3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
Projection of line $\frac{x+1}{2} = \frac{y+1}{-1} = \frac{z+3}{4}$ on the plane $x + 2y + z = 6$ has equation:
x + 2y + z - 6 = 0, 9x - 2y - 5z = 8
x + 2y + z + 6 = 0, 9x - 2y + 5z = 4
x-1/4 = y-3/-7 = z+1/10
x+3/4 = y-2/7 = z-7/-10
Step-by-Step Solution
Key Concept: The required line connects the line-plane intersection with the foot of the perpendicular from an external point to the plane.
Find the point of intersection $A$ of the line and the given plane. Then find the foot of the normal $B$ from point $(-1, -1, -3)$ to the plane. The line $AB$ connecting these two points is the required line, as it passes through the plane and is perpendicular to it at point $A$.
Correct Answer: I need to find the projection of the given line on the plane.
**Step 1: Find point A (intersection of line and plane)**
Line: $\frac{x+1}{2} = \frac{y+1}{-1} = \frac{z+3}{4} = t$
Parametric form:
- $x