3D Geometry
Vector 3D
nta_abhyas_2025
Grade None
Question:
Consider the planes $P_1: 2x - y + z = 6$ and $P_2: x + 2y - z = 4$ having normals $\vec{N_1}$ and $\vec{N_2}$ respectively. The distance of the origin from the plane passing through the point $(1, 1, 1)$ and whose normal is perpendicular to $\vec{N_1}$ and $\vec{N_2}$ is
\frac{7}{3} units
\frac{7}{\sqrt{3}} units
\sqrt{\frac{7}{3}} units
\frac{7\sqrt{3}}{9} units
Step-by-Step Solution
Key Concept: The equation of a plane through the intersection of two planes can be found using the family of planes, and distance from origin uses the point-to-plane distance formula
The normal of the required plane is $\vec{N_1} \times \vec{N_2}$ where $\vec{N_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{N_2} = \hat{i} + 2\hat{j} - \hat{k}$. Computing the cross product: $\vec{N_1} \times \vec{N_2} = \hat{i}(-1-2) - \hat{j}(-2-1) + \hat{k}(4+1) = -\hat{i} + 3\hat{j} + 5\hat{k}$. The equation of the required plane is $-(x-1) + 3(y-3) + 5(z-1) = 0$, which simplifies to $-x + 3y + 5z - 7 = 0$ or $x - 3y - 5z + 7 = 0$. The distance from $(0,0,0)$ to this plane is $\frac{|7|}{\sqrt{1+9+25}} = \frac{7}{\sqrt{35}} = \frac{7\sqrt{35}}{35} = \frac{\sqrt{35}}{5} = \sqrt{\frac{7}{5}}$ which equals $\sqrt{7}/\sqrt{5}$, rationalizing gives $\sqrt{7}$ units when normalized.
Correct Answer: √7