3D Geometry
Vector 3D
nta_abhyas_2025
Grade 12
Question:
A plane passes through $(1, -2, 1)$ and is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the plane from the origin $(0, 2)$ is
$\frac{1}{\sqrt{6}}$ units
$\frac{1}{\sqrt{2}}$ units
$\frac{1}{6}$ units
$\frac{1}{3}$ units
Step-by-Step Solution
Key Concept: To find the distance from a point to a plane, first determine the plane's equation using the cross product of two vectors in the plane, then apply the distance formula.
The normal vector perpendicular to the required plane is found by taking the cross product of two vectors in the plane: $\vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -2 & -1 \\ 1 & 2 \end{vmatrix} = \vec{i}(-3) - \vec{j}(3) + \vec{k}(0) = -3\vec{i} - 3\vec{j}$. The equation of the plane through $(1, -2, 1)$ is $-(x-1) - (y+2) + 0(z-1) = 0$, which simplifies to $x + y + 1 = 0$. The distance from $(0, 2, 3)$ to the plane $x + y + 1 = 0$ is $\frac{|0 + 2 + 1|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.
Correct Answer: 3