HCA Mathematical Expositions: Determinants, cross products and the scalar triple product

© Gerard Sayson, 2025 1 Determinants, cross products and the scalar triple product Determinants, cross products and the scalar triple product Exploring the cross product in greater detail Gerard Sayson <geruls@broskiclan.org > January 27, 2025 Preface *looks at MF27* What is this crappy formula for the cross product? GERARD SAYSON Ask most H2 Mathematics students to give you the formula for a cross product. You’ll be seeing one of two types of these students: • The student can memorize and regurgitate the required answer. • They either scavenge through their formula booklet or are unable to come up with the answer. According to MF27, the official formula booklet, we have the following: [ 𝑎1 𝑎2 𝑎3 ] × [ 𝑏1 𝑏2 𝑐3 ] = [ 𝑎2𝑏3 − 𝑎3𝑏2 𝑎3𝑏1 − 𝑎1𝑏3 𝑎1𝑏2 − 𝑎2𝑏1 ] Would you look at that?! It’s hard to memorize , hard to derive and just outright confusing! Instead of that mess, what if I told you that there’s a way to memorize it much more efficiently? Don’t believe me? Have a look below: [ 𝑎1 𝑎2 𝑎3 ] × [ 𝑏1 𝑏2 𝑐3 ] = | 𝐢 𝐣 𝐤 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 | Now, you may be wondering: what is that? What are those vertical lines? No, they’re not the absolute value; you can’t take that for a matrix! Instead, we have what’s called the determinant. With the determinant, we can calculate something explicitly excluded from the syllabus: the scalar triple product , 𝐚 ⋅ (𝐛 × 𝐜). In fact, with the determinant, we can give it a geometric meaning and enhance our knowledge of vectors! This document was written in Microsoft Word instead of the usual LaTeX, because making diagrams in LaTeX is a pain! Also, check out https://lib.gsn.bz for more stuff.

© Gerard Sayson, 2025 2 Determinants, cross products and the scalar triple product What is a determinant? To put it simply, a determinant of a square matrix, denoted by det 𝐀, is a function of the square matrix which has some special properties. If 𝐀 is two-dimensional, we have det 𝐀 = |𝑎 𝑏 𝑐 𝑑| = 𝑎𝑑 − 𝑏𝑐 where 𝐀 = [𝑎 𝑏 𝑐 𝑑]. But what is the geometric significance? Suppose we have a vector 𝐯 = [𝑥 𝑦]. We define a linear transformation 𝑇 as a function that, for any vectors 𝐮 and 𝐯, the properties 𝑇(𝐮 + 𝐯) = 𝑇(𝐮) + 𝑇(𝐯) 𝑇(𝑐𝐮) = 𝑐𝑇(𝐮) are true. For example, the transformation 𝑥′ = 𝑥 cos(𝜃) − 𝑦 sin(𝜃) 𝑦′ = 𝑥 sin(𝜃) + 𝑦 cos(𝜃) is linear. The above transformation corresponds to rotating a point about the origin by 𝜃: Note that 𝑟 = √𝑥2 + 𝑦2 and tan(𝛼) = 𝑦 𝑥. This allows us to consider the angle and the length of the line from the origin. See that 𝑥′ = 𝑟 cos(𝛼 + 𝜃) = 𝑟 cos(𝛼) cos(𝜃) − 𝑟 sin(𝛼) sin(𝜃) = 𝑥 cos(𝜃) − 𝑦 sin(𝜃) 𝑦′ = 𝑟 sin(𝛼 + 𝜃) = 𝑟 sin(𝛼) cos(𝜃) + 𝑟 cos(𝛼) sin(𝜃) = 𝑥 sin(𝜃) + 𝑦 cos(𝜃). Is this a linear transformation? Yes, it is. Exercise 1. Verify that this is a linear transformation. Note that we can represent any linear transformation as a matrix: 𝑅𝜃 = [cos(𝜃) − sin(𝜃) sin(𝜃) cos(𝜃) ] ⟺ 𝑅𝜃 [𝑥 𝑦] = [𝑥 cos(𝜃) − 𝑦 sin(𝜃) 𝑥 sin(𝜃) + 𝑦 cos(𝜃)] This key result holds due to matrix multiplication. 𝜃 𝛼 𝑟 cos(𝛼) 𝑟 sin(𝛼)

