Q74 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors such that $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is not parallel to c. If $\alpha$ and $\beta$ are the angles between $\mathbf{a}, \mathbf{b}$ and a,c respectively then $\alpha-\beta=$
- A$\frac{3 \pi}{4}$
- B$\frac{\pi}{4}$
- Correct answerC$\frac{\pi}{2}$
- D0
Solution
Given $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors
$\therefore \quad|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$
Now, $\quad \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$
$\Rightarrow \quad(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$
$\Rightarrow \quad(|\mathbf{a}||\mathbf{c}| \cos \beta) \mathbf{b}-(|\mathbf{a}||\mathbf{b}| \cos \alpha) \mathbf{c}=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$
$\Rightarrow \quad \cos \beta \mathbf{b}-\cos \alpha \mathbf{c}=\frac{1}{\sqrt{2}} \mathbf{b}+\frac{1}{\sqrt{2}} \mathbf{c}$
$\Rightarrow \quad \cos \beta=\frac{1}{\sqrt{2}}$ and $\cos \alpha=\frac{-1}{\sqrt{2}}$
$\Rightarrow \quad \cos \beta=\cos \frac{\pi}{4}$ and $\cos \alpha=\cos \frac{3 \pi}{4}$
$\Rightarrow \quad \beta=\frac{\pi}{4}$ and $\alpha=\frac{3 \pi}{4}$
$\therefore \quad \alpha-\beta=\frac{3 \pi}{4}-\frac{\pi}{4}=\frac{2 \pi}{4}=\frac{\pi}{2}$
Hence, the correct answer is option: 'C'.