Math Problem Statement

Write the equation of the plane with a normal vector n

⟨ 1 , − 5 , 1 ⟩ and the point (1, 2, 3) lying on the plane.

Solution

To write the equation of a plane, we use the general formula for the equation of a plane:

Ax+By+Cz=DAx + By + Cz = D

Where:

  • n=A,B,C\vec{n} = \langle A, B, C \rangle is the normal vector to the plane.
  • (x1,y1,z1)(x_1, y_1, z_1) is a point on the plane.

Given that the normal vector n=1,5,1\vec{n} = \langle 1, -5, 1 \rangle and the point (1,2,3)(1, 2, 3) lies on the plane, we can substitute A=1A = 1, B=5B = -5, and C=1C = 1 into the plane equation. The point (1,2,3)(1, 2, 3) will help us find DD.

The equation becomes:

1(x1)5(y2)+1(z3)=01(x - 1) - 5(y - 2) + 1(z - 3) = 0

Now simplify it:

(x1)5(y2)+(z3)=0(x - 1) - 5(y - 2) + (z - 3) = 0

Expand the terms:

x15y+10+z3=0x - 1 - 5y + 10 + z - 3 = 0

Combine like terms:

x5y+z+6=0x - 5y + z + 6 = 0

Thus, the equation of the plane is:

x5y+z+6=0x - 5y + z + 6 = 0

Would you like more details or have any questions?


Here are 5 related questions:

  1. How do you find the normal vector from a given plane equation?
  2. What is the equation of a plane parallel to another plane?
  3. How can you determine if two planes are perpendicular?
  4. What happens when a point is substituted into the plane equation?
  5. How do you find the distance between a point and a plane?

Tip: Always check that the normal vector is perpendicular to the plane for any given plane equation.

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Math Problem Analysis

Mathematical Concepts

Geometry
Planes
Vectors

Formulas

Equation of a plane: Ax + By + Cz = D

Theorems

Plane equation theorem based on a normal vector and a point on the plane

Suitable Grade Level

Grades 10-12