Find the equation of the plane that passes through the point $(2, - 1, 4)$ and has normal vector $⟨ 3, - 2, 5 ⟩$ .

easy MCQ 2 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

3x - 2y + 5z - 28 = 0

The correct answer is 3x - 2y + 5z - 28 = 0 — Combine like terms to obtain $3 x - 2 y + 5 z - 28 = 0$ .

Step-by-step solution

Write the general plane equation using the normal vector $n = (3, - 2, 5)$ as $3 (x - x_{0}) - 2 (y - y_{0}) + 5 (z - z_{0}) = 0$ .
Substitute the given point $(x_{0}, y_{0}, z_{0}) = (2, - 1, 4)$ into the equation.
Simplify: $3 (x - 2) - 2 (y + 1) + 5 (z - 4) = 0 \Rightarrow 3 x - 6 - 2 y - 2 + 5 z - 20 = 0$ .
Combine like terms to obtain $3 x - 2 y + 5 z - 28 = 0$ .

Common mistake: Using the point incorrectly, leading to a sign error in the constant term.

Concept tested: Equation of a plane