3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The equation of line intersecting and perpendicular to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and passing through $(-2, -5, 7)$ is :
\frac{x+2}{14} = \frac{y+5}{123} = \frac{z-7}{104}
\frac{x+2}{14} = \frac{y+5}{137} = \frac{z-7}{-204}
\frac{x+2}{76} = \frac{y+5}{137} = \frac{z-7}{-254}
None of these
Step-by-Step Solution
Key Concept: Convert the line to parametric form and substitute into the plane equation to find the parameter value at intersection.
Setting $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$ gives parametric equations $x = 2\lambda$, $y = 2 + 3\lambda$, $z = 3 + 4\lambda$. Substituting into the plane equation $2x + 2 + (7 + 3\lambda)3 + (-4 + 4\lambda)4 = 0$ and simplifying yields $29\lambda + 9 = 0$, so $\lambda = -\frac{9}{29}$.
Correct Answer: I need to find the line that intersects and is perpendicular to the given line and passes through (-2, -5, 7).
**Step 1: Direction vector of the given line**
The given line is $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$