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Calculus at Moravian University

Michael Fraboni (Editor)

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Section 11.4 The Cross Product

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β€œOrthogonality” is immensely important. A quick scan of your current environment will undoubtedly reveal numerous surfaces and edges that are perpendicular to each other (including the edges of this page). The dot product provides a quick test for orthogonality: vectors uβ†’ and vβ†’ are perpendicular if, and only if, uβ†’β‹…vβ†’=0.
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Given two non-parallel, nonzero vectors u→ and v→ in space, it is very useful to find a vector w→ that is perpendicular to both u→ and v→. There is an operation, called the cross product, that creates such a vector. This section defines the cross product, then explores its properties and applications.
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Definition 11.4.1. Cross Product.

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Let uβ†’=⟨u1,u2,u3⟩ and vβ†’=⟨v1,v2,v3⟩ be vectors in R3. The cross product of uβ†’ and vβ†’, denoted uβ†’Γ—vβ†’, is the vector
uβ†’Γ—vβ†’=⟨u2v3βˆ’u3v2,βˆ’(u1v3βˆ’u3v1),u1v2βˆ’u2v1⟩.
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This definition can be a bit cumbersome to remember. After an example we will give a convenient method for computing the cross product. For now, careful examination of the products and differences given in the definition should reveal a pattern that is not too difficult to remember. (For instance, in the first component only 2 and 3 appear as subscripts; in the second component, only 1 and 3 appear as subscripts. Further study reveals the order in which they appear.)
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Let’s practice using this definition by computing a cross product.
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Example 11.4.2. Computing a cross product.

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Let uβ†’=⟨2,βˆ’1,4⟩ and vβ†’=⟨3,2,5⟩. Find uβ†’Γ—vβ†’, and verify that it is orthogonal to both uβ†’ and vβ†’.
Solution.
Using Definition 11.4.1, we have
uβ†’Γ—vβ†’=⟨(βˆ’1)5βˆ’(4)2,βˆ’((2)5βˆ’(4)3),(2)2βˆ’(βˆ’1)3⟩=βŸ¨βˆ’13,2,7⟩.
(We encourage the reader to compute this product on their own, then verify their result.)
We test whether or not u→×v→ is orthogonal to u→ and v→ using the dot product:
(uβ†’Γ—vβ†’)β‹…uβ†’=βŸ¨βˆ’13,2,7βŸ©β‹…βŸ¨2,βˆ’1,4⟩=0,
(uβ†’Γ—vβ†’)β‹…vβ†’=βŸ¨βˆ’13,2,7βŸ©β‹…βŸ¨3,2,5⟩=0.
Since both dot products are zero, u→×v→ is indeed orthogonal to both u→ and v→.
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A convenient method of computing the cross product starts with forming a particular 3Γ—3 matrix, or rectangular array. The first row comprises the standard unit vectors Δ±^, Θ·^, and k^. The second and third rows are the vectors uβ†’ and vβ†’, respectively. Using uβ†’ and vβ†’ from Example 11.4.2, we begin with:
Δ±^Θ·^k^2βˆ’14325
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Now repeat the first two columns after the original three:
Δ±^Θ·^k^Δ±^Θ·^2βˆ’142βˆ’132532
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This gives three full β€œupper left to lower right” diagonals, and three full β€œupper right to lower left” diagonals, as shown. Compute the products along each diagonal, then add the products on the right and subtract the products on the left:
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uβ†’Γ—vβ†’=(βˆ’5Δ±^+12Θ·^+4k^)βˆ’(βˆ’3k^+8Δ±^+10Θ·^)=βˆ’13Δ±^+2Θ·^+7k^=βŸ¨βˆ’13,2,7⟩.
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We practice using this method.
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Example 11.4.3. Computing a cross product.

