Find the equation of the plane that passes through point $P (1, 2, 3)$ and is perpendicular to the line $\frac{x - 4}{2} = \frac{y + 1}{- 3} = \frac{z}{5}$ .

hard 4 marks Class 12 · Mathematics · Vector Algebra

Answer

2x - 3y + 5z - 11 = 0

The correct answer is 2x - 3y + 5z - 11 = 0 — Thus the required plane equation is $2 x - 3 y + 5 z - 11 = 0$ .

Step-by-step solution

Identify the direction vector of the given line: $d = ⟨ 2, - 3, 5 ⟩$ .
Because the plane is perpendicular to the line, $d$ serves as the normal vector of the plane.
Write the plane equation using the normal vector and point $P$ : $2 (x - 1) - 3 (y - 2) + 5 (z - 3) = 0$ .
Distribute and combine constants: $2 x - 2 - 3 y + 6 + 5 z - 15 = 0$ .
Simplify to obtain $2 x - 3 y + 5 z - 11 = 0$ .
Thus the required plane equation is $2 x - 3 y + 5 z - 11 = 0$ .

Common mistake: Using the line’s direction as a tangent to the plane instead of as its normal vector

Concept tested: Plane perpendicular to a line (normal vector)

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked