Linear Mappings and Their Matrices
3.10 The Cross Product, Lines, and Planes in R 3
Generally in Rn there is no natural way to associate to a pair of vectors u and v a third vector. In R3, however, the plane specified by u and v has only one orthogonal direction, i.e., dimension 3 is special because 3− 2 = 1. In R3 a normal vector to u and v can be specified by making suitable conventions on its orientation viz a viz the other two vectors, and on its length. This will give a vector-valued product of two vectors that is special to three-dimensional space, called the cross product. The first part of this section develops these ideas.
Given any two vectors u, v∈ R3, we want their cross product u× v ∈ R3 to be orthogonal to u and v,
u× v ⊥ u and u × v ⊥ v. (3.11)
There is the question of which way u×v should point along the line orthogonal to the plane spanned by u and v. The natural answer is that the direction should be chosen to make the ordered triple of vectors{u, v, u × v} positive unless it is degenerate,
det(u, v, u× v) ≥ 0. (3.12)
Also there is the question of how long u× v should be. With hindsight, we assert that specifying the length to be the area of the parallelogram spanned by u and v will work well. That is,
3.10 The Cross Product, Lines, and Planes in R3 127
|u × v| = area P(u, v). (3.13)
The three desired geometric properties (3.11) through (3.13) seem to describe the cross product completely. (See figure 3.22.)
u v
Figure 3.22.The cross product of u and v
The three geometric properties also seem disparate. However, they combine into a uniform algebraic property, as follows. Since the determinant in (3.12) is nonnegative, it is the volume of the parallelepiped spanned by u, v, and u× v.
The volume is the base times the height, and since u× v is normal to u and v the base is the area ofP(u, v) and the height is |u × v|. Thus
det(u, v, u× v) = area P(u, v) |u × v|.
It follows from the previous display and (3.13) that
|u × v|2= det(u, v, u× v).
Since orthogonal vectors have inner product 0, since the determinant is 0 when two rows agree, and since the square of the absolute value is the vector’s inner product with itself, we can rewrite (3.11) and this last display (obtained from (3.12) and (3.13)) uniformly as equalities of the formhu × v, wi = det(u, v, w) for various w,
hu × v, ui = det(u, v, u), hu × v, vi = det(u, v, v),
hu × v, u × vi = det(u, v, u × v).
(3.14)
Instead of saying what the cross product is, as an equality of the form u× v = f (u, v) would, the three equalities of (3.14) say how the cross product interacts with certain vectors—including itself—via the inner product. Again, the idea is to characterize rather than construct.
(The reader may object to the argument just given that det(u, v, u× v) = areaP(u, v) |u × v|, on the grounds that we don’t really understand the area of a 2-dimensional parallelogram in 3-dimensional space to start with, that
in R3 we measure volume rather than area, and the parallelogram surely has volume zero. In fact, the argument can be viewed as motivating the formula as the definition of the area. This idea will be discussed more generally in section 9.1.)
Based on (3.14), we leap boldly to an intrinsic algebraic characterization of the cross product.
Definition 3.10.1 (Cross Product). Let u and v be any two vectors in R3. Their cross product u× v is defined by the property
hu × v, wi = det(u, v, w) for all w ∈ R3.
That is, u× v is the unique vector x ∈ R3 such that hx, wi = det(u, v, w) for all w∈ R3.
As with the determinant earlier, we do not yet know that the characterizing property determines the cross product uniquely, or even that a cross product that satisfies the characterizing property exists at all. But also as with the determinant, we defer those issues and first reap the consequences of the characterizing property with no reference to an unpleasant formula for the cross product. Of course the cross product will exist and be unique, but for now the point is that graceful arguments with its characterizing property show that it has all the further properties that we want it to have.
Proposition 3.10.2 (Properties of the Cross Product).
(CP1) The cross product is skew-symmetric: v× u = −u × v for all u, v ∈ R3. (CP2) The cross product is bilinear: For all scalars a, a′, b, b′ ∈ R and all
vectors u, u′, v, v′∈ R3,
(au + a′u′)× v = a(u × v) + a′(u′× v), u× (bv + b′v′) = b(u× v) + b′(u× v′).
(CP3) The cross product u× v is orthogonal to u and v.
(CP4) u× v = 0 if and only if u and v are collinear (meaning that u = av or v = au for some a∈ R).
(CP5) If u and v are not collinear then the triple{u, v, u × v} is right-handed.
(CP6) The magnitude |u × v| is the area of the parallelogram spanned by u and v.
Proof. (1) This follows from the skew-symmetry of the determinant. For any w∈ R3,
hv × u, wi = det(v, u, w) = − det(u, v, w) = −hu × v, wi = h−u × v, wi.
Since w is arbitrary, v× u = −u × v.
