5.3 The Cross Product in R3
Definition 5.3.1 Let ~u = [u1, u2, u3] and ~v = [v1, v2, v3]. Then the vector given by [u2v3− u3v2, u3v1− u1v3, u1v2− u2v1]
is called the cross product (or vector product) of ~u and ~v and denoted ~u× ~v.
Example 5.3.2 Suppose ~u = [1, 2,−1] and ~v = [2, 1, 0]. Find ~u × ~v and show it is orthogonal to both ~u and ~v.
Solution:
~u× ~v = [u2v3− u3v2, u3v1− u1v3, u1v2− u2v1]
= [2(0)− (−1)(1), −1(2) − (1)(0), 1(1) − 2(2)]
= [1,−2, −3]
~u.(~u× ~v) = [1, 2, −1].[1, −2, −3] = 1 − 4 + 3 = 0
~v.(~u× ~v) = [2, 1, 0].[1, −2, −3] = 2 − 2 + 0 = 0 Thus ~u⊥ ~u × ~v and ~v ⊥ ~u × ~v.
This is a general feature of the cross product : it is often used to produce a vector which is orthogonal to two given ones.
Note: If ~u and ~v have the same initial point, there is exactly one plane inR3 which contains the line segments representing both of them. Since ~u× ~v is orthogonal to both ~u and ~v, it is normal to this plane. Hence the cross product provides us with the means to finish Example 6.2.6∗.
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~u
~v
~u× ~v
Back to Example 5.2.6: P contains the points A(1, 2, 1), B(2, 4, 1) and C(−1, 0, 3), so the vec- tors ~AB = [1, 2, 0] and ~AC = [−2, −2, 2] lie within P . A vector ~n which is normal to P is orthogonal to both ~AB and ~AC.
Let ~n = ~AB× ~AC = [1, 2, 0]× [−2, −2, 2].
~u× ~v = [u2v3− u3v2, u3v1− u1v3, u1v2− u2v1]
=⇒ [1, 2, 0] × [−2, −2, 2] = [2(2) − 0(−2), 0(−2) − 1(2), 1(−2) − 2(−2)]
= [4,−2, 2]
(Check this is orthogonal to both ~AB and ~AC).
So ~n = [4,−2, 2] is normal to P :
P : 4x− 2y + 2z = d
A(1, 2, 1) belongs to P =⇒ 4(1) − 2(2) + 2(1) = d : d = 2; P : 4x − 2y + 2z = 2 Equation of P : 2x− y + z = 1
Check that the coordinates of the three original points A, B and C do indeed satisfy this equation.
Properties of the Cross Product Let ~u = [u1, u2, u3] and ~v = [v1, v2, v3] be vectors inR3. Then :
1. ~u.(~u× ~v) = 0 and ~v.(~u × ~v) = 0 Proof :
~u.(~u× ~v) = [u1, u2, u3].[u2v3− u3v2, u3v1− u1v3, u1v2− u2v1]
= u1(u2v3− u3v2) + u2(u3v1− u1v3) + u3(u1v2− u2v1)
= u1u2v3− u1u3v2+ u2u3v1− u2u1v3) + u3u1v2− u3u2v1
= 0
Similarly for ~v.(~u× ~v).
2. The cross product is not commutative. In fact ~v× ~u = −(~u × ~v). So ~u × ~v and ~v × ~u have opposite directions.
This is easily checked from the formula (Exercise).
3. ~u× ~u = ~0 (= [0, 0, 0]). The cross product of any vector with itself is the zero vector.
Again this is easily seen from the formula.
4. If k is a scalar, then ~u× (k~v) = k(~u × ~v); e.g. ~u × (2~v) = 2(~u × ~v), etc.
Note: In particular this means : if k is a scalar then ~u×(k~u) = k(~u×~u) = k~0 by Property 3 : i.e., if two vectors inR3 have the same (or opposite) direction, their cross product is the zero vector.
5. If ~u and ~v are non-zero vectors inR3, and ~v is not a scalar multiple of ~u, then ~u× ~v is a non-zero vector orthogonal to both ~u and ~v.
6. Distributivity of the Cross Product over Vector Addition : Let ~u, ~v and ~w be vectors in R3. Then
(a) ~u× (~v + ~w) = (~u× ~v) + (~u × ~w) (b) (~u + ~v)× ~w = (~u× ~w) + (~v× ~w)
The Standard Basis Vectors in R3 Let ~v = [2, 3, 4]. Then we could write
~v = [2, 0, 0] + [0, 3, 0] + [0, 0, 4]
~v = 2[1, 0, 0] + 3[0, 1, 0] + 4[0, 0, 1]
Note that [1, 0, 0], [0, 1, 0] and [0, 0, 1] are unit vectors pointing along the positive X, Y and Z axes respectively. The above example indicates that any vector inR3can be written as the sum of three vectors, each a scalar multiple of [1, 0, 0], [0, 1, 0] or [0, 0, 1].
