JEE Main 2026 — Mathematics Vectors & 3D Geometry

Let $\vec{a}_k = \tan\theta_k\,\hat{i} + \hat{j}$ak​=tanθk​i^+j^​ and $\vec{b}_k = \hat{i} - \cot\theta_k\,\hat{j}$bk​=i^−cotθk​j^​, where $\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}$θk​=2n+12k−1π​.

$$|\vec{a}_k|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k$$∣ak​∣2=tan2θk​+1=sec2θk​

$$|\vec{b}_k|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k$$∣bk​∣2=1+cot2θk​=csc2θk​

So the required ratio is:

$$\dfrac{\displaystyle\sum_{k=1}^n |\vec{a}_k|^2}{\displaystyle\sum_{k=1}^n |\vec{b}_k|^2} = \dfrac{\displaystyle\sum_{k=1}^n \sec^2\theta_k}{\displaystyle\sum_{k=1}^n \csc^2\theta_k}$$k=1∑n​∣bk​∣2k=1∑n​∣ak​∣2​=k=1∑n​csc2θk​k=1∑n​sec2θk​​

Relation between successive angles:

$$\theta_{k+1} = \dfrac{2^k\pi}{2^n + 1} = 2\theta_k$$θk+1​=2n+12kπ​=2θk​

Key identity:

$$\sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta} = \dfrac{4}{\sin^2 2\theta} = 4\csc^2 2\theta$$sec2θ+csc2θ=cos2θ1​+sin2θ1​=sin2θcos2θ1​=sin22θ4​=4csc22θ

Hence:

$$\sec^2\theta = 4\csc^2 2\theta - \csc^2\theta$$sec2θ=4csc22θ−csc2θ

Evaluating the numerator using this identity and $2\theta_k = \theta_{k+1}$2θk​=θk+1​:

$$\displaystyle\sum_{k=1}^n \sec^2\theta_k = \sum_{k=1}^n \left(4\csc^2\theta_{k+1} - \csc^2\theta_k\right) = 4\sum_{k=1}^n \csc^2\theta_{k+1} - \sum_{k=1}^n \csc^2\theta_k$$k=1∑n​sec2θk​=k=1∑n​(4csc2θk+1​−csc2θk​)=4k=1∑n​csc2θk+1​−k=1∑n​csc2θk​

Let $\displaystyle S_b = \displaystyle\sum_{k=1}^n \csc^2\theta_k$Sb​=k=1∑n​csc2θk​. Then:

$$\displaystyle\sum_{k=1}^n \csc^2\theta_{k+1} = S_b - \csc^2\theta_1 + \csc^2\theta_{n+1}$$k=1∑n​csc2θk+1​=Sb​−csc2θ1​+csc2θn+1​

So:

$$\displaystyle\sum_{k=1}^n \sec^2\theta_k = 4\left(S_b - \csc^2\theta_1 + \csc^2\theta_{n+1}\right) - S_b = 3S_b - 4\csc^2\theta_1 + 4\csc^2\theta_{n+1}$$k=1∑n​sec2θk​=4(Sb​−csc2θ1​+csc2θn+1​)−Sb​=3Sb​−4csc2θ1​+4csc2θn+1​

Boundary terms:

$\theta_1 = \dfrac{\pi}{2^n + 1}$θ1​=2n+1π​, $\theta_{n+1} = \dfrac{2^n\pi}{2^n + 1}$θn+1​=2n+12nπ​

$$\theta_1 + \theta_{n+1} = \dfrac{\pi + 2^n\pi}{2^n + 1} = \dfrac{(2^n + 1)\pi}{2^n + 1} = \pi$$θ1​+θn+1​=2n+1π+2nπ​=2n+1(2n+1)π​=π

So $\theta_{n+1} = \pi - \theta_1$θn+1​=π−θ1​, and since $\csc(\pi - x) = \csc x$csc(π−x)=cscx:

$$\csc^2\theta_{n+1} = \csc^2\theta_1$$csc2θn+1​=csc2θ1​

Therefore, $-4\csc^2\theta_1 + 4\csc^2\theta_{n+1} = 0$−4csc2θ1​+4csc2θn+1​=0, and:

$$\displaystyle\sum_{k=1}^n \sec^2\theta_k = 3S_b = 3\sum_{k=1}^n \csc^2\theta_k$$k=1∑n​sec2θk​=3Sb​=3k=1∑n​csc2θk​

Hence, the required ratio is:

$$\dfrac{3S_b}{S_b} = 3$$Sb​3Sb​​=3

Therefore, the answer is $3$3.