Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) §11.3 Cross Products and Determinants, #44 · Multivariable Calculus · Vectors & geometry

Lifting the triangle of Problem 42 up to the plane … gives…

These problems compute the areas of triangles.

Lifting the triangle of Problem 42 up to the plane $z = 1$ gives corners $(a_{1}, b_{1}, 1), (a_{2}, b_{2}, 1), (a_{3}, b_{3}, 1)$ . The area of the triangle times $\frac{1}{3}$ is the volume of the upside-down pyramid from $(0, 0, 0)$ to these corners. This pyramid volume is $\frac{1}{6}$ the box volume, so $\frac{1}{3}$ (area of triangle) $= \frac{1}{6}$ (volume of box):
$area of triangle = \frac{1}{2} a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} 111 .$
Find the area $A$ in Problem 43 from this determinant.

This exercise builds on another exercise in its source book, which isn't shown here.

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Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §11.3 Cross Products and Determinants, #44

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