Question

Determine whether the planes are parallel, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle of intersection. $$ \begin{array}{l} 3 x+y-4 z=3 \\ -9 x-3 y+12 z=4 \end{array} $$

Short answer

The planes are parallel to each other, therefore they do not intersect.

Step-by-step solution

Recognize the normal vectors

The normal vector of a plane can be found from the coefficients of \(x\), \(y\), and \(z\). So the normal vectors of the given planes are \( \mathbf{N1} = (3, 1, -4) \) and \( \mathbf{N2} = (-9, -3, 12) \).

Determine whether the planes are parallel

Two planes are parallel if their normal vectors are parallel. This will be the case if one is a scalar multiple of the other. To determine this, compare the normal vectors. We can see \( \mathbf{N2} = -3\mathbf{N1} \), so the planes are parallel.

Confirm the planes are not orthogonal

Two planes are orthogonal if their normal vectors are orthogonal. The dot product of two orthogonal vectors is zero. Compute the dot product of the two normal vectors. The dot product is \( -27 - 3 - 48 \neq 0 \), so the planes are not orthogonal.

Angle of Intersection

Since the planes are parallel, they don't intersect. Therefore, there's no angle of intersection.