Final answer:
The equation of the plane through the point (6, -2, 6) and perpendicular to the vector -i + 4j + 5k is -x + 4y + 5z - 16 = 0.
Step-by-step explanation:
To find an equation of the plane that passes through the point (6, -2, 6) and is perpendicular to the vector -i + 4j + 5k, we can use the general equation of a plane:
A(x - x0) + B(y - y0) + C(z - z0) = 0
Where (x0, y0, z0) is a point on the plane, and the vector represents the normal to the plane. Here, the normal vector is given by the coefficients of the vector -i + 4j + 5k, which means A = -1, B = 4, and C = 5.
Substituting the point (6, -2, 6) into the equation and using the values of A, B, and C, we obtain:
-1(x - 6) + 4(y + 2) + 5(z - 6) = 0
Simplifying, we get:
-x + 6 + 4y + 8 + 5z - 30 = 0
Which further simplifies to:
-x + 4y + 5z - 16 = 0.
This is the equation of the plane in scalar form.