Q74 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
a, b and $\mathbf{c}$ are three unit vectors such that no two of them are collinear. If $\mathbf{b}=2\{\mathbf{a} \times(\mathbf{b} \times \mathbf{c})\}$ and $\alpha$ is the angle between $\mathbf{a}, \mathbf{c}$ and $\beta$ is the angle between $\mathbf{a}, \mathbf{b}$, then $\cos (\alpha+\beta)=$
- A$\frac{\sqrt{3}}{2}$
- Correct answerB$-\frac{\sqrt{3}}{2}$
- C$\frac{1}{2}$
- D$-\frac{1}{2}$
Solution
We have,
$$
\begin{aligned}
& \mathbf{b}=2\{\mathbf{a} \times(\mathbf{b} \times \mathbf{c})\} \\
& \mathbf{b}=2\{(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\} \\
& \mathbf{b}=2(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-2(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}
\end{aligned}
$$
On comparing $\mathbf{b}$ and $\mathbf{c}$, we get
$$
\therefore \quad 2(\mathbf{a} \cdot \mathbf{c})=1 \text { and } \mathbf{a} \cdot \mathbf{b}=0
$$
$\alpha$ is a angle between $\mathbf{a}, \mathbf{c}$ and $\beta$ is the angle between $\mathbf{a}, \mathbf{b}$.
$$
\begin{aligned}
|\mathbf{a}||\mathbf{c}| \cos \alpha & =\frac{1}{2}, \mathbf{a} \cdot \mathbf{b}=0 \\
\cos \alpha & =\cos \frac{\pi}{3}, \text { and } \cos \beta=\cos \frac{\pi}{2} \\
\therefore \quad \alpha & =\pi / 3 \text { and } \beta=\frac{\pi}{2} \\
\Rightarrow \cos (\alpha+\beta) & =-\sin \pi / 3=-\sqrt{3} / 2
\end{aligned}
$$
Hence, the correct answer is option: 'B'.