3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The direction cosines of the projection of the line $\frac{1}{2}(x-1) = -y = z+2$ on the plane $2x + y - 3z = 4$ are :
\left(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
\left(\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
\left(\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right)
None of these
Step-by-Step Solution
Key Concept: A line parallel to a plane projects onto itself with unchanged direction cosines.
The line $\frac{x-1}{2} = \frac{y}{-1} = \frac{z+2}{1}$ has direction vector $(2, -1, 1)$. The plane has normal vector $(2, -1, 1)$, which is parallel to the line's direction. Since the line does not satisfy the plane equation, the line is parallel to the plane with no intersection, so the direction cosines of the projected line remain the same.
Correct Answer: I need to find the direction cosines of the projection of the given line onto the plane.
**Given:**
- Line: $\frac{1}{2}(x-1) = -y = z+2$ or $\frac{x-1}{2} = \frac{y}{-1} = \frac{z+2}{