3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The point of intersection of the line, passing through $(0, 0, 1)$ and intersecting the lines $x + 2y + z = 1, -x + y - 2z = 2$ and $x + y = 2, x + z = 2$ with plane is:
\left(\frac{5}{3}, -\frac{1}{3}, 0\right)
(1, 1, 0)
\left(\frac{2}{3}, -\frac{1}{3}, 0\right)
\left(-\frac{5}{3}, \frac{1}{3}, 0\right)
Step-by-Step Solution
Key Concept: A point lies on both lines when it satisfies the parametric equations of both.
The equation of line 1 is $x + 2y + z - 1 = λ(−x + y − 2z − 2) = 0$. The second line equation is $x + y − 2 + μ(x + z − 2) = 0$. The point $(0,0,1)$ lies on line 1, giving $−λ = 0$, so $μ = -2$. To find intersection points, set $z = 0$ and solve both equations simultaneously.
Correct Answer: Let me work through this step-by-step.
**Step 1: Find the line passing through (0,0,1) that intersects both given lines.**
Line 1 is the intersection of planes: $x + 2y + z = 1$ and $-x + y - 2z = 2$
Line 2 is the