JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a_k} = (\tan\theta_k)\hat{i} + \hat{j}\) and \(\vec{b_k} = \hat{i} - (\cot\theta_k)\hat{j}\), where \(\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}\), for some \(n \in \mathbb{N}\), \(n > 5\). Then the value of \(\dfrac{\displaystyle\sum_{k=1}^{n}|\vec{a_k}|^2}{\displaystyle\sum_{k=1}^{n}|\vec{b_k}|^2}\) is _____.
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
Let \(\vec{a}_k = \tan\theta_k\,\hat{i} + \hat{j}\) and \(\vec{b}_k = \hat{i} - \cot\theta_k\,\hat{j}\), where \(\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}\). \(|\vec{a}_k|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k\) \(|\vec{b}_k|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k\) So the…
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