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Let uβ†’=⟨1,3,6⟩ and vβ†’=βŸ¨βˆ’1,2,1⟩. Compute both uβ†’Γ—vβ†’ and vβ†’Γ—uβ†’.
Solution.
To compute u→×v→, we form the matrix as prescribed above, complete with repeated first columns:
 Δ±^  Θ·^  k^  Δ±^  Θ·^ 13613βˆ’121βˆ’12
We let the reader compute the products of the diagonals; we give the result:
uβ†’Γ—vβ†’=(3Δ±^βˆ’6Θ·^+2k^)βˆ’(βˆ’3k^+12Δ±^+Θ·^)=βŸ¨βˆ’9,βˆ’7,5⟩.
To compute v→×u→, we switch the second and third rows of the above matrix, then multiply along diagonals and subtract:
 Δ±^  Θ·^  k^  Δ±^  Θ·^ βˆ’121βˆ’1213613
Note how with the rows being switched, the products that once appeared on the right now appear on the left, and vice-versa. Thus the result is:
vβ†’Γ—uβ†’=(12Δ±^+Θ·^βˆ’3k^)βˆ’(2k^+3Δ±^βˆ’6Θ·^)=⟨9,7,βˆ’5⟩,
which is the opposite of u→×v→. We leave it to the reader to verify that each of these vectors is orthogonal to u→ and v→.
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Subsection 11.4.1 Properties of the Cross Product

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It is not coincidence that vβ†’Γ—uβ†’=βˆ’(uβ†’Γ—vβ†’) in the preceding example; one can show using Definition 11.4.1 that this will always be the case. The following theorem states several useful properties of the cross product, each of which can be verified by referring to the definition.
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We introduced the cross product as a way to find a vector orthogonal to two given vectors, but we did not give a proof that the construction given in Definition 11.4.1 satisfies this property. Theorem 11.4.4 asserts this property holds; we leave it as a problem in the Exercise section to verify this.
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Property 5 from the theorem is also left to the reader to prove in the Exercise section, but it reveals something more interesting than β€œthe cross product of a vector with itself is 0β†’.” Let uβ†’ and vβ†’ be parallel vectors; that is, let there be a scalar c such that vβ†’=cuβ†’. Consider their cross product:
uβ†’Γ—vβ†’=uβ†’Γ—(cuβ†’)=c(uβ†’Γ—uβ†’) (by Property 3 of Theorem 11.4.4)=0β†’ (by Property 5 of Theorem 11.4.4).
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We have just shown that the cross product of parallel vectors is 0β†’. This hints at something deeper. Theorem 11.3.5 related the angle between two vectors and their dot product; there is a similar relationship relating the cross product of two vectors and the angle between them, given by the following theorem.
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Note that this theorem makes a statement about the magnitude of the cross product. When the angle between uβ†’ and vβ†’ is 0 or Ο€ (i.e., the vectors are parallel), the magnitude of the cross product is 0. The only vector with a magnitude of 0 is 0β†’ (see Property 9 of Theorem 11.2.15), hence the cross product of parallel vectors is 0β†’.
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We demonstrate the truth of this theorem in the following example.
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Example 11.4.6. The cross product and angles.

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Let uβ†’=⟨1,3,6⟩ and vβ†’=βŸ¨βˆ’1,2,1⟩ as in Example 11.4.3. Verify Theorem 11.4.5 by finding ΞΈ, the angle between uβ†’ and vβ†’, and the magnitude of uβ†’Γ—vβ†’.
Solution.
We use Theorem 11.3.5 to find the angle between uβ†’ and vβ†’.
ΞΈ=cosβˆ’1⁑(uβ†’β‹…vβ†’β€–uβ†’β€–β€–vβ†’β€–)=cosβˆ’1⁑(11466)β‰ˆ0.8471=48.54∘.
Our work in Example 11.4.3 showed that uβ†’Γ—vβ†’=βŸ¨βˆ’9,βˆ’7,5⟩, hence β€–uβ†’Γ—vβ†’β€–=155. Is β€–uβ†’Γ—vβ†’β€–=β€–uβ†’β€–β€–vβ†’β€–sin⁑(ΞΈ)? Using numerical approximations, we find:
β€–uβ†’Γ—vβ†’β€–=155β€–uβ†’β€–β€–vβ†’β€–sin⁑(ΞΈ)=466sin⁑(0.8471)β‰ˆ12.45.β‰ˆ12.45.
Numerically, they seem equal. Using a right triangle, one can show that
sin⁑(cosβˆ’1⁑(11466))=155466,
which allows us to verify the theorem exactly.
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Right Hand Rule.

