Areas from the vector product, and the scalar triple product and the volume of a parallelepiped
Take the parallelogram with adjacent sides A and B. Its base is ‖A‖ and its height is ‖B‖ sin θ, so its area is ‖A‖‖B‖ sin θ = ‖A × B‖. The triangle on the same two sides is half of it: area = ½‖A × B‖. For a triangle PQR use A = PQ and B = PR. Now take three vectors A, B, C from one point. The scalar triple product (in Arabic الضرب الثلاثي القياسي) is A · (B × C): first do B × C, which gives a vector, then take its dot product with A, which gives a number. No brackets are needed in A · B × C because B × C must be done first. It is the 3 by 3 determinant with the three vectors as rows. Cyclic order does not change it: A · (B × C) = B · (C × A) = C · (A × B), but swapping two vectors changes the sign. Geometry: the three vectors are three edges of a parallelepiped (in Arabic متوازي السطوح), and its volume is |A · (B × C)|: the base area ‖B × C‖ times the height, which is the component of A along the direction of B × C. If the three vectors lie in one plane the solid is flat, so the triple product is 0, and this is the test for coplanar vectors. The scalar triple product is used in statics and in design, where volumes and moments are needed.
Volume of a brick. A brick of 2 by 3 by 5 units has edges along i, j, k: A = 2i, B = 3j, C = 5k. Check with the determinant that A · (B × C) = 30, which is the length times the width times the height. Then slant the top by changing C to (1, 1, 5): does the volume change? (It does not: the base and the height are the same.)