Line in space (point A, direction vector →d): →p = →a + t→d (t is a real parameter)
Equation of a plane (normal vector →n=(a,b,c), point P₀(x₀,y₀,z₀)):
a(x−x₀) + b(y−y₀) + c(z−z₀) = 0
Example ① Find the equation of the plane with normal vector →n=(2,−1,3) passing through point (1,0,2).
Solution: 2(x−1) − (y−0) + 3(z−2) = 0 → 2x − y + 3z = 8
Solution: 2(x−1) − (y−0) + 3(z−2) = 0 → 2x − y + 3z = 8
Key Points
- The normal vector of the plane ax+by+cz+d=0 is (a,b,c)
Practice Problems
- Equation of the plane passing through the three points A(1,0,0), B(0,2,0), and C(0,0,3).
Answers and Explanations
- Answer: 6x+3y+2z=6 (or x+y/2+z/3=1)