Question

Describe a method for determining when two planes \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) are (a) parallel and (b) perpendicular. Explain your reasoning.

Short answer

Parallelism of two planes can be determined if their normal vectors are proportional. This condition can be expressed with the equation \( \vec{n1}=k \cdot \vec{n2} \). Perpendicularity of two planes can be determined if the dot (scalar) product of their normal vectors equals zero, which is expressed as \( \vec{n1} \cdot \vec{n2} = 0 \).

Step-by-step solution

Derive Normal Vectors

In 3D geometry, the vector normal to a plane is represented by the coefficients of x, y, and z in the equation of the plane. So, for the two given equations the normal vectors can be given as \( \vec{n1} = (a_{1}, b_{1}, c_{1}) \) for the first plane and \( \vec{n2} = (a_{2}, b_{2}, c_{2}) \) for the second plane.

Condition for Parallel Planes

Two planes are parallel if their normal vectors are proportional, i.e., one vector is a scalar multiple of the other. This can be expressed as \( \vec{n1}=k \cdot \vec{n2} \) for some scalar k. So, the conditions for the planes to be parallel are \( a_{1}=k \cdot a_{2} \), \( b_{1}=k \cdot b_{2} \), and \( c_{1}=k \cdot c_{2} \). These conditions need to be checked for the coefficients of the plane equations.

Condition for Perpendicular Planes

Two planes are perpendicular if the dot (scalar) product of their normal vectors equals zero. This can be expressed as \( \vec{n1} \cdot \vec{n2} = 0 \). So, the condition for the planes to be perpendicular is \( a_{1} \cdot a_{2} + b_{1} \cdot b_{2} + c_{1} \cdot c_{2} = 0 \). This is the equation to check if the planes are perpendicular.