Section 12.4: Cross Product

SOLUTION In Example 3 we computed that \(\vec{PQ} \times \vec{PR} = \langle -40, -15, 15 \rangle\). The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product: \[ |\vec{PQ} \times \vec{PR}| = \sqrt{(-40)^2 + (-15)^2 + 15^2} = \sqrt{1600 + 225 + 225} = \sqrt{2050} = 5\sqrt{82} \] The area A of the triangle PQR is half the area of this parallelogram, that is, \(\frac{5}{2}\sqrt{82}\).