3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
A plane meets a set of three mutually perpendicular planes in the sides of a triangle whose angles are $A$, $B$ and $C$ respectively. The squares of cosines of angles which first plane makes with the other planes are:
$\cot B \cot C, \cot C \cot A, \cot A \cot B$
$\tan B \tan C, \tan C \tan A, \tan A \tan B$
$\cosec B \cosec C, \cosec C \cosec A, \cosec A \cosec B$
None of these
Step-by-Step Solution
Key Concept: The angle between a plane and a coordinate plane is found using the normal vector of the plane and the normal of the coordinate plane.
For the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ with three mutually perpendicular planes $x = 0$, $y = 0$, $z = 0$, the direction ratios of edges are $AB: (a, -b, 0)$, $AC: (a, 0, -c)$, with $\cos A = \frac{a^2}{\sqrt{a^2 + \sum a^2b^2}}$ and similarly for other angles. If $\alpha$ is the angle between the plane and $x = 0$, then $\cos \alpha = \frac{a}{\sqrt{\sum a^2b^2}}$.
Correct Answer: I need to find the squares of cosines of angles which the plane makes with the three mutually perpendicular planes.
Let me work through this systematically.
Consider a plane meeting three mutually perpendicular planes (the coordinate planes x=0, y=0, z=0) in the sides of a triangle with angles A, B, C.
For a plane