Final answer:
The equation of the plane through the point (-1, 1/2, 3) and with a normal vector (i + 4j + k) is x + 4y + z - 4 = 0.
Step-by-step explanation:
To find an equation of a plane through a point (-1, 1/2, 3) with a normal vector (i + 4j + k), we use the point-normal form of a plane equation. We plug in the point for (x0, y0, z0) and the normal vector for (a, b, c) into the equation a(x - x0) + b(y - y0) + c(z - z0) = 0.
In this case, the normal vector gives us a = 1, b = 4, and c = 1. The point provides x0 = -1, y0 = 1/2, and z0 = 3. Thus, the equation of the plane becomes:
1(x - (-1)) + 4(y - (1/2)) + 1(z - 3) = 0
Simplifying this we get:
x + 4y + z + 1 - 2 - 3 = 0
Which further simplifies to the final form:
x + 4y + z - 4 = 0