Let A = (2, 2, 0) B = (1, – 1, 2) and C = (3, 1, 1) be three points in \mathbb{R} ^ 3 Denote P to be the plane passing through A, B, C.

(a) Find a vector vec n that is orthogonal (perpendicular) to the plane P.

(b) Find an equation of the plane P in the form

ax + by + cz = d

(c) Let D = (2, 1, 4) . Use projection to find the point E on the plane P closest to D.

Here’s a step-by-step solution:

(a) Finding the normal vector:

  1. Vectors in the plane: We can find two vectors lying in the plane P by taking the differences between the given points. Let’s use vectors AB and AC:
    AB = B – A = (1-2, -1-2, 2-0) = (-1, -3, 2)
    AC = C – A = (3-2, 1-2, 1-0) = (1, -1, 1)

  2. Cross product: A vector orthogonal to both AB and AC (and therefore to the plane P) is given by their cross product:
    n = AB x AC = (-1, -3, 2) x (1, -1, 1)
    Calculating the cross product:
    n = ((-3)(1) – (2)(-1), (2)(1) – (-1)(1), (-1)(-1) – (-3)(1))
    n = (-3 + 2, 2 + 1, 1 + 3) = (-1, 3, 4)

Therefore, the vector n = (-1, 3, 4) is orthogonal to the plane P.

(b) Finding the equation of the plane:

  1. Using the normal vector and a point: The equation of a plane can be written as ax + by + cz = d, where (a, b, c) is the normal vector and (x, y, z) represents any point on the plane. We can use any of the points A, B, or C. Let’s use A = (2, 2, 0):

  2. Substituting: Using n = (-1, 3, 4) and A = (2, 2, 0), we have:
    (-1)x + (3)y + (4)z = d
    -2 + 6 + 0 = d
    d = 4

So the equation of the plane is -x + 3y + 4z = 4.

(c) Finding the closest point E on the plane to D:

  1. Vector AD: We need the vector from point A (which lies in the plane) to point D:
    AD = D – A = (2-2, 1-2, 4-0) = (0, -1, 4)

  2. Projection of AD onto n: The projection of AD onto the normal vector n gives the vector along n from the plane to point D. We’ll call this projected vector “proj_n AD”:
    proj_n AD = (AD ⋅ n) / ||n||² * n
    AD ⋅ n = (0)(-1) + (-1)(3) + (4)(4) = -3 + 16 = 13
    ||n||² = (-1)² + 3² + 4² = 1 + 9 + 16 = 26
    proj_n AD = (13/26) * (-1, 3, 4) = (1/2)(-1, 3, 4) = (-1/2, 3/2, 2)

  3. Finding E: To find the point E on the plane closest to D, we subtract the projection from D:
    E = D – proj_n AD = (2, 1, 4) – (-1/2, 3/2, 2) = (2 + 1/2, 1 – 3/2, 4 – 2) = (5/2, -1/2, 2)

Final Answer: The final answers are $\boxed{n = (-1, 3, 4)}$, $\boxed{-x + 3y + 4z = 4}$, and $\boxed{E = (5/2, -1/2, 2)}$.