Question

Determine whether the planes are parallel, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle of intersection. $$ \begin{array}{l} x-3 y+6 z=4 \\ 5 x+y-z=4 \end{array} $$

Short answer

The given planes are neither parallel nor orthogonal. The angle of intersection between the two planes is approximately 97.3 degrees.

Step-by-step solution

Get The Normal Vectors of Each Plane

Finding the normal vectors of each plane is the first step. For the first plane, the normal vector \( \mathbf{N1} \) can be obtained from coefficients of x, y and z: \( \mathbf{N1} = \langle 1, -3, 6 \rangle \). Likewise, for the second plane, we derive the normal vector \( \mathbf{N2} \) as: \( \mathbf{N2} = \langle 5, 1, -1 \rangle \).

Check If The Planes Are Parallel

Planes are parallel if their normal vectors are scalar multiples of each other. It's plain to see that the normal vector of the second plane is not a scalar multiple of the first, thus these planes are not parallel.

Check If The Planes Are Orthogonal

If the dot product of the normal vectors of two planes is zero, then the planes are orthogonal. So we calculate the dot product \( \mathbf{N1} \cdot \mathbf{N2} = (1*5)+(-3*1)+(6*(-1)) = -4 \) which is not equal to zero. Therefore, the planes are not orthogonal.

Calculation of the Angle of Intersection

To find the angle of intersection theta \( \theta \) between the two planes, we can use the formula cos\( \theta \) = \( \frac{\mathbf{N1} \cdot \mathbf{N2}}{||\mathbf{N1}||\ ||\mathbf{N2}||} \). First, we need to calculate the magnitude of both normal vectors: \( ||\mathbf{N1}|| = \sqrt{(1)^2+(-3)^2+(6)^2} = \sqrt{46} \) and \( ||\mathbf{N2}|| = \sqrt{(5)^2+(1)^2+(-1)^2} = \sqrt{27} \). Now, substitute these values into the formula: cos\( \theta \) = \( \frac{-4}{\sqrt{46} \cdot \sqrt{27}} \) = -0.127. Taking the inverse cosine (arccos) of this value gives \( \theta \) in degrees, which is approximately 97.3 degrees.