3D Geometry

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JEE Main 2024 (Online) 27th January Morning ShiftQuestion 106

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

3D Geometry

$12$$14$$18$$21$

JEE Main 2025 (Online) 4th April Evening ShiftQuestion 107

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

$\frac{3}{2}$$\frac{2}{3}$$18$$9$

JEE Main 2023 (Online) 13th April Morning ShiftQuestion 108

The distance of the point $(-1,2,3)$ from the plane $\vec{r}\cdot\left(\hat{i}-2\hat{j}+3\hat{k}\right)=10$ parallel to the line of the shortest distance between the lines $\vec{r}=\left(\hat{i}-\hat{j}\right)+\lambda\left(2\hat{i}+\hat{k}\right)$ and $\vec{r}=\left(2\hat{i}-\hat{j}\right)+\mu\left(\hat{i}-\hat{j}+\hat{k}\right)$ is

3D Geometry

$3\sqrt{5}$$2\sqrt{6}$$3\sqrt{6}$$2\sqrt{5}$

JEE Main 2023 (Online) 13th April Evening ShiftQuestion 109

Let $N$ be the foot of perpendicular from the point $P(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2x-2y+z+5=0$ is

3D Geometry

$6$$7$$8$$9$

JEE Main 2022 (Online) 26th July Morning ShiftQuestion 110

The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2y+3z-5$ is :

3D Geometry

$\sqrt{\frac{21}{2}}$$\sqrt{\frac92}$$\sqrt{\frac{73}{2}}$$1$