Question

determine whether the following planes below are parallel, perpendicular, or neither.

1. ( x + 2y - 6z = 0 ) and ( -2x - 4y + 12z = -3 ).
2. ( x - 3y + z = 0 ) and ( -x - 4y + z = 5 ).
3. ( x + 6z = 0 ) and ( 6x - z = 2 ).

submit answer next item

Explanation:

Problem 1: Planes \( x + 2y - 6z = 0 \) and \( -2x - 4y + 12z = -3 \)

Step 1: Find normal vectors

For plane \( ax + by + cz = d \), normal vector is \( \langle a, b, c
angle \).
First plane: \( \mathbf{n_1} = \langle 1, 2, -6
angle \)
Second plane: \( \mathbf{n_2} = \langle -2, -4, 12
angle \)

Step 2: Check scalar multiple

See if \( \mathbf{n_2} = k\mathbf{n_1} \).
\( -2 = k(1) \implies k = -2 \)
\( -4 = k(2) \implies k = -2 \)
\( 12 = k(-6) \implies k = -2 \)
All components match \( k = -2 \), so normal vectors are parallel. Thus, planes are parallel.

Problem 2: Planes \( x - 3y + z = 0 \) and \( -x - 4y + z = 5 \)

Step 1: Find normal vectors

First plane: \( \mathbf{n_1} = \langle 1, -3, 1
angle \)
Second plane: \( \mathbf{n_2} = \langle -1, -4, 1
angle \)

Step 2: Check parallel (scalar multiple)

Check if \( \mathbf{n_2} = k\mathbf{n_1} \).
\( -1 = k(1) \implies k = -1 \)
\( -4 = k(-3) \implies k = \frac{4}{3} \) (not equal to \( -1 \)), so not parallel.

Step 3: Check perpendicular (dot product)

Dot product: \( \mathbf{n_1} \cdot \mathbf{n_2} = (1)(-1) + (-3)(-4) + (1)(1) = -1 + 12 + 1 = 12
eq 0 \).
Not perpendicular. Thus, neither.

Problem 3: Planes \( x + 6z = 0 \) (or \( x + 0y + 6z = 0 \)) and \( 6x - z = 2 \) (or \( 6x + 0y - z = 2 \))

Step 1: Find normal vectors

First plane: \( \mathbf{n_1} = \langle 1, 0, 6
angle \)
Second plane: \( \mathbf{n_2} = \langle 6, 0, -1
angle \)

Step 2: Check perpendicular (dot product)

Dot product: \( \mathbf{n_1} \cdot \mathbf{n_2} = (1)(6) + (0)(0) + (6)(-1) = 6 - 6 = 0 \).
Dot product is 0, so normal vectors are perpendicular. Thus, planes are perpendicular.

Answer:

s:

1. Parallel
2. Neither
3. Perpendicular