Final answer:
The equation of the plane through the points (3, -1, 1), (4, 0, 2), and (6, 3, 1) can be found by calculating normal vector using cross product of vectors formed by these points. With the normal vector and using one of the points, we arrive at the equation of the plane: -4x - 3y + 3z + 7 = 0.
Step-by-step explanation:
To find an equation of the plane through the points (3, -1, 1), (4, 0, 2), and (6, 3, 1), we first need to find two vectors that lie in the plane. These vectors can be found by subtracting the coordinates of the points in pairs. Let's use the points to find vectors AB and AC:
- Vector AB = (4, 0, 2) - (3, -1, 1) = (1, 1, 1)
- Vector AC = (6, 3, 1) - (3, -1, 1) = (3, 4, 0)
Now, we can find a normal vector to the plane by taking the cross product of AB and AC:
Normal vector, n = AB x AC = i(1*0 - 1*4) - j(1*3 - 1*0) + k(1*4 - 1*1) = i(-4) - j(3) + k(3)
We now have the normal vector n = (-4, -3, 3). Any point on the plane can be used in conjunction with the normal vector to find the equation of the plane. Using point A (3, -1, 1), the equation is -4(x - 3) -3(y + 1) +3(z - 1) = 0.
Expanding this equation and simplifying gives us the equation of the plane: -4x - 3y + 3z + 7 = 0.