# Determine whether the planes are parallel, perpendicular, or neither. If the planes are neither parallel nor perpendicular, find the angle between them.

2x + 3y + z =                                                                                           3x – 2y – z = 3 The question aims to find whether the planes are parallel, perpendicular, or they are neither parallel nor perpendicular. In that case, the angle between them needs to be calculated.

Angle

Parallel line

The question depends on the concept of plane geometry and trigonometric ratios. The angle between two non-parallel planes is given as: $\cos \theta = \Big[ \dfrac{ A1A2 + B1B2 + C1C2 }{ \sqrt{ A1^2 + B1^2 + C1^2 } \sqrt{ A2^2 + B2^2 + C2^2 }} \Big]$

We have two equations of two planes, we need to check if they are parallel. We can check that by taking the ratio of coefficients of the respective variables. It is given as: $\dfrac{ A_1 }{ A_2 } = \dfrac{ B_1 }{ B_2 } = \dfrac{ C_1 }{ C_2 }$ Substituting the values, we get: $\dfrac{ 2 }{ 3 } = \dfrac{ 3 }{ -2 } = \dfrac{ 1 }{ -1 }$ $0.67 \ne – 1.5 \ne -1$ Thus, the planes are not parallel. To check if the planes are perpendicular to each other, we use the following formula: $n_1 . n_2 = 0$ Substituting the values, we get: $< 2, 3, 1 > . < 3, -2, -1 > = 0$ $2 . 3 + 3 . -2 + 1 . -1 = 0$ $6\ -\ 6\ -\ 1 = 0$ $-1 \ne 0$ The above equation proves that the planes are not perpendicular to each other. The angle between two planes is given as: $\cos \theta = \Big[ \dfrac{ A1A2 + B1B2 + C1C2 }{ \sqrt{ A1^2 + B1^2 + C1^2 } \sqrt{ A2^2 + B2^2 + C2^2 }} \Big]$ $\theta = \cos^{-1} \Big[ \dfrac{ A1A2 + B1B2 + C1C2 }{ \sqrt{ A1^2 + B1^2 + C1^2 } \sqrt{ A2^2 + B2^2 + C2^2 }} \Big]$ Substituting the values, we get: $\theta = \cos^{-1} \Big[ \dfrac{ (2)(3) + (3)(-2) + (1)(-1) }{ \sqrt{ 2^2 + 3^2 + 1^2 } \sqrt{ 3^2 + (-2)^2 + (-1)^2 }} \Big]$ $\theta = \cos^{-1} \Big[ \dfrac{ 6\ -\ 6\ -\ 1 }{ \sqrt{ 4 + 9 + 1 } \sqrt{ 9 + 4 + 1 }} \Big]$ $\theta = \cos^{-1} \Big[ \dfrac{ -1 }{ 2 \sqrt{ 14 }} \Big]$ $\theta = \cos^{-1} ( -0.1336)$ $\theta = 97.68^ {\circ}$

## Numerical Result

The given planes < 2, 3, 1 > and < 3, -2, -1 > are neither parallel nor perpendicular, and the angle between them is calculated to be: $\theta = 97.68^ {\circ}$

## Example

Find the angle between two planes given as follows: $x + y + 3z = 5$ $x\ + 2y\ -\ 3z = 4$ The formula for calculating the angle between two planes is given as: $\theta = \cos^ {-1} \Big[ \dfrac{ (1)(1) + (1)(2) + (3)(-3) }{ \sqrt{ 1^2 + 1^2 + 3^2 } \sqrt{ 1^2 + (2)^2 + (-3)^2 }} \Big]$ $\theta = \cos^ {-1} \Big[ \dfrac{ 1 + 2\ -\ 9 }{ 2 \sqrt {11} }$ $\theta = \cos^ {-1} (-0.9045)$ $\theta = 154.7^{\circ}$