(2) For the first variable, this follows from the linearity of the determinant in its first row-vector variable and the linearity of the inner product in its first vector variable. Fix a, a′ ∈ R, u, u′, v∈ R3. For any w∈ R3,
3.10 The Cross Product, Lines, and Planes in R3 129 h(au + a′u′)× v, wi = det(au + a′u′, v, w)
= a det(u, v, w) + a′det(u′, v, w)
= ahu × v, wi + a′hu′× v, wi
=ha(u × v) + a′(u′× v), wi.
Since w is arbitrary, (au + a′u′)× v = a(u × v) + a′(u′× v). The proof for the second variable follows from the result for the first variable and from (1).
(3) hu × v, ui = det(u, v, u) = 0 since the determinant of a matrix with two equal rows vanishes. Similarly,hu × v, vi = 0.
(4) If u = av then for any w∈ R3,
hu × v, wi = hav × v, wi = det(av, v, w) = a det(v, v, w) = 0.
Since w is arbitrary, u× v = 0. And similarly if v = au.
Conversely, suppose that u and v are not collinear. Then they are linearly independent, and so no element of R3 can be written as a linear combination of u and v in more than one way. The set{u, v} is not a basis of R3, because every basis consists of three elements. Since no elements of R3can be written as a linear combination of u and v in more than one way, and since {u, v}
is not a basis, the only possibility is that some w ∈ R3 can not be written as a linear combination of u and v at all. Thus the set {u, v, w} is a linearly independent set of three elements, making it a basis of R3. Compute that since{u, v, w} is a basis,
hu × v, wi = det(u, v, w) 6= 0.
Therefore u× v 6= 0.
(5) By (4), u× v 6= 0, so 0 < hu × v, u × vi = det(u, v, u × v). By the results on determinants and orientation,{u, v, u × v} is right-handed.
(6) By definition,|u × v|2=hu × v, u × vi = det(u, v, u × v). As discussed earlier in the section, det(u, v, u× v) = areaP(u, v) |u × v|. The result follows from dividing by|u × v| if it is nonzero, and from (4) otherwise. ⊓⊔ Now we show that the characterizing property determines the cross prod-uct uniquely. The idea is that a vector’s inner prodprod-ucts with all other vectors completely describe the vector itself. The observation to make is that for any vector x∈ Rn (n need not be 3 in this paragraph),
ifhx, wi = 0 for all w ∈ Rn then x = 0n.
To justify the observation, specialize w to x to show thathx, xi = 0, giving the result since 0n is the only vector whose inner product with itself is 0. (Here we use the nontrivial direction of the degeneracy condition in the positive definiteness property of the inner product.) In consequence of the observation, for any two vectors x, x′∈ Rn,
ifhx, wi = hx′, wi for all w ∈ Rn then x = x′.
That is, the inner product valueshx, wi for all w ∈ Rnspecify x, as anticipated.
To prove that the cross product exists, it suffices to write a formula for it that satisfies the characterizing property in Definition 3.10.1. Since we need
hu × v, e1i = det(u, v, e1), hu × v, e2i = det(u, v, e2), hu × v, e3i = det(u, v, e3), the only possible formula is
u× v = (det(u, v, e1), det(u, v, e2), det(u, v, e3)).
This formula indeed satisfies the definition because by definition of the inner product and then by the linearity of the determinant in its third argument we have for any w = (w1, w2, w3)∈ R3,
hu × v, wi = det(u, v, e1)· w1+ det(u, v, e2)· w2+ det(u, v, e3)· w3
= det(u, v, w1e1+ w2e2+ w3e3)
= det(u, v, w).
In coordinates, the formula for the cross product is
u× v = (det
A bit more conceptually, the cross product formula in coordinates is
u× v = det
The previous display is only a mnemonic device: strictly speaking, it doesn’t lie within our grammar because the entries of the bottom row are vectors rather than scalars. But even so, its two terms u1v2e3− u2v1e3 do give the third entry of the cross product, and similarly for the others. In chapter 9, where we will have to compromise our philosophy of working intrinsically rather than in coordinates, this formula will be cited and generalized. In the meantime its details are not important except for mechanical calculations, and we want to use it as little as possible, as with the determinant earlier.
Indeed, the display shows that the cross product is essentially a special case of the determinant.
It is worth knowing the cross products of the standard basis pairs,
3.10 The Cross Product, Lines, and Planes in R3 131 e1× e1= 03, e1× e2= e3, e1× e3=−e2,
e2× e1=−e3, e2× e2= 03, e2× e3= e1, e3× e1= e2, e3× e2=−e1, e3× e3= 03.
Here ei× ej is 03 if i = j, and ei× ej is the remaining standard basis vector if i6= j and i and j are in order in the diagram
2
1
//
3
YY
and ei× ej is minus the remaining standard basis vector if i6= j and i and j are out of order in the diagram.