Definition 5.3.3 The vectors [1, 0, 0], [0, 1, 0] and [0, 0, 1] are called the standard basis vectors inR3 and are denoted by ~ı, ~ and ~k respectively.
~ı = [1, 0, 0], ~ = [0, 1, 0], ~k = [0, 0, 1]
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X Y
Z
~ı ~
~k
Then for example [2, 3, 4] = 2~ı+ 3~+ 4~k and in general the vector [a, b, c] may also be written a~ı + b~ + c~k. The “~ı, ~, ~k” notation will be convenient for computing cross products.
Computing Cross Products using 3× 3 Determinants Another method for computing ~u× ~v. Take ~u = [2, 1, 1] and ~v = [1, 0, 2].
Step 1 Write a 3× 3 “matrix” whose rows consist of
1. ~ı ~ ~k ~ı ~ ~k
2. Components of ~u 2 1 1
3. Components of ~v 1 0 2
Step 2 ~u× ~v is the determinant of this matrix, and can be computed for example using the
“basket-weave” method (or cofactor expansion along the first row):
~ı ~ ~k ~ı ~
2 1 1 2 1
1 0 2 1 0
~u× ~v is given by
(~ı(1)(2) + ~(1)(1) + ~k(2)(0))
| {z }
products of terms along top right→bottom left diagonals
− (~k(1)(1) +~ı(1)(0) + ~k(1)(1))
| {z }
products of terms along top left→bottom right diagonals
~u× ~v = 2~ı + ~ − ~k − 4~ = 2~ı − 3~ − ~k = [2, −3, −1]
Example 5.3.4 Find [3, 1,−1] × [3, −1, 1]
Solution:
~ı ~ ~k ~ı ~
3 1 −1 3 1
3 −1 1 3 −1
[3, 1,−1] × [3, −1, 1] = ~ı(1)(1) = ~(−1)(3) + ~k(3)(−1) − ~k(1)(3) −~ı(−1)(−1) − ~(3)(1)
= ~ı− 3~ − 3~k − 3~k −~ı − 3~
= −6~ − 6~k [3, 1,−1] × [3, −1, 1] = [0, −6, −6]
Remark: The cross product is not associative : i.e., if ~u, ~v and ~w are vectors, (~u× ~v) × ~w need not be equal to ~u× (~v × ~w). For example
~ı× (~ı × ~) = ~ı × ~k = −~ (Check) (~ı×~ı) × ~ = ~0 × ~ = ~0 6= −~
Length of the Cross Product
So far our discussion of the cross product has focussed on its direction. The length of ~u× ~v also has significance, relating to the angle θ between ~u and ~v.
Fact 5.3.5 (Lagrange’s Identity) For any vectors ~u and ~v inR3
||~u × ~v||2=||~u||2||~v||2− (~u.~v)2
This can be proved by writing each term in terms of components of ~u and ~v.
Replacing ~u.~v by||~u|| ||~v|| cos θ, Lagrange’s Identity becomes :
||~u × ~v||2 = ||~u||2||~v||2− (||~u|| ||~v|| cos θ)2
= ||~u||2||~v||2− ||~u||2||~v||2cos2θ
= ||~u||2||~v||2(1− cos2θ)
= ||~u||2||~v||2(sin2θ)
=⇒ ||~u × ~v|| = ||~u|| ||~v|| sin θ (Note that sin θ≥ 0 since θ is between 0 and π (180◦)).
Application: Area of a Parallelogram
Suppose ~u and ~v are vectors representing adjacent sides of a parallelogram P . The area of P is||~u|| × h, where ~u is regarded as the base, and h denotes the perpendicular height of P above ~u.
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~u
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h θ
Then
sin θ = h
||~v|| =⇒ h = ||~v|| sin θ Area of P = ||~u||h = ||~u|| ||~v|| sin θ = ||~u × ~v||
Example 5.3.6 (Summer 1999 Q6) Find the area of the parallelogram inR3having the vectors [1, 2, 3] and [7, 6,−7] as adjacent sides.
Solution: Area is||[1, 2, 3] × [7, 6, −7]||
~ı ~ ~k ~ı ~
1 2 3 1 2
7 6 −7 7 6
[1, 2, 3]× [7, 6, −7] = ~ı(2)(−7) + ~(3)(7) + ~k(1)(6) − ~k(2)(7) −~ı(3)(6) − ~(1)(−7)
= −32~ı + 28~ − 8~k
= [−32, 28, −8]
Area of parallelogram =||[−32, 28, −8]|| = 4||[−8, 7, −2]|| = 4√
64 + 49 + 4 = 4√
117 = 12√ 13.
