Final answer:
The equation of the plane passing through point (1,0,4) and perpendicular to the line with direction vector (-2, -4, -1) is -2x - 4y - z + 6 = 0.
Step-by-step explanation:
The student is asking to find the equation of a plane that passes through a given point and is perpendicular to a given line. The direction vector of the line, which is perpendicular to the plane, can be found from the parametric equations of the line x=1-2t, y=5-4t, z=9-t. The coefficients of t in the line equations give us the direction vector of the line, which is (-2, -4, -1).
To find the equation of the plane, we can use the point-normal form since we have a normal vector (the direction vector of the line) and a point through which the plane passes, which is (1,0,4). The normal vector of the plane is the same as the direction vector of the line and thus is N = (-2, -4, -1). The plane equation in the point-normal form will be:
N · (X - X0) = 0
Substituting the normal vector and the point, we get:
(-2, -4, -1) · (x - 1, y - 0, z - 4) = 0
Which expands to:
-2(x - 1) - 4(y - 0) - 1(z - 4) = 0
By distributing and combining like terms, the final equation of the plane is:
-2x - 4y - z + 2 + 4 = 0
Or simplifying, we get:
-2x - 4y - z + 6 = 0