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The anticommutative property of the cross product demonstrates that u→×v→ and v→×u→ differ only by a sign — these vectors have the same magnitude but point in the opposite direction. When seeking a vector perpendicular to u→ and v→, we essentially have two directions to choose from, one in the direction of u→×v→ and one in the direction of v→×u→. Does it matter which we choose? How can we tell which one we will get without graphing, etc.?
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Another wonderful property of the cross product, as defined, is that it follows the right hand rule. Given uβ†’ and vβ†’ in R3 with the same initial point, point the index finger of your right hand in the direction of uβ†’ and let your middle finger point in the direction of vβ†’ (much as we did when establishing the right hand rule for the 3-dimensional coordinate system). Your thumb will naturally extend in the direction of uβ†’Γ—vβ†’. One can β€œpractice” this using Figure 11.4.7. If you switch, and point the index finder in the direction of vβ†’ and the middle finger in the direction of uβ†’, your thumb will now point in the opposite direction, allowing you to β€œvisualize” the anticommutative property of the cross product.
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Figure 11.4.7. Illustrating the Right Hand Rule of the cross product
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Subsection 11.4.2 Applications of the Cross Product

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There are a number of ways in which the cross product is useful in mathematics, physics and other areas of science beyond β€œjust” finding a vector perpendicular to two others. We highlight a few here.
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Area of a Parallelogram.

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It is a standard geometry fact that the area of a parallelogram is A=bh, where b is the length of the base and h is the height of the parallelogram, as illustrated in Figure 11.4.8.(a). As shown when defining the Parallelogram Law of vector addition, two vectors uβ†’ and vβ†’ define a parallelogram when drawn from the same initial point, as illustrated in Figure 11.4.8.(b). Trigonometry tells us that h=β€–uβ†’β€–sin⁑(ΞΈ), hence the area of the parallelogram is
(11.4.1)A=β€–uβ†’β€–β€–vβ†’β€–sin⁑(ΞΈ)=β€–uβ†’Γ—vβ†’β€–,
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where the second equality comes from Theorem 11.4.5.
(a)
(b)
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Figure 11.4.8. Using the cross product to find the area of a parallelogram
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We illustrate using Equation (11.4.1) in the following example.
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Example 11.4.9. Finding the area of a parallelogram.
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    Find the area of the parallelogram defined by the vectors uβ†’=⟨2,1⟩ and vβ†’=⟨1,3⟩.
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    Verify that the points A=(1,1,1), B=(2,3,2), C=(4,5,3) and D=(3,3,2) are the vertices of a parallelogram. Find the area of the parallelogram.
Solution.
  1. Figure 11.4.10.(a) sketches the parallelogram defined by the vectors uβ†’ and vβ†’. We have a slight problem in that our vectors exist in R2, not R3, and the cross product is only defined on vectors in R3. We skirt this issue by viewing uβ†’ and vβ†’ as vectors in the xβˆ’y plane of R3, and rewrite them as uβ†’=⟨2,1,0⟩ and vβ†’=⟨1,3,0⟩. We can now compute the cross product. It is easy to show that uβ†’Γ—vβ†’=⟨0,0,5⟩; therefore the area of the parallelogram is A=β€–uβ†’Γ—vβ†’β€–=5.
    (a)
    (b)
    Figure 11.4.10. Sketching the parallelograms in Example 11.4.9
  2. To show that the quadrilateral ABCD is a parallelogram (shown in Figure 11.4.10.(b)), we need to show that the opposite sides are parallel. We can quickly show that ABβ†’=DCβ†’=⟨1,2,1⟩ and BCβ†’=ADβ†’=⟨2,2,1⟩. We find the area by computing the magnitude of the cross product of ABβ†’ and BCβ†’:
    ABβ†’Γ—BCβ†’=⟨0,1,βˆ’2βŸ©β‡’β€–ABβ†’Γ—BCβ†’β€–=5β‰ˆ2.236.
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This application is perhaps more useful in finding the area of a triangle (in short, triangles are used more often than parallelograms). We illustrate this in the following example.
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Example 11.4.11. Area of a triangle.
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Find the area of the triangle with vertices A=(1,2), B=(2,3) and C=(3,1), as pictured in Figure 11.4.12.
Figure 11.4.12. Finding the area of a triangle in Example 11.4.11
Solution.
We found the area of this triangle in Example 7.1.10 to be 1.5 using integration. There we discussed the fact that finding the area of a triangle can be inconvenient using the β€œ12bh” formula as one has to compute the height, which generally involves finding angles, etc. Using a cross product is much more direct.
We can choose any two sides of the triangle to use to form vectors; we choose ABβ†’=⟨1,1⟩ and ACβ†’=⟨2,βˆ’1⟩. As in the previous example, we will rewrite these vectors with a third component of 0 so that we can apply the cross product. The area of the triangle is
12β€–ABβ†’Γ—ACβ†’β€–=12β€–βŸ¨1,1,0βŸ©Γ—βŸ¨2,βˆ’1,0βŸ©β€–=12β€–βŸ¨0,0,βˆ’3βŸ©β€–=32.
We arrive at the same answer as before with less work.
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Volume of a Parallelepiped.

