determine the angle between the planes 2x + 2y + 7z = 8 and 3x - 4y + 4z = 5
To determine the angle between two planes, we need to find the normal vectors of both planes and then calculate the angle between these two vectors.
Step 1: Convert the equations of the planes to the general form Ax + By + Cz + D = 0.
For the first plane, 2x + 2y + 7z = 8, the coefficients are A = 2, B = 2, C = 7, and D = -8.
For the second plane, 3x - 4y + 4z = 5, the coefficients are A = 3, B = -4, C = 4, and D = -5.
Step 2: Find the normal vectors of both planes.
For a plane in general form, the normal vector is given by (A, B, C). Therefore, the normal vector of the first plane is (2, 2, 7), and the normal vector of the second plane is (3, -4, 4).
Step 3: Calculate the angle between the two normal vectors.
The angle between two vectors can be found using the dot product formula:
cos(theta) = (v1 · v2) / (|v1| |v2|)
where v1 and v2 are the two vectors being compared, · denotes the dot product, and |v1| and |v2| represent the magnitudes of v1 and v2, respectively.
Let's substitute the values for v1 and v2:
v1 = (2, 2, 7)
v2 = (3, -4, 4)
Now, we can calculate the dot product:
v1 · v2 = (2 * 3) + (2 * -4) + (7 * 4) = 6 - 8 + 28 = 26
The magnitudes of the vectors are:
|v1| = sqrt(2^2 + 2^2 + 7^2) = sqrt(4 + 4 + 49) = sqrt(57)
|v2| = sqrt(3^2 + (-4)^2 + 4^2) = sqrt(9 + 16 + 16) = sqrt(41)
Substituting these values into the equation, we get:
cos(theta) = 26 / (sqrt(57) * sqrt(41))
Finally, we can calculate the angle theta by taking the inverse cosine (arccos) of cos(theta):
theta = arccos(26 / (sqrt(57) * sqrt(41)))
So, the angle between the planes 2x + 2y + 7z = 8 and 3x - 4y + 4z = 5 is equal to arccos(26 / (sqrt(57) * sqrt(41))).