91Ó°ÊÓ

A normal vector is a fundamental concept when discussing planes in 3D geometry. It is essentially a vector that is perpendicular to a plane.
In mathematical terms, if we have a plane described by the equation
\(Ax + By + Cz = D\),
then
\((A, B, C)\)
is the normal vector of the plane.
This vector essentially dictates the orientation of the plane in three-dimensional space.

• A change in the components of the normal vector alters the plane's tilt and direction.
• The magnitude of the normal vector does not affect the angle of the plane, only its orientation matters.

The normal vector plays a crucial role in determining if two planes are perpendicular. When two planes are perpendicular, their normal vectors must also be perpendicular. This is where the dot product comes in handy.

The dot product is a way to multiply two vectors to get a scalar (a single number). In three dimensions, the dot product of vectors
\(\mathbf{a} = (a_1, a_2, a_3)\)
and
\(\mathbf{b} = (b_1, b_2, b_3)\)
is calculated as follows:

• \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

A key property of the dot product is that it can tell us about the angle between two vectors. If the dot product is zero, the vectors are perpendicular.
This is useful for plane geometry, where checking the perpendicularity of two normal vectors ensures the planes are at a right angle to one another.
In our problem, we found vectors
\(\mathbf{n} = (A, B, C)\)
and \(\mathbf{n}_M = (2, 3, 1)\)
to be perpendicular by confirming \(2A + 3B + C = 0\).
This relationship ensures that the plane with normal vector \(\mathbf{n}\)
is perpendicular to the plane \(M\).

The equation of a plane in three-dimensional space is pivotal to understanding how a plane is defined and how it interacts with other geometric entities.
A standard form equation of a plane is given by:

• \(Ax + By + Cz = D\)

where \((A, B, C)\)
represents the normal vector and \(D\)
is a constant that signifies how far the plane is from the origin.
In scenarios where the plane passes through the origin, like in our exercise, \(D = 0\).

• This special case results in a simplified equation: \(Ax + By + Cz = 0\).

Finding such an equation often requires identifying a normal vector that satisfies certain criteria, such as being perpendicular to another plane's normal vector.
This is achieved using the dot product as seen earlier. Once \(A, B,\) and \(C\) are determined, the equation can be easily written.
For instance, we determined \((3, -2, 0)\)
as a suitable normal vector,
resulting in the plane \(3x - 2y = 0\),
which is the perpendicular desired plane.