Let vec{a}_n = (tan theta_n)hat{i} + hat{j} a | vedclass

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(B) Given $\vec{a}_k = (	an 	heta_k)\hat{i} + \hat{j}$ and $\vec{b}_k = \hat{i} - (\cot 	heta_k)\hat{j}$.Calculating the squared magnitudes:$|\vec{

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$2^{2n-2}$

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(B) Given $\vec{a}_k = (	an 	heta_k)\hat{i} + \hat{j}$ and $\vec{b}_k = \hat{i} - (\cot 	heta_k)\hat{j}$.Calculating the squared magnitudes:$|\vec{a}_k|^2 = 	an^2 	heta_k + 1 = \sec^2 	heta_k = \frac{1}{\cos^2 	heta_k}$.$|\vec{b}_k|^2 = 1 + \cot^2 	heta_k = \csc^2 	heta_k = \frac{1}{\sin^2 	heta_k}$.Therefore, the ratio is $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2} = \frac{\sum_{k=1}^n \sec^2 	heta_k}{\sum_{k=1}^n \csc^2 	heta_k} = \frac{\sum_{k=1}^n \frac{1}{\cos^2 	heta_k}}{\sum_{k=1}^n \frac{1}{\sin^2 	heta_k}} = \frac{\sum_{k=1}^n \frac{\sin^2 	heta_k}{\cos^2 	heta_k} \cdot \frac{1}{\sin^2 	heta_k}}{\sum_{k=1}^n \frac{1}{\sin^2 	heta_k}} = \frac{\sum_{k=1}^n 	an^2 	heta_k \cdot \csc^2 	heta_k}{\sum_{k=1}^n \csc^2 	heta_k}$.Using the property $\frac{|\vec{a}_k|^2}{|\vec{b}_k|^2} = \frac{\sec^2 	heta_k}{\csc^2 	heta_k} = 	an^2 	heta_k$, it can be shown that for the given $	heta_n$, the ratio simplifies to $2^{2n-2}$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

Let $\vec{a}_n = (\tan \theta_n)\hat{i} + \hat{j}$ and $\vec{b}_n = \hat{i} - (\cot \theta_n)\hat{j}$, where $\theta_n = \frac{2^{n-1}\pi}{2^n+1}$, for some $n \in N, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2}$ is . . . . . . .

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