Q74 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
If $\mathbf{b}, \mathbf{c}$ are non collinear vectors, $|\mathbf{c}| \neq 0$, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$ and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$, then the scalars $\alpha$ and $\beta$ are
- A$\alpha=\frac{\pi}{2}+\frac{n \pi}{3}, n \in Z ; \beta=1$
- Correct answerB$\alpha=\frac{\pi}{2}+2 n \pi, n \in Z ; \beta=1$
- C$\alpha=\frac{\pi}{2}+(2 n+1) \frac{\pi}{2}, n \in Z, \beta=2$
- D$\alpha=(2 n+1) \frac{\pi}{2}, n \in Z, \beta=\frac{3}{2}$
Solution
For non-collinear vectors $\mathbf{b}$ and $\mathbf{c},|\mathbf{c}| \neq 0$, it is given that, $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c} \Rightarrow \mathbf{a} \cdot \mathbf{c}=1$ $\ldots(\mathrm{i})$
and $\mathbf{a}\times(\mathbf{b} \times \mathbf{c})+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \quad=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$
$\Rightarrow(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}=(4-2 \beta-\sin \alpha)$ $\mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$
So, $\mathbf{a} \cdot \mathbf{c}+\mathbf{a} \cdot \mathbf{b}=4-2 \beta-\sin \alpha$ $\ldots(\mathrm{ii})$
and $\quad-\mathbf{a} \cdot \mathbf{b}=\beta^2-1$ $\ldots(\mathrm{iii})$
From Eqs. (i), (ii) and (iii), we get
$1=4-2 \beta-\sin \alpha+\beta^2-1$
$\Rightarrow \quad(\beta-1)^2-\sin \alpha=-1$
So, if $\beta=1$, then $\sin \alpha=1 \Rightarrow \alpha=2 n \pi+\frac{\pi}{2}, n \in \mathbf{Z}$ and other options does not satisfy.
Hence, the correct answer is option: 'B'.