Final answer:
The equation of the plane containing the points (0,6,0) and (-2,-3,4) and parallel to the ray with direction ratios (7,0,6) is -2x - 9y + 4z + 54 = 0.
Step-by-step explanation:
To find the equation of the plane that contains the points (0,6,0) and (-2,-3,4) and is parallel to the ray with direction ratios (7,0,6), we need to determine a normal vector to the plane. Since the plane is parallel to the given ray, the normal vector of the plane should be perpendicular to the direction ratios of the ray. We start by finding a vector that is parallel to the plane by subtracting the two given points:
- Vector P1P2: (-2 - 0, -3 - 6, 4 - 0) = (-2, -9, 4)
To find a normal vector, we can take the cross product of the Vector P1P2 and the given direction vector of the ray (7,0,6). However, in this problem, we see that the ray is already perpendicular to vector P1P2 since their scalar (dot) product is zero: (-2)×7 + (-9)×0 + 4×6 = 0. Therefore, we can directly use P1P2 as the normal vector.
Now we can use the normal vector and one of the given points to write the equation of the plane. With normal vector n = (-2, -9, 4) and point P1 (0,6,0), the equation of the plane is given by:
n·(r - P1) = 0
Where r = (x, y, z) represents any point on the plane, leading to:
-2(x - 0) - 9(y - 6) + 4(z - 0) = 0
Which simplifies to the plane equation:
-2x - 9y + 4z + 54 = 0