Finding a vector perpendicular to two planes is typically interpreted as finding a vector that is parallel to the line where the two planes intersect. This vector is mathematically perpendicular to the normal vectors of both planes.
The most direct way to find such a vector is by using the cross product of the normal vectors of the two planes.
The Method: Using the Cross Product
The normal vector of a plane given by the equation Ax + By + Cz = D is the vector <A, B, C>. This vector is perpendicular to the plane itself. To find a vector parallel to the line of intersection of two planes, you find a vector that is perpendicular to both of their normal vectors. The cross product of two vectors yields a vector that is perpendicular to both of the original vectors.
Here are the steps:
- Identify the normal vector of the first plane. If the plane's equation is A₁x + B₁y + C₁z = D₁, its normal vector n₁ is <A₁, B₁, C₁>.
- Identify the normal vector of the second plane. If the plane's equation is A₂x + B₂y + C₂z = D₂, its normal vector n₂ is <A₂, B₂y + C₂z = D₂>, its normal vector n₂ is <A₂, B₂, C₂>.
- Calculate the cross product of the two normal vectors. The cross product n₁ × n₂ will result in a vector that is perpendicular to both n₁ and n₂. This resulting vector is therefore parallel to the line of intersection of the two planes.
Mathematically, the cross product of n₁ = <A₁, B₁, C₁> and n₂ = <A₂, B₂, C₂> is calculated as:
n₁ × n₂ = <(B₁C₂ - B₂C₁), (C₁A₂ - C₂A₁), (A₁B₂ - A₂B₁)>
Incorporating Information from the Reference
As described in the provided reference (video timestamp ~2:04-4:23), the process involves using the cross product in relation to the two planes. The reference states, "So to find the normal vector n i'm going to take the cross. Product of the coordinates of these two planes. And the cross product will give me a perpendicular vector." This aligns with the method of taking the cross product of the normal vectors (represented by the coefficients/coordinates in the plane equations) to find a vector that is perpendicular to those normal vectors, yielding the direction vector parallel to the planes' intersection line.
Example
Let's find a vector perpendicular to the planes:
- Plane 1: x + 2y - z = 5
- Plane 2: 2x - y + 3z = 8
- The normal vector for Plane 1 is n₁ = <1, 2, -1>.
- The normal vector for Plane 2 is n₂ = <2, -1, 3>.
- Calculate the cross product n₁ × n₂:
n₁ × n₂ = <(2 3 - (-1 -1)), (-1 2 - 3 1), (1 -1 - 2 2)>
n₁ × n₂ = <(6 - 1), (-2 - 3), (-1 - 4)>
n₁ × n₂ = <5, -5, -5>
The vector <5, -5, -5> is perpendicular to both normal vectors and is therefore parallel to the line of intersection of the two planes. Any scalar multiple of this vector (e.g., <1, -1, -1>) is also a valid vector perpendicular to the two normal vectors and parallel to the intersection line.
Summary
In summary, to find a vector perpendicular to two planes (specifically, parallel to their line of intersection and perpendicular to their normal vectors), you calculate the cross product of their respective normal vectors.
