3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The direction cosines of the shortest distance lie between the planes $y + z = 0$ and $z + x = 0$ is:
$-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$
$-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}$
$-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
None of these
Step-by-Step Solution
Key Concept: The shortest distance between two skew lines is found by finding direction cosines perpendicular to both lines using the perpendicularity conditions.
The line of intersection of planes $y + z = 0$ and $z + x = 0$ is $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1}$, and the line of intersection of $x + y = 0$ and $x + y + z = a$ is $\frac{x}{1} = \frac{y}{-1} = \frac{z-a}{0}$. The direction cosines of the shortest distance line satisfy $l - l + m - l - n = 0$ and $l - l - m + 0 = 0$, giving $l = -\frac{1}{\sqrt{6}}$, $m = \frac{1}{\sqrt{6}}$, $n = -\frac{1}{\sqrt{6}}$.
Correct Answer: I need to find the direction cosines of the shortest distance between the two planes y + z = 0 and z + x = 0.
The shortest distance between two planes lies along the line perpendicular to both planes.
For plane y + z = 0, the normal vector is **nā** = (0, 1, 1)
For