Question

Determine whether the planes are parallel, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle of intersection. $$ \begin{array}{l} 3 x+2 y-z=7 \\ x-4 y+2 z=0 \end{array} $$

Short answer

The planes are not parallel or orthogonal. The angle of intersection is about 113.2 degrees.

Step-by-step solution

Determine the Normal Vectors

First, identify the coefficients of \(x\), \(y\), and \(z\) in each plane equation, they form the normal vector of each plane. For the first equation, the normal vector \(N_1\) is (3,2,-1) and for the second equation, the normal vector \(N_2\) is (1,-4,2).

Check if the Planes are Parallel

Two planes are parallel if their normal vectors are collinear, meaning that one vector is a scalar multiple of the other. Compare \(N_1\) and \(N_2\), it's clear that \(N_1\) is not a scalar multiple of \(N_2\), so the planes are not parallel.

Check if the Planes are Orthogonal

Two planes are orthogonal if their normal vectors are perpendicular, which means that the dot product of the normal vectors is zero. The dot product of \(N_1\) and \(N_2\) is \(3*1 + 2*(-4) + (-1)*2 = -5\). Since this is not zero, the planes are not orthogonal.

Find the Angle of Intersection

Since the planes are neither parallel nor orthogonal, we can calculate the angle of intersection using the dot product formula: \[ \cos(\Theta) = \frac{{N_1 \cdot N_2}}{{||N_1|| ||N_2||}} \]. The magnitudes of the normal vectors are \(||N_1|| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{14}\), and \( ||N_2|| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{21}\). Substituting the values in the dot product formula we get: \[ \cos(\Theta) = \frac{-5}{\sqrt{14} \sqrt{21}} \approx -0.392\]. Finally, find the angle by taking the inverse cosine (arccos) of the value, i.e., \( \Theta = \arccos(-0.392) = 113.2 ^{\circ}\).