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The three dimensional analogue to the parallelogram is the parallelepiped. Each face is parallel to the opposite face, as illustrated in Figure 11.4.13. By crossing vβ†’ and wβ†’, one gets a vector whose magnitude is the area of the base. Dotting this vector with uβ†’ computes the volume of parallelepiped! (Up to a sign; take the absolute value.)
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Figure 11.4.13. A parallelepiped is the three dimensional analogue to the parallelogram
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Thus the volume of a parallelepiped defined by vectors u→, v→ and w→ is
(11.4.2)V=|u→⋅(v→×w→)|.
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Note how this is the Triple Scalar Product, first seen in Theorem 11.4.4. Applying the identities given in the theorem shows that we can apply the Triple Scalar Product in any β€œorder” we choose to find the volume. That is,
V=|uβ†’β‹…(vβ†’Γ—wβ†’)|=|uβ†’β‹…(wβ†’Γ—vβ†’)|=|(uβ†’Γ—vβ†’)β‹…wβ†’|, etc. 
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Example 11.4.14. Finding the volume of parallelepiped.
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Find the volume of the parallelepiped defined by the vectors uβ†’=⟨1,1,0⟩, vβ†’=βŸ¨βˆ’1,1,0⟩ and wβ†’=⟨0,1,1⟩.
Solution.
We apply Equation (11.4.2). We first find vβ†’Γ—wβ†’=⟨1,1,βˆ’1⟩. Then
|uβ†’β‹…(vβ†’Γ—wβ†’)|=|⟨1,1,0βŸ©β‹…βŸ¨1,1,βˆ’1⟩|=2.
So the volume of the parallelepiped is 2 cubic units.
Figure 11.4.15. A parallelepiped in Example 11.4.14
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While this application of the Triple Scalar Product is interesting, it is not used all that often: parallelepipeds are not a common shape in physics and engineering. The last application of the cross product is very applicable in engineering.
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Torque.

