The Area of a Parallelogram = 4.1 AREAS, VOLUMES, AND CROSS PRODUCTS 239 The parallelogram determined by two nonzero and nonparallel vectors a = [a1, a2] and b vertex at the origin, and we regard the arrows representing a and b as forming [b1, b2] in R2 is shown in Figure 4.1. This parallelogram has a the two sides of the parallelogram having the origin as a common vertex. We can find the area of this parallelogram by multiplying the length ||a|| of its base by the altitude h, obtaining Area = - ||a|| h ||a|| ||b||(sin 0) = ||a|| ||b|| V1 - cos20. Recall from page 24 of Section 1.2 that a b = ||a|| ||b||(cos 0). Squaring our area equation, we have (Area)² = ||a||2||b||2 — ||a||||b||² cos²0 = ||a||||b|| (a b) = (a₁² + a₂²)(b₁² + b₂²) - (a,b, + a2b₂)² = (a,b₂- a₂b₁)². (1) The last equality should be checked using pencil and paper. On taking square roots, we obtain = Areala,ba₂b₁l. The number within the absolute value bars is known as the determinant of the matrix A = [b₁ b₁₂ and is denoted by |A| or det(A), so that x2 h det(A) = = a, a₂ b₁ b₂ ° x1 FIGURE 4.1 The parallelogram determined by a and b.