3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The equation of the straight line through the origin parallel to the line $(b+c)x + (c+a)y + (a+b)z = k = (b-c)x + (c-a)y + (a-b)z$ is:
$\frac{x}{b^2-c^2} = \frac{y}{c^2-a^2} = \frac{z}{a^2-b^2}$
$\frac{x}{b} = \frac{y}{c} = \frac{z}{a}$
$\frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab}$
None of these
Step-by-Step Solution
Key Concept: Setting up and solving two simultaneous linear constraints on direction ratios yields the unique line satisfying both plane conditions.
The equation of the line through the origin is $\frac{x}{l} = \frac{y}{m} = \frac{z}{n}$. Using the conditions $l(b+c) + m(c+a) + n(a+b) = 0$ and $l(b-c) + m(c-a) + n(a-b) = 0$, we solve for direction ratios to get $\frac{x}{a^2 - bc} = \frac{y}{b^2 - ca} = \frac{z}{c^2 - ab}$.
Correct Answer: 3