© Gerard Sayson, 2025 3 Determinants, cross products and the scalar triple product Now, we are finally able to address the geometric significance of the determinant. Consider the unit square bounded by the vectors 𝐢 and 𝐣, as shown below. Applying a linear transformation 𝑇 to this, we see a parallelogram. What is the area of this parallelogram? Obviously, it is 2 × 1 2 |𝑇(𝐢)||𝑇(𝐣)| sin(𝜃) = |𝑇(𝐢)||𝑇(𝐣)| sin(𝜃). Now, suppose that 𝑇(𝐢) is the position vector of the point (𝑎, 𝑏) and 𝑇(𝐣) that of the point (𝑐, 𝑑). Due to our result from earlier, we can express: 𝑇 = [𝑎 𝑏 𝑐 𝑑]. Exercise 2. Verify that applying this to the unit vectors 𝐢 and 𝐣 yields the position vectors mentioned above. Considering the perpendicular vector ⊥ 𝑇(𝐢) = [−𝑏 𝑎 ], we can express the above area using the dot product, since |𝑇(𝐢)||𝑇(𝐣)| sin(𝜃) = |⊥ 𝑇(𝐢)||𝑇(𝐣)| sin(90° − 𝜃) = |⊥ 𝑇(𝐢)||𝑇(𝐣)| cos(𝜃) = (⊥ 𝑇(𝐢)) ⋅ 𝑇(𝐣) = [−𝑏 𝑎 ] ⋅ [𝑐 𝑑] = 𝑎𝑑 − 𝑏𝑐. Notice that we can now see the determinant representing the (signed) area of a parallelogram created by the vectors 𝑇(𝐢) and 𝑇(𝐣); furthermore, since the original area created by the two unit vectors was |1||1| sin(90°) = 1, the determinant can represent the change in area caused by a linear transformation. Exercise 3. Deduce that swapping the rows of a 2 × 2 matrix negates its determinant. Now, we define the determinant for a 3 × 3 matrix as follows: | 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 𝑔 ℎ 𝑖 | = 𝑎 |𝑒 𝑓 ℎ 𝑖 | − 𝑏 |𝑑 𝑓 𝑔 𝑖 | + 𝑐 |𝑑 𝑒 𝑔 ℎ|. Sorry if this definition is unmotivated; you can find a derivation of this online. 𝐢 𝐣 𝜃

© Gerard Sayson, 2025 4 Determinants, cross products and the scalar triple product The cross product with the determinant Recall that the cross product of two three -dimensional vectors 𝐚 and 𝐛 is defined as the vector perpendicular to both vectors such that the vector’s magnitude is the area of the parallelogram created by 𝐚 and 𝐛. We’ll have the following formula: 𝐚 × 𝐛 = |𝐚||𝐛| sin(θ) 𝐧̂. Here, 𝐧̂ is the unit vector perpendicular to the plane containing the two vectors, which follows the right hand rule (shown below). If we look at the unit vectors 𝐢, 𝐣, and 𝐤, we usually have the following: 𝐢 × 𝐣 = 𝐤 𝐣 × 𝐤 = 𝐢 𝐤 × 𝐢 = 𝐣. Since any vector can be defined as a linear combination (a sum) of these vectors, we’ll have 𝐚 × 𝐛 = (𝑎1𝐢 + a2𝐣 + a3𝐤) × (𝑏1𝐢 + b2𝐣 + b3𝐤) = (𝑎2𝑏3 − 𝑎3𝑏2)𝐢 − (𝑎1𝑏3 − 𝑎3𝑏1)𝐣 + (𝑎1𝑏2 − 𝑎2𝑏1)𝐤 = |𝑎2 𝑎3 𝑏2 𝑏3 | 𝐢 − |𝑎1 𝑎3 𝑏1 𝑏3 | 𝐣 + |𝑎1 𝑎2 𝑏1 𝑏2 | 𝐤 = | 𝐢 𝐣 𝐤 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 |. Remark. The full working is not shown here, for brevity. Because of Exercise 3, we can observe that 𝐚 × 𝐛 = −𝐛 × 𝐚. thumb: 𝐚 × 𝐛