Example 5.3.7 (Summer 2001 Q3) Find the area of the triangle inR3with vertices A(1, 2, 1), B(2, 4, 1) and C(−1, 0, 3)
Solution: ~AB = [1, 2, 0], ~AC = [−2, −2, 2]
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AC~ AB~
From the diagram, we want half of the area of the parallelogram having ~AB and ~AC as adjacent sides; i.e.
Area of triangle = 1
2|| ~AB× ~AC|| = 1
2||[4, −2, 2]|| (from Problem 6.2.6∗) = 1 2
√24 =√ 6 Summary:
• Area of parallelogram with adjacent sides ~u and ~v : ||~u × ~v||
• Area of triangle with adjacent sides ~u and ~v : 12||~u × ~v||
Another Application: Volume of a Parallelepiped.
A parallelepiped inR3is a six-faced object, in which pairs of opposite faces consist of similar parallelograms.
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A Parallelpiped
If ~u, ~v and ~w are vectors inR3 having different directions and initial points at O, they form three adjacent sides of a unique parallelepiped.
Example 5.3.8 (Summer 1999 Q6 (b)) Find the volume of the parallelepiped P having ~u = [1, 2, 3], ~v = [7, 6,−7] and ~w = [4, 5,−3] as adjacent sides.
Solution: Suppose the parallelogram with ~u and ~v as sides forms the “base” of P .
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~u
~v
~ w
~u× ~v
h
Then
V = Volume of P = A× h
where A is the area of the base and h is the (perpendicular) height of P above this base.
A =||~u × ~v||
The vector ~u× ~v is perpendicular to the base of P and from the diagram we see that h = ||proj~u×~vw~||
=⇒ h = | ~w.(~u× ~v)|
||~u × ~v||
Then
V = A× h = ||~u × ~v||| ~w.(~u× ~v)|
||~u × ~v||
Volume of P = | ~w.(~u× ~v)|
From Example 5.3.6 ~u× ~v = [1, 2, 3] × [7, 6, −7] = [−32, 28, −8]. Then
Volume of P = [4, 5,−3].[−32, 28, −8] = 4(−32) + 5(28) − 3(−8) = 36
Remarks:
1. If ~w, ~u and ~v are adjacent sides of a parallelepiped P , the volume of P is given by
| ~w.(~u× ~v)|.
2. ~w.(~u× ~v) is called the scalar triple product of ~w, ~u and ~v. Although this definition looks non-symmetric, it turns out that| ~w.(~u× ~v)| does not depend on the order in which ~u, ~v and ~w are written; i.e.
| ~w.(~u× ~v)| = |~u.(~v × ~w)| = |~v.( ~w× ~u)| = | ~w.(~v× ~u)| = |~v.(~u × ~w)| = |~u.( ~w× ~v)|, for any vectors ~u, ~v, ~w inR3. (Typically three of the six expressions inside the absolute value signs above will be negative and three positive, but all will have the same absolute value).
In Example 5.3.8 there was no particular reason to choose the parallelogram defined by
~u and ~v as the base : choosing a different face, for example the one defined by ~u and ~w would have resulted in|~v.(~u × ~w)| as the volume formula.
Exercise: Check that|~v.(~u × ~w)| = | ~w.(~u× ~v)| for this example.
3. Scalar Triple Products and the 3× 3 Determinant.
In Example 5.3.8 suppose we want to compute
~u.( ~w× ~v) = [1, 2, 3].([4, 5, −3] × [7, 6, −7]).
This is just the determinant of the matrix
1 2 3
4 5 −3
7 6 −7
Components of ~u Components of ~w Components of ~v.
This determinant can be computed for example by the “basket-weave” method:
1 2 3 1 2
4 5 −3 4 5
7 6 −7 7 6
~u.( ~w× ~v) = 1(5)(−7) + 2(−3)(7) + 3(4)(6) − 3(5)(7) − 1(−3)(6) − 2(4)(−7)
= −35 − 42 + 72 − 105 + 18 + 56
= −36
So ~u.( ~w× ~v) = −36. (As expected |~u.( ~w× ~v)| = 36 = Volume of P .) This completes our discussion of the cross product. Summary of Section 5.3 : 1. Definition of Cross Product
2. Computation of cross product using, formula, standard basis vectors or “basket-weave”
method.
3. Use of cross product to find the equation of a plane given three points.
4. Area of parallelograms and triangles.
5. Volume of a parallelepiped and the scalar triple product.
