TS EAMCET · Maths · Vector Algebra
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 \(\alpha=\frac{\pi}{2}+\frac{n \pi}{3}, n \in Z ; \beta=1\)
- B \(\alpha=\frac{\pi}{2}+2 n \pi, n \in Z ; \beta=1\)
- C \(\alpha=\frac{\pi}{2}+(2 n+1) \frac{\pi}{2}, n \in Z, \beta=2\)
- D \(\alpha=(2 n+1) \frac{\pi}{2}, n \in Z, \beta=\frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(\alpha=\frac{\pi}{2}+2 n \pi, n \in Z ; \beta=1\)
Step-by-step Solution
Detailed explanation
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…
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