JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Three Dimensional Geometry

Let P be the plane such that it contains the straight line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+2}{1}$ and is perpendicular to the plane $x+2y+3z=4$. Let $P_1$ be the plane which passes through the point $(4,2,2)$ and is parallel to $P$. T …

Let L be the straight line joining the points P(1,2, -1) and Q(2,3,1). Let S be the foot of the perpendicular drawn from the point $\mathrm{R}(4,-1,5)$ to the line L. Another line passing through R intersects L at a point T such that the point S divides the line segment PT intern …

Let $L_1$ be the line of intersection of the planes given by the equations

Let $L_2$ be the line passing through the point $P(2,-1,3)$ and parallel to $L_1$. Let $M$ denote the plane given by the equation …

A straight line drawn from the point $P(1,3,2)$, parallel to the line $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$, intersects the plane $L_1:x-y+3z=6$ at the point $Q$. Another straight line which passes through $Q$ and is perpendicular to the plane $L_1$ intersects the plan …

Let ${\mathbb{R}}^3$ denote the three-dimensional space. Take two points $\mathrm{P}=(1,2,3)$ and $\mathrm{Q}=(4,2,7)$.
Let dist (X, Y) denote the distance between two points X and Y in ${\mathbb{R}}^3$. Let …

Let $\gamma \in \mathbb{R}$ be such that the lines ${\mathrm{L}}_1:\frac{\mathrm{x}+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{z}+29}{3}$ and ${\mathrm{L}}_2:\frac{\mathrm{x}+16}{3}=\frac{\mathrm{y}+11}{2}=\frac{\mathrm{z}+4}{\gamma }$ intersect. Let ${\mathrm{R}}_1$ b …