The area of the triangle ABC is r⁵/2fgh where OP=r and the correct option is B.
The area of the triangle ABC is given by:
Area = 1/2 * base * height
Where base is the length of the side BC and height is the length of the altitude from A to BC.
The side BC is the projection of OP onto the xy-plane. The length of the projection of OP onto the xy-plane is given by:
Projection of OP onto xy-plane = OP * cos(theta)
Where theta is the angle between OP and the xy-plane. The angle between OP and the xy-plane is equal to the angle between the vector OP and the vector normal to the xy-plane. The vector normal to the xy-plane is (1,0,0). The vector OP is (f,g,h). The dot product of these two vectors is given by:
OP . (1,0,0) = f
The cosine of the angle between two vectors is equal to the dot product of the vectors divided by the magnitude of the vectors. The magnitude of the vector OP is given by:
||OP|| = sqrt(f^2 + g^2 + h^2)
Therefore, the length of the projection of OP onto the xy-plane is given by:
Projection of OP onto xy-plane = f / sqrt(f^2 + g^2 + h^2)
The height of the triangle ABC is the length of the line segment from A to BC. The line segment from A to BC is parallel to the z-axis. Therefore, the length of the line segment from A to BC is equal to the z-coordinate of A. The z-coordinate of A is h. Therefore, the height of the triangle ABC is h.
Therefore, the area of the triangle ABC is given by:
Area = 1/2 * (f / sqrt(f^2 + g^2 + h^2)) * h
Simplifying, we get:
Area = fh / 2sqrt(f^2 + g^2 + h^2)
Therefore, the answer is B) r⁵/2fgh where OP=r.