The remainder of this section describes lines and planes in R3.
A line in R3 is determined by a point p and a direction vector d. (See figure 3.23.) A point q lies on the line exactly when it is a translation from p by some multiple of d. Therefore,
ℓ(p, d) ={p + td : t ∈ R}.
In coordinates, a point (x, y, z) lies on ℓ((xp, yp, zp), (xd, yd, zd)) exactly when x = xp+ txd, y = yp+ tyd, z = zp+ tzd for some t∈ R.
If the components of d are all nonzero then the relation between the coordi-nates can be expressed without the parameter t,
x− xp
xd =y− yp
yd =z− zp zd .
For example, the line through (1, 1, 1) in the direction (1, 2, 3) consists of all points (x, y, z) satisfying x = 1 + t, y = 1 + 2t, z = 1 + 3t for t ∈ R, or, equivalently, satisfying x− 1 = (y − 1)/2 = (z − 1)/3.
A plane in R3 is determined by a point p and a normal (orthogonal) vector n. (See figure 3.24.) A point x lies on the plane exactly when the vector from p to x is orthogonal to n. Therefore,
P (p, n) ={x ∈ R3:hx − p, ni = 0}.
In coordinates, a point (x, y, z) lies on P ((xp, yp, zp), (xn, yn, zn)) exactly when (x− xp)xn+ (y− yp)yn+ (z− zp)zn= 0.
p + d p
Figure 3.23.Line in R3
p n
Figure 3.24.Plane in R3
Exercises
3.10.1. Evaluate (2, 0,−1) × (1, −3, 2).
3.10.2. Suppose that a vector v ∈ R3 takes the form v = u1× e1 = u2× e2 for some u1 and u2. Describe v.
3.10.3. True or false: For all u, v, w in R3, (u× v) × w = u × (v × w).
3.10.4. Express (u + v)× (u − v) as a scalar multiple of u × v.
3.10.5. Investigate the extent to which a cancellation law holds for the cross product, as follows: for fixed u, v in R3 with u 6= 0, describe the vectors w satisfying the condition u× v = u × w.
3.10.6. What is the line specified by two points p and p′?
3.10.7. Give conditions on the points p, p′ and the directions d, d′ so that ℓ(p, d) = ℓ(p′, d′).
3.10.8. Express the relation between the coordinates of a point on ℓ(p, d) if the x-component of d is 0.
3.11 Summary 133 3.10.9. What can you conclude about the lines
x− xp
xd = y− yp
yd = z− zp
zd and x− xp
xD = y− yp
yD = z− zp zD
given that xdxD+ ydyD+ zdzD = 0? What can you conclude if xd/xD = yd/yD= zd/zD?
3.10.10. Show that ℓ(p, d) and ℓ(p′, d′) intersect if and only if the linear equa-tion Dt = ∆p is solvable, where D∈ M3,2(R) has columns d and d′, t is the column vectort1
t2
, and ∆p = p′−p. For what points p and p′do ℓ(p, (1, 2, 2)) and ℓ(p′, (2,−1, 4)) intersect?
3.10.11. Use vector geometry to show that the distance from the point q to the line ℓ(p, d) is
|(q − p) × d|
|d| .
(Hint: what is the area of the parallelogram spanned by q− p and d?) Find the distance from the point (3, 4, 5) to the line ℓ((1, 1, 1), (1, 2, 3)).
3.10.12. Show that the time of nearest approach of two particles whose po-sitions are s(t) = p + tv, ˜s(t) = ˜p + t˜v is t = −h∆p, ∆vi/|∆v|2. (You may assume that the particles are at their nearest approach when the difference of their velocities is orthogonal to the difference of their positions.)
3.10.13. Write the equation of the plane through (1, 2, 3) with normal direc-tion (1, 1, 1).
3.10.14. Where does the plane x/a + y/b + z/c = 1 intersect each axis?
3.10.15. Specify the plane containing the point p and spanned by directions d and d′. Specify the plane containing the three points p, q, and r.
3.10.16. Use vector geometry to show that the distance from the point q to the plane P (p, n) is
|hq − p, ni|
|n| .
(Hint: Resolve q − p into components parallel and normal to n.) Find the distance from the point (3, 4, 5) to the plane P ((1, 1, 1), (1, 2, 3)).
3.11 Summary
Linear algebra is easy in the abstract since the vector space operations pass through linear mappings, and it is easy in the concrete since mechanical ma-trix manipulations are straightforward. While the mama-trix methods from this chapter are handy computational tools, it is also crucial to understand the intrinsic notion of a linear mapping: this is the idea that we will use to define multivariable differentiation in the next chapter.