Questions: Determine whether the following planes below are parallel, perpendicular, or neither. choose one 1. x+2y-6z=0 and -3x-6y+18z=-3. choose one 2. x-3y+z=0 and -x-2y+z=5. choose one 3. x+9z=0 and 9x-z=1. Note: You only have two attempts at this problem.

Solution Steps

To determine whether two planes are parallel, perpendicular, or neither, we need to compare their normal vectors. The normal vector of a plane given by \(ax + by + cz = d\) is \((a, b, c)\).

1. For the planes \(x + 2y - 6z = 0\) and \(-3x - 6y + 18z = -3\):

   • Extract the normal vectors and check if they are scalar multiples (parallel) or if their dot product is zero (perpendicular).

2. For the planes \(x - 3y + z = 0\) and \(-x - 2y + z = 5\):

   • Extract the normal vectors and check if they are scalar multiples (parallel) or if their dot product is zero (perpendicular).

3. For the planes \(x + 9z = 0\) and \(9x - z = 1\):

   • Extract the normal vectors and check if they are scalar multiples (parallel) or if their dot product is zero (perpendicular).

For the planes given by the equations \(x + 2y - 6z = 0\) and \(-3x - 6y + 18z = -3\), we extract their normal vectors:

• Normal vector of Plane 1: \(\mathbf{n_1} = (1, 2, -6)\)
• Normal vector of Plane 2: \(\mathbf{n_2} = (-3, -6, 18)\)

To check if the planes are parallel, we see if \(\mathbf{n_2} = k \cdot \mathbf{n_1}\) for some scalar \(k\). Here, we find that: \[ \mathbf{n_2} = -3 \cdot \mathbf{n_1} \] Thus, the planes are parallel.

For the planes given by the equations \(x - 3y + z = 0\) and \(-x - 2y + z = 5\), we extract their normal vectors:

• Normal vector of Plane 3: \(\mathbf{n_3} = (1, -3, 1)\)
• Normal vector of Plane 4: \(\mathbf{n_4} = (-1, -2, 1)\)

To check if the planes are parallel, we see if \(\mathbf{n_4} = k \cdot \mathbf{n_3}\). In this case, there is no scalar \(k\) such that \(\mathbf{n_4}\) is a multiple of \(\mathbf{n_3}\). Additionally, we compute the dot product: \[ \mathbf{n_3} \cdot \mathbf{n_4} = 1 \cdot (-1) + (-3) \cdot (-2) + 1 \cdot 1 = -1 + 6 + 1 = 6 \neq 0 \] Thus, the planes are neither parallel nor perpendicular.

For the planes given by the equations \(x + 9z = 0\) and \(9x - z = 1\), we extract their normal vectors:

• Normal vector of Plane 5: \(\mathbf{n_5} = (1, 0, 9)\)
• Normal vector of Plane 6: \(\mathbf{n_6} = (9, 0, -1)\)

To check if the planes are perpendicular, we compute the dot product: \[ \mathbf{n_5} \cdot \mathbf{n_6} = 1 \cdot 9 + 0 \cdot 0 + 9 \cdot (-1) = 9 - 9 = 0 \] Since the dot product is zero, the planes are perpendicular.

Final Answer