To find the angle of intersection between two spheres at a given point, we need to find the normal vectors to the spheres at the given point and then use the dot product formula to find the angle between them.
The first sphere is centered at the origin and has a radius of sqrt(29). The equation of this sphere can be written as:
x^2 + y^2 + z^2 = 29
The second sphere has its center at (-2, 3, 4) and a radius of sqrt(29). We can write its equation as:
(x + 2)^2 + (y - 3)^2 + (z - 4)^2 = 29
To find the normal vector to each sphere at the point (-4, 3, 2), we can take the gradient of the left-hand side of each equation and evaluate it at that point. The gradient of a sphere centered at (a, b, c) with radius r is simply the vector (2(x-a), 2(y-b), 2(z-c)).
So, for the first sphere, the gradient at (-4, 3, 2) is:
(2(-4), 2(3), 2(2)) = (-8, 6, 4)
For the second sphere, the gradient at (-4, 3, 2) is:
(2(-4 - 2), 2(3 - 3), 2(2 - 4)) = (-12, 0, -4)
Now we can use the dot product formula to find the angle between these two normal vectors:
cos(theta) = (n1 . n2) / (|n1| |n2|)
where n1 and n2 are the normal vectors and |n1| and |n2| are their magnitudes.
n1 . n2 = (-8)(-12) + (6)(0) + (4)(-4) = 80
|n1| = sqrt((-8)^2 + 6^2 + 4^2) = 10
|n2| = sqrt((-12)^2 + 0^2 + (-4)^2) = 4*sqrt(2)
cos(theta) = 80 / (10 * 4sqrt(2)) = 2sqrt(2) / 5
Therefore, the angle of intersection between the two spheres at (-4, 3, 2) is:
theta = cos^-1(2*sqrt(2) / 5) = 49.47 degrees (rounded to two decimal places)