3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The equation of a line in $xz$ plane equally inclined with $x$ and $z$ axes which is at a unit distance from the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z}{1}$ is:
$\frac{x - \sqrt{6} + 1}{1} = \frac{y}{0} = \frac{z}{-1}$
$\frac{x - 1 - \sqrt{2}}{1} = \frac{y}{0} = \frac{z}{1}$
$\frac{x-1}{1} = \frac{y}{0} = \frac{z}{1}$
None of these
Step-by-Step Solution
Key Concept: Use perpendicularity of lines (dot product of direction vectors = 0) combined with normalization to find the required line.
The required line passes through $(h, 0, 0)$ with equation $\frac{x - h}{l} = \frac{y - 0}{m} = \frac{z - 0}{n}$. The perpendicularity condition with the given line yields $l + m \times 1 + n \times 1 = 0$. Solving with $l^2 + m^2 + n^2 = 1$ gives $l = \frac{1}{\sqrt{6}}, m = -\frac{2}{\sqrt{6}}, n = \frac{1}{\sqrt{6}}$. This leads to the final line equations: $\frac{x - \sqrt{6} + 1}{1} = \frac{y - 0}{0} = \frac{z}{-1}$ and $\frac{x - 1 - \sqrt{2}}{1} = \frac{y - 0}{0} = \frac{z - 0}{1}$.
Correct Answer: 1,2