Final answer:
To find the equation for the plane perpendicular to n = [1,-2,4] and passing through q(-4,5,6), use the point-normal form of a plane's equation. Plug the values into the formula 1(x + 4) - 2(y - 5) + 4(z - 6) = 0 and simplify to get the plane's equation x - 2y + 4z + 9 = 0.
Step-by-step explanation:
The student's question is seeking to find an equation for the plane that is perpendicular to n = [1,-2,4] and passes through the point q(-4,5,6). The vector n represents the normal vector to the plane, which gives the direction in which the plane is perpendicular. Given a point q and a normal vector n, the equation of the plane can be found using the point-normal form of the equation of a plane.
To write the equation of the plane, we will use the formula:
A(x - x0) + B(y - y0) + C(z - z0) = 0
where (A, B, C) are the components of the normal vector n, and (x0, y0, z0) are the coordinates of the point q the plane passes through.
Using the given normal vector n = [1, -2, 4] and point q(-4, 5, 6), we plug in the values into the formula to get:
1(x + 4) - 2(y - 5) + 4(z - 6) = 0
This simplifies to:
x - 2y + 4z + 9 = 0
This is the equation for the plane that is required.