3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade None
Question:
Direction cosines of normal to the plane containing lines $x = y = z$ and $x - 1 = y - 1 = \frac{z-1}{d}$ (where $d \in \mathbb{R} - \{1\}$), are :
\left\{\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right\}
\left\{\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right\}
\left\{0, \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\}
None of these
Step-by-Step Solution
Key Concept: Using constraint equations on direction ratios eliminates variables, yielding specific direction vectors for plane normals.
Let $l, m, n$ be the direction ratios of the normal to the plane. From conditions $l + m + n = 0$ and $l + m + nd = 0$, we get $n(1-d) = 0$, so $n = 0$. This gives two independent normal vectors: $\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$ and $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$.
Correct Answer: 1