3D Geometry

Showing 51-55 of 461 questions.

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JEE Main 2023 (Online) 11th April Evening ShiftQuestion 51

Let the line passing through the points $P(2,-1,2)$ and $Q(5,3,4)$ meet the plane $x-y+z=4$ at the point $R$. Then the distance of the point $R$ from the plane $x+2y+3z+2=0$ measured parallel to the line $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is equal to

3D Geometry

$3$$\sqrt{31}$$\sqrt{61}$$\sqrt{189}$

JEE Main 2023 (Online) 10th April Evening ShiftQuestion 52

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

3D Geometry

$\frac{10}{3}$$\frac{11}{3}$$4$$3$

JEE Main 2023 (Online) 6th April Morning ShiftQuestion 53

One vertex of a rectangular parallelopiped is at the origin $O$ and the lengths of its edges along $x$, $y$ and $z$ axes are $3,4$ and $5$ units respectively. Let $P$ be the vertex $(3,4,5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to $z$ axis, not passing through $O$ or $P$ is:

3D Geometry

$\frac{12}{5}$$\frac{12}{5\sqrt{5}}$$12\sqrt{5}$$\frac{12}{\sqrt{5}}$

JEE Main 2021 (Online) 16th March Evening ShiftQuestion 54

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_1:\frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4}$, $l\neq0$ is $(3,5,7)$, then the shortest distance between the line $L_1$ and line $L_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to :

3D Geometry

$\frac{1}{\sqrt{6}}$$\frac{1}{2}$$\frac{1}{\sqrt{3}}$$\sqrt{\frac{2}{3}}$

JEE Main 2023 (Online) 8th April Morning ShiftQuestion 55

The shortest distance between the lines $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ is

3D Geometry

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