Question

Consider the points below.

P(0, −2, 0), Q(5, 1, −3), R(5, 2, 1).
(a) Find a nonzero vector orthogonal to the plane through the points �, �, and �.
(b) Find the area of the triangle △PQR.

Answer 1

A non-zero vector orthogonal to the plane through the points P, Q, and R is (-15, -20, 5). The area of the triangle △PQR is approximately 12.72 square units.

Step-by-step explanation:

To solve part (a) and find a non-zero vector orthogonal to the plane through the points P, Q, and R, we first need two vectors that lie within the plane. We can take the vector PQ and vector PR, which we get by subtracting the coordinates of P from Q and R, respectively. The vector PQ is (5 - 0, 1 - (-2), -3 - 0) = (5, 3, -3), and PR is (5 - 0, 2 - (-2), 1 - 0) = (5, 4, 1). The cross product of these two vectors will give us a vector orthogonal to the plane, so PQ × PR = (3*-1 - (-3)*4, -3*5 - 1*5, 5*4 - 3*5) = (-3 - 12, -15 - 5, 20 - 15), which simplifies to (-15, -20, 5).

To find the area of the triangle △PQR in part (b), we can use half the magnitude of the cross product, since the area of the parallelogram formed by PQ and PR is the magnitude of the cross product, and a triangle is half a parallelogram. The magnitude of the cross product (PQ × PR) is √((-15)^2 + (-20)^2 + 5^2) = √(225 + 400 + 25) = √(650). So, the area of the triangle is ½*√(650), which roughly equals 12.72 square units.