Question

To determine

Whether the planes \(x + y + z = 1\) and \(x - y + z = 1\) are parallel, perpendicular, or neither.

Short answer

The planes \(x + y + z = 1\) and \(x - y + z = 1\) is neither parallel nor perpendicular.

Step-by-step solution

Concept of Parallel plane, Perpendicular plane, scalar multiple and dot product

The normal vector of one of the planes is the scalar multiple of the normal vector of the other plane if the two planes are parallel.

The dot product of the normal vectors of the two planes is\({\bf{0}}\)if they are perpendicular to each other (the normal vectors of the planes must be orthogonal).

Scalar multiple of vector

Multiply each component of a vector by the scalar to multiply a vector by a scalar. If\({\bf{u}} = {\bf{u1}},{\rm{ }}{\bf{u2}}\)has magnitude\(\left| {\bf{u}} \right|\)and direction\(d\), then\({\bf{nu}} = {\bf{nu1}},{\rm{ }}{\bf{u2}} = {\bf{nu1}},{\rm{ }}{\bf{nu2}}\), where\({\bf{n}}\)is a positive real integer, magnitude\(\left| {{\bf{nu}}} \right|\), and direction\(d\).

Dot product of vector

The product of the magnitudes of the two vectors and the cosine of the angle between the two vectors is called the dot product of vectors. When two vectors are dot product, the resulting is in the same plane as the two vectors. The true worth of a dot product can be either positive or negative.

Explanation of Parallel plane with the help of Scalar multiple.

The planes' equations are as follows: \(x + y + z = 1\) and \(x - y + z = 1\).

The normal vector from the plane equation \(x + y + z = 1\) is \({{\rm{n}}_1} = \langle 1,1,1\rangle \).

The normal vector from the plane equation \(x - y + z = 1\) is \({{\rm{n}}_2} = \langle 1, - 1,1\rangle \).

From the normal vectors of the planes, it is clear that the normal vector of one of the planes is not the scalar multiple of another plane.

\({{\rm{n}}_1} \ne k{{\rm{n}}_2}\)

Therefore, the planes are not parallel.

Explanation of perpendicular plane with the help of dot product

Calculate the dot product of the normal vectors \({{\rm{n}}_1}\) and \({{\rm{n}}_2}\).

\(\begin{array}{l}{{\rm{n}}_1} \cdot {{\rm{n}}_2} = \langle 1,1,1\rangle \cdot \langle 1, - 1,1\rangle \\{{\rm{n}}_1} \cdot {{\rm{n}}_2} = (1)(1) + (1)( - 1) + (1)(1)\\{{\rm{n}}_1} \cdot {{\rm{n}}_2} = 1 - 1 + 1\\{{\rm{n}}_1} \cdot {{\rm{n}}_2} = 1 \ne 0\end{array}\)

As the dot product of the normal vectors of two planes is not zero, the planes are not perpendicular to each other

Thus, the planes \(x + y + z = 1\) and \(x - y + z = 1\) is neither parallel nor perpendicular.