Find the equation of the plane passing through points P(1, 1, 1), Q(2, 3, 1), and R(3, 1, 4).
Answer and explanation
Correct answer: B. 6x - 3y - 4*z + 1 = 0
- Step 1: Vectors in the plane: PQ = (1, 2, 0) and PR = (2, 0, 3).
- Step 2: Normal vector n = PQ x PR = (23 - 0, 02 - 13, 10 - 2*2) = (6, -3, -4).
- Step 3: Plane equation: 6(x - 1) - 3(y - 1) - 4*(z - 1) = 0.
- Step 4: Expand: 6x - 6 - 3y + 3 - 4z + 4 = 0, which simplifies to 6x - 3y - 4z + 1 = 0.
Answer
6x - 3y - 4*z + 1 = 0.
Why not the others: 2x + y - z - 2 = 0 does not contain R (6 + 1 - 4 - 2 = 1 != 0); 3x - 2y + z - 2 = 0 does not contain Q (6 - 6 + 1 - 2 = -1 != 0); x + 2y + 3*z - 6 = 0 does not contain Q (2 + 6 + 3 - 6 = 5 != 0).
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