Final answer:
To find the angle between two planes, we need to find the angle between their normal vectors. The angle between the two planes x − 12y − 9z = 1 and 3x + 6y + 9z = 11 is approximately 92.86 degrees.
Step-by-step explanation:
To find the angle between two planes, we need to find the angle between their normal vectors.
The normal vector of the first plane is (1, -12, -9) and the normal vector of the second plane is (3, 6, 9).
Using the dot product formula, we can find the angle between the two normal vectors. The dot product of the normal vectors is (1)(3) + (-12)(6) + (-9)(9) = 3 - 72 - 81 = -150.
The magnitude of the first normal vector is sqrt(1^2 + (-12)^2 + (-9)^2) = sqrt(1 + 144 + 81) = sqrt(226) and the magnitude of the second normal vector is sqrt(3^2 + 6^2 + 9^2) = sqrt(9 + 36 + 81) = sqrt(126).
Using the formula cos(theta) = dot product / (magnitude of first vector * magnitude of second vector), we can find the angle theta.
cos(theta) = -150 / (sqrt(226) * sqrt(126)) = -150 / (sqrt(226) * 3 * sqrt(14)) = -150 / (3 * 1 * sqrt(14)) = -50 / sqrt(14)
Therefore, the angle between the two planes is cos^(-1)(-50 / sqrt(14)) ≈ 92.86 degrees.