Final answer:
To find a vector orthogonal to the plane through points P, Q, and R, calculate the cross product of vectors PQ and PR. The resultant vector, -i + 7j + 6k, is orthogonal to the plane.
Step-by-step explanation:
The student is asking to find a nonzero vector that is orthogonal to the plane going through three given points P(0, 0, -3), Q(4, 2, 0), and R(3, 3, 1). To solve this, we first need to find two vectors that are parallel to the plane by subtracting the coordinates of the points:
- Vector PQ: Q - P = (4, 2, 0) - (0, 0, -3) = (4, 2, 3)
- Vector PR: R - P = (3, 3, 1) - (0, 0, -3) = (3, 3, 4)
Next, we calculate the cross product of these two vectors to find a vector that is orthogonal to both, and thus to the plane:
PQ × PR = i(2 × 4 - 3 × 3) - j(4 × 4 - 3 × 3) + k(4 × 3 - 2 × 3)
= i(8 - 9) - j(16 - 9) + k(12 - 6)
= -i + 7j + 6k
Therefore, the nonzero vector orthogonal to the plane is -i + 7j + 6k.