Find the angle of intersection of the plane y-3x-2z=-5 with the plane 4x-3y-2z=0 . Answer in radians: and in degrees:

To find the angle of intersection between two planes, we first need to find the normal vectors of the planes. The normal vector of a plane ax + by + cz = d is (a, b, c).

1. For the plane y - 3x - 2z = -5, the normal vector is (-3, 1, -2).
2. For the plane 4x - 3y - 2z = 0, the normal vector is (4, -3, -2).

The angle θ between two vectors u = (u1, u2, u3) and v = (v1, v2, v3) can be found using the dot product formula:

cos(θ) = (u . v) / (||u|| ||v||)

where:

• u . v = u1v1 + u2v2 + u3v3 is the dot product of u and v,
• ||u|| = sqrt(u1^2 + u2^2 + u3^2) is the magnitude of u,
• ||v|| = sqrt(v1^2 + v2^2 + v3^2) is the magnitude of v.

3. Calculate the dot product of the normal vectors: u . v = (-3)(4) + (1)(-3) + (-2)(-2) = -12 - 3 + 4 = -11

4. Calculate the magnitudes of the normal vectors: ||u|| = sqrt((-3)^2 + 1^2 + (-2)^2) = sqrt(9 + 1 + 4) = sqrt(14) ||v|| = sqrt(4^2 + (-3)^2 + (-2)^2) = sqrt(16 + 9 + 4) = sqrt(29)

5. Substitute these values into the dot product formula to find cos(θ): cos(θ) = (-11) / (sqrt(14) * sqrt(29)) = -11 / sqrt(406)

6. Use the arccos function to find θ: θ = arccos(-11 / sqrt(406))

In radians, this is approximately 2.36.

To convert this to degrees, use the formula:

degrees = radians * (180 / π)

So, θ ≈ 2.36 * (180 / π) ≈ 135.18 degrees.