Final answer:
To find the angle of intersection between the planes 5x - 3y + 4 = 0 and 3x + 2y - 2 = 4, we need to find the normal vectors of both planes and use the dot product formula.
Step-by-step explanation:
To find the angle of intersection between the planes 5x - 3y + 4 = 0 and 3x + 2y - 2 = 4, we need to find the normal vectors of both planes.
The normal vector of the first plane is [5, -3, 0] and the normal vector of the second plane is [3, 2, 0].
The angle between the two planes can be found using the dot product formula: cos(theta) = (A · B) / (|A| |B|), where A and B are the normal vectors.
Substituting the values, cos(theta) = ([5, -3, 0] · [3, 2, 0]) / (|5, -3, 0| |3, 2, 0|).
Calculating the dot product and the magnitudes, cos(theta) = (15 + (-6)) / (sqrt(34) sqrt(13)).
Simplifying further, cos(theta) = 9 / (sqrt(442)).
Finally, taking the inverse cosine (arccos) of cos(theta), we get theta = 80.81 degrees.