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Torque is a measure of the turning force applied to an object. A classic scenario involving torque is the application of a wrench to a bolt. When a force is applied to the wrench, the bolt turns. When we represent the force and wrench with vectors Fβ†’ and β„“β†’, we see that the bolt moves (because of the threads) in a direction orthogonal to Fβ†’ and β„“β†’. Torque is usually represented by the Greek letter Ο„, or tau, and has units of Nβ‹…m, a Newton–meter, or ftβ‹…lb, a foot–pound.
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While a full understanding of torque is beyond the purposes of this book, when a force F→ is applied to a lever arm ℓ→, the resulting torque is
(11.4.3)τ→=ℓ→×F→.
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Example 11.4.16. Computing torque.
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A lever of length 2ft makes an angle with the horizontal of 45∘. Find the resulting torque when a force of 10lb is applied to the end of the level where:
Figure 11.4.17. Showing a force being applied to a lever in Example 11.4.16
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    the force is perpendicular to the lever, and
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    the force makes an angle of 60∘ with the lever, as shown in Figure 11.4.17.
Solution.
  1. We start by determining vectors for the force and lever arm. Since the lever arm makes a 45∘ angle with the horizontal and is 2ft long, we can state that β„“β†’=2⟨cos⁑(45∘),sin⁑(45∘)⟩=⟨2,2⟩. Since the force vector is perpendicular to the lever arm (as seen in the left hand side of Figure 11.4.17), we can conclude it is making an angle of βˆ’45∘ with the horizontal. As it has a magnitude of 10lb, we can state Fβ†’=10⟨cos⁑(βˆ’45∘),sin⁑(βˆ’45∘)⟩=⟨52,βˆ’52⟩. Using Equation (11.4.3) to find the torque requires a cross product. We again let the third component of each vector be 0 and compute the cross product:
    Ο„β†’=β„“β†’Γ—Fβ†’=⟨2,2,0βŸ©Γ—βŸ¨52,βˆ’52,0⟩=⟨0,0,βˆ’20⟩
    This clearly has a magnitude of 20 ft-lb. We can view the force and lever arm vectors as lying β€œon the page”; our computation of Ο„β†’ shows that the torque goes β€œinto the page.” This follows the Right Hand Rule of the cross product, and it also matches well with the example of the wrench turning the bolt. Turning a bolt clockwise moves it in.
  2. Our lever arm can still be represented by β„“β†’=⟨2,2⟩. As our force vector makes a 60∘ angle with β„“β†’, we can see (referencing the right hand side of the figure) that Fβ†’ makes a βˆ’15∘ angle with the horizontal. Thus
    Fβ†’=10⟨cosβˆ’15∘,sinβˆ’15∘⟩=⟨5(1+3)2,5(βˆ’1+3)2βŸ©β‰ˆβŸ¨9.659,βˆ’2.588⟩.
    We again make the third component 0 and take the cross product to find the torque:
    Ο„β†’=β„“β†’Γ—Fβ†’=⟨2,2,0βŸ©Γ—βŸ¨5(1+3)2,5(βˆ’1+3)2,0⟩=⟨0,0,βˆ’103βŸ©β‰ˆβŸ¨0,0,βˆ’17.321⟩.
    As one might expect, when the force and lever arm vectors are orthogonal, the magnitude of force is greater than when the vectors are not orthogonal.
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While the cross product has a variety of applications (as noted in this chapter), its fundamental use is finding a vector perpendicular to two others. Knowing a vector is orthogonal to two others is of incredible importance, as it allows us to find the equations of lines and planes in a variety of contexts. The importance of the cross product, in some sense, relies on the importance of lines and planes, which see widespread use throughout engineering, physics and mathematics. We study lines and planes in the next two sections.
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Exercises 11.4.3 Exercises

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Terms and Concepts

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4.
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T/F: A fundamental principle of the cross product is that u→×v→ is orthogonal to u→ and v→.
Answer.
True
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Problems