© Gerard Sayson, 2025 5 Determinants, cross products and the scalar triple product The scalar triple product Finally, we get to the key highlight of this document: the scalar triple product. Let’s look at the scalar triple product first: 𝐚 ⋅ (𝐛 × 𝐜) = 𝐚 ⋅ | 𝐢 𝐣 𝐤 𝑏1 𝑏2 𝑏3 𝑐1 𝑐2 𝑐3 | = | 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 𝑐1 𝑐2 𝑐3 |. That’s cool and all, but what exactly does this mean? Well, we know that the cross product will give us the volume of the parallelogram created by 𝐛 and 𝐜 and extended into a prism by the vector 𝐚. Such a solid is called a parallelepiped, which is shown below: Using the dot (or scalar) product, see that 𝐚 ⋅ (𝐛 × 𝐜) = |𝐚||𝐛 × 𝐜| cos(𝜃) where 𝜃 is the angle between 𝐚 and the perpendicular to the plane containing 𝐛 and 𝐜 (that’s the definition of the cross product). As the diagram shows above, see that we can circularly shift the vectors and get the same volume! Remark. If you swap the vectors above (e.g. 𝐚 ⋅ (𝐜 × 𝐛)), the cross product has the property 𝐚 × 𝐛 = −𝐛 × 𝐚, implying that 𝐚 ⋅ (𝐜 × 𝐛) = −𝐚 ⋅ (𝐛 × 𝐜). 𝐛 𝐚 The vectors follow a right-hand rule. We have the result 𝐚 ⋅ (𝐛 × 𝐜) = 𝐜 ⋅ (𝐚 × 𝐛) = 𝐛 ⋅ (𝐜 × 𝐚) 𝜃 𝐜 𝐚 𝐜 𝐛 𝐛 𝐚 Base |𝐛 × 𝐜| |𝐚| cos(𝜃)

© Gerard Sayson, 2025 6 Determinants, cross products and the scalar triple product Notice that if 𝐚, 𝐚, and 𝐚 represented coplanar points, the volume of the parallelepiped that the three vectors create is zero. In fact, using the determinant, one has 𝐚 ⋅ (𝐛 × 𝐜) = | 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 𝑐1 𝑐2 𝑐3 | = 0. Using this fact, we’ll be able to find the Cartesian equation for the plane containing three points (𝑎1, 𝑎2, 𝑎3), (𝑏1, 𝑏2, 𝑏3) and (𝑐1, 𝑐2, 𝑐3). Let’s represent these coplanar points with the vectors 𝐚, 𝐛, and 𝐜, respectively. See that since the three points are coplanar, and if the point represented by the position vector 𝐫 lies on the plane, the vectors create a parallelepiped with zero volume. Particularly, (𝐫 − 𝐚) ⋅ [(𝐛 − 𝐚) × (𝐜 − 𝐚)] = 0. Remark. Another interpretation of this can be that the vector (𝐫 − 𝐚) is perpendicular to the cross product (𝐛 − 𝐚) × (𝐜 − 𝐚). In fact, at this point, the order of the vectors above does not matter, since they will still create a volume of zero! Therefore, the Cartesian equation of the plane containing the three po