To determine the relationship between two planes, we need to find their normal vectors and then check their dot product. Let's write the equations of the planes in standard form:
Plane 1: 5x + 12y + 13z - 2 = 0 (normal vector = <5, 12, 13>)
Plane 2: -4x + 2y + 6z = 0 (normal vector = <-4, 2, 6>)
Now let's take the dot product of the two normal vectors:
<5, 12, 13> · <-4, 2, 6> = (5)(-4) + (12)(2) + (13)(6) = -20 + 24 + 78 = 82
Since the dot product is not zero, the planes are not perpendicular. To determine if they are parallel, we need to check if one normal vector is a scalar multiple of the other. Let's divide the dot product by the length of each normal vector:
|<5, 12, 13>| = sqrt(5^2 + 12^2 + 13^2) = sqrt(338)
|<-4, 2, 6>| = sqrt((-4)^2 + 2^2 + 6^2) = sqrt(56)
(82) / (sqrt(338) * sqrt(56)) ≈ 0.998
Since the result is very close to 1, we can conclude that the planes are almost parallel. The angle between them is approximately:
cos⁻¹(0.998) ≈ 0.06 radians ≈ 3.4 degrees