Plane Parallel to a Given Plane (Cartesian Form)

In three-dimensional geometry, understanding the concept of parallel planes is crucial. Two planes are said to be parallel if they do not intersect at any point, no matter how far they are extended. This concept is similar to parallel lines in two-dimensional geometry. In Cartesian form, a plane is represented by a linear equation.

Equation of a Plane in Cartesian Form

The general equation of a plane in Cartesian coordinates is given by:

$$ Ax + By + Cz + D = 0 $$

where ( A ), ( B ), and ( C ) are the coefficients that represent the normal vector to the plane, and ( D ) is the constant term.

Conditions for Parallelism

For two planes to be parallel, their normal vectors must be parallel. This means that the normal vectors are either identical or scalar multiples of each other. If we have two planes:

$$ \text{Plane 1: } A_1x + B_1y + C_1z + D_1 = 0 $$ $$ \text{Plane 2: } A_2x + B_2y + C_2z + D_2 = 0 $$

Then, Plane 1 is parallel to Plane 2 if and only if:

$$ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} $$

However, the constant terms ( D_1 ) and ( D_2 ) do not need to be equal. If ( D_1 = D_2 ), the planes would not only be parallel but coincident (the same plane).

Table of Differences and Important Points

Aspect	Plane 1	Plane 2	Note on Parallelism
Normal Vector	( \vec{n_1} = (A_1, B_1, C_1) )	( \vec{n_2} = (A_2, B_2, C_2) )	Must be parallel (scalar multiples)
Constant Term	( D_1 )	( D_2 )	Can be different for parallel planes
Equation	( A_1x + B_1y + C_1z + D_1 = 0 )	( A_2x + B_2y + C_2z + D_2 = 0 )	Must have proportional coefficients

Examples

Example 1: Checking for Parallelism

Determine if the following planes are parallel:

Plane 1: ( 2x - 3y + 4z + 5 = 0 )

Plane 2: ( 4x - 6y + 8z + 10 = 0 )

Solution:

The normal vector of Plane 1 is ( (2, -3, 4) ), and the normal vector of Plane 2 is ( (4, -6, 8) ). We can see that ( \vec{n_2} = 2\vec{n_1} ), which means the normal vectors are parallel. Therefore, Plane 1 is parallel to Plane 2.

Example 2: Finding a Parallel Plane

Find the equation of a plane that is parallel to the plane ( 3x + 2y - z + 4 = 0 ) and passes through the point ( (1, -2, 3) ).

Solution:

Since the plane must be parallel, the normal vector will be the same, which is ( (3, 2, -1) ). We use the point to find the constant term ( D ):

$$ 3(1) + 2(-2) - 1(3) + D = 0 \ 3 - 4 - 3 + D = 0 \ D = 4 $$

The equation of the parallel plane is:

$$ 3x + 2y - z + 4 = 0 $$

Notice that the constant term ( D ) is the same because we used the equation of the given plane to find it. However, if we wanted the plane to pass through a different point, ( D ) would be different.

Formulas

• Equation of a plane: ( Ax + By + Cz + D = 0 )
• Condition for parallel planes: ( \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} )

By understanding these concepts and using the provided examples, one should be well-prepared to tackle problems involving planes parallel to a given plane in Cartesian form for exams.