Final answer:
The equation of the plane passing through the point (6, -2, 6) and perpendicular to the vector (-i + 4j + 5k) is –x + 4y + 5z = 20.
Step-by-step explanation:
The student is looking to find the equation of a plane passing through a given point and perpendicular to a given vector.
The equation of a plane can be expressed in the form ax + by + cz = d, where a, b, and c are the components of a normal vector to the plane, and d is a constant that can be found using the given point.
In this case, the normal vector is given as (−1, 4, 5), so our equation will start off as −x + 4y + 5z = d.
To find the value of d, we plug in the coordinates of the point (6, −2, 6) which gives us:
−(6) + 4(−2) + 5(6) = d, solving this we find that d = 20.
Therefore, the equation of the plane is −x + 4y + 5z = 20.