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Exercise Group.
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In the following exercises, vectors u→ and v→ are given. Compute u→×v→ and check this is orthogonal to both u→ and v→.
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8.
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uβ†’=⟨5,βˆ’4,3⟩, vβ†’=⟨2,βˆ’5,1⟩
Answer.
⟨11,1,βˆ’17⟩
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9.
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uβ†’=⟨4,βˆ’5,βˆ’5⟩, vβ†’=⟨3,3,4⟩
Answer.
βŸ¨βˆ’5,βˆ’31,27⟩
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10.
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uβ†’=βŸ¨βˆ’4,7,βˆ’10⟩, vβ†’=⟨4,4,1⟩
Answer.
⟨47,βˆ’36,βˆ’44⟩
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12.
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uβ†’=⟨1,5,βˆ’4⟩, vβ†’=βŸ¨βˆ’2,βˆ’10,8⟩
Answer.
⟨0,0,0⟩
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13.
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uβ†’=⟨a,b,0⟩, vβ†’=⟨c,d,0⟩
Answer.
uβ†’Γ—vβ†’=⟨0,0,adβˆ’bc⟩
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17.
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Pick any vectors u→, v→ and w→ in R3 and show that u→×(v→+w→)=u→×v→+u→×w→.
Answer.
Answers will vary.
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18.
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Pick any vectors u→, v→ and w→ in R3 and show that u→⋅(v→×w→)=(u→×v→)⋅w→.
Answer.
Answers will vary.
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Exercise Group.
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In the following exercises, the magnitudes of vectors u→ and v→ in R3 are given, along with the angle θ between them. Use this information to find the magnitude of u→×v→.
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Exercise Group.
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In the following exercises, find the area of the parallelogram defined by the given vectors.
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Exercise Group.
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In the following exercises, find the area of the triangle with the given vertices.
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Exercise Group.
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In the following exercises, find the area of the quadrilateral with the given vertices. (Hint: break the quadrilateral into two triangles.)
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Exercise Group.
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In the following exercises, find the volume of the parallelepiped defined by the given vectors.
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33.
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uβ†’=⟨1,1,1⟩, vβ†’=⟨1,2,3⟩, and wβ†’=⟨1,0,1⟩
Answer.
2
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34.
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uβ†’=βŸ¨βˆ’1,2,1⟩, vβ†’=⟨2,2,1⟩, and wβ†’=⟨3,1,3⟩
Answer.
15
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Exercise Group.
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In the following exercises, find a unit vector orthogonal to both u→ and v→.
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35.
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uβ†’=⟨1,1,1⟩, and vβ†’=⟨2,0,1⟩
Answer.
±⟨1/6,1/6,βˆ’2/6⟩
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36.
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uβ†’=⟨1,βˆ’2,1⟩, and vβ†’=⟨3,2,1⟩.
Answer.
Β±βŸ¨βˆ’2/21,1/21,4/21⟩
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38.
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uβ†’=⟨1,βˆ’2,1⟩, and vβ†’=βŸ¨βˆ’2,4,βˆ’2⟩
Answer.
Since uβ†’ and vβ†’ are parallel, any unit vector orthogonal to uβ†’ works (such as 12⟨1,0,βˆ’1⟩).
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39.
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A bicycle rider applies 150lb of force, straight down, onto a pedal that extends 7in horizontally from the crankshaft. Find the magnitude of the torque applied to the crankshaft.
Answer.
87.5ft–lb
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40.
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A bicycle rider applies 150lb of force, straight down, onto a pedal that extends 7in from the crankshaft, making a 30∘ angle with the horizontal. Find the magnitude of the torque applied to the crankshaft.
Answer.
43.753β‰ˆ75.78ft–lb
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41.
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To turn a stubborn bolt, 80lb of force is applied to a 10in wrench. What is the maximum amount of torque that can be applied to the bolt?
Answer.
200/3β‰ˆ66.67ft–lb
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42.
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To turn a stubborn bolt, 80lb of force is applied to a 10in wrench in a confined space, where the direction of applied force makes a 10∘ angle with the wrench. How much torque is subsequently applied to the wrench?
Answer.
11.58ft–lb
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43.
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Show, using the definition of the Cross Product, that u→⋅(u→×v→)=0; that is, that u→ is orthogonal to the cross product of u→ and v→.
Solution.
With uβ†’=⟨u1,u2,u3⟩ and vβ†’=⟨v1,v2,v3⟩, we have
uβ†’β‹…(uβ†’Γ—vβ†’)=⟨u1,u2,u3βŸ©β‹…(⟨u2v3βˆ’u3v2,βˆ’(u1v3βˆ’u3v1),u1v2βˆ’u2v1⟩)=u1(u2v3βˆ’u3v2)βˆ’u2(u1v3βˆ’u3v1)+u3(u1v2βˆ’u2v1)=0.
πŸ”—
44.
πŸ”—
Show, using the definition of the Cross Product, that u→×u→=0→.
Solution.
With uβ†’=⟨u1,u2,u3⟩, we have
uβ†’Γ—uβ†’=⟨u2u3βˆ’u3u2,βˆ’(u1u3βˆ’u3u1),u1u2βˆ’u2u1⟩)=⟨0,0,0⟩=0β†’.
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