overline{ b }=-\sqrt{3} \overline{ c } \\ \Rightarrow|\overline{ a }+\overline{ b }|=\sqrt{3}|\overline{ c }| \\ \Rightarrow|\overline{ a }+\overline{ b }|^2=3|\overline{ c }|^2 \\ \Rightarrow|\overline{ a }|^2+|\overline{ b }|^2+2|\overline{ a }||\overline{ b }| \cos \theta=3|\overline{ c }|^2$
where $\theta$ is the angle between $\overline{ a }$ and $\overline{ b }$
$\Rightarrow 1+1+2 \cos \theta=3$
$\Rightarrow \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$
$\overline{ a }+\overline{ b }+\sqrt{3} \overline{ c }=\overline{0}$
$\Rightarrow \overline{ a }+\overline{ b }=-\sqrt{3} \overline{ c } \\ \Rightarrow|\overline{ a }+\overline{ b }|=\sqrt{3}|\overline{ c }| \\ \Rightarrow|\overline{ a }+\overline{ b }|^2=3|\overline{ c }|^2 \\ \Rightarrow|\overline{ a }|^2+|\overline{ b }|^2+2|\overline{ a }||\overline{ b }| \cos \theta=3|\overline{ c }|^2$
where $\theta$ is the angle between $\overline{ a }$ and $\overline{ b }$
$\Rightarrow 1+1+2 \cos \theta=3$
$\Rightarrow \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$
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