JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
Let the values of $p$, for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\vec{r}=\left(p\hat{i}+2\hat{j}+\hat{k}\right)+\lambda\left(2\hat{i}+3\hat{j}+4\hat{k}\right)$ is $\frac{1}{\sqrt{6}}$, be $a,b$, $(a<b)$. Then the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is :
3D Geometry
Solution
$\vec d_1\times\vec d_2=(1,-2,1)$, $|\cdot|=\sqrt6$; joining vector $(p+1,2,1)$ gives $|p-2|/\sqrt6=1/\sqrt6$, so $p=1,3$. Ellipse $x^2+\frac{y^2}{9}=1$ has latus rectum $\frac{2\cdot1^2}{3}=\frac23$.