Let the line be given by
(\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{2} = \lambda)
Then the coordinates of any point on the line are given by:
x = 3λ + 2
y = 4λ - 1
z = 2λ + 2
This point lies on the plane x - y + z - 5 = 0. Substituting the coordinates, we get:
(3λ + 2) - (4λ - 1) + (2λ + 2) - 5 = 0
3λ + 2 - 4λ + 1 + 2λ + 2 - 5 = 0
λ = 0
Therefore, the coordinates of the point of intersection are:
x = 3(0) + 2 = 2
y = 4(0) - 1 = -1
z = 2(0) + 2 = 2
The point of intersection is (2, -1, 2).
To find the angle between the line and the plane, we first find the direction ratios of the line, which are (3, 4, 2). The normal to the plane x - y + z - 5 = 0 is given by (1, -1, 1).
Let θ be the angle between the line and the normal to the plane. Then
cos θ = (\frac{(3, 4, 2) \cdot (1, -1, 1)}{\sqrt{3^2 + 4^2 + 2^2}\sqrt{1^2 + (-1)^2 + 1^2}})
cos θ = (\frac{3 - 4 + 2}{\sqrt{29}\sqrt{3}}) = (\frac{1}{\sqrt{87}})
θ = cos⁻¹(\frac{1}{\sqrt{87}})
Let ϕ be the angle between the line and the plane. Then
ϕ = 90° - θ
ϕ = 90° - cos⁻¹(\frac{1}{\sqrt{87}})