Section 10.4 The Cross Product
Definition 10.4.1. Cross Product.
Let βu=β¨u1,u2,u3β© and βv=β¨v1,v2,v3β© be vectors in R3. The cross product of βu and βv, denoted βuΓβv, is the vector
Example 10.4.2. Computing a cross product.
Let βu=β¨2,β1,4β© and βv=β¨3,2,5β©. Find βuΓβv, and verify that it is orthogonal to both βu and βv.
Using Definition 10.4.1, we have
(We encourage the reader to compute this product on their own, then verify their result.)
We test whether or not \(\crossp uv\) is orthogonal to \(\vec u\) and \(\vec v\) using the dot product:
Since both dot products are zero, \(\crossp uv\) is indeed orthogonal to both \(\vec u\) and \(\vec v\text{.}\)
Example 10.4.3. Computing a cross product.
Let βu=β¨1,3,6β© and βv=β¨β1,2,1β©. Compute both βuΓβv and βvΓβu.
To compute \(\crossp uv\text{,}\) we form the matrix as prescribed above, complete with repeated first columns:
We let the reader compute the products of the diagonals; we give the result:
To compute \(\crossp vu\text{,}\) we switch the second and third rows of the above matrix, then multiply along diagonals and subtract:
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:
which is the opposite of \(\crossp uv\text{.}\) We leave it to the reader to verify that each of these vectors is orthogonal to \(\vec u\) and \(\vec v\text{.}\)
Subsection 10.4.1 Properties of the Cross Product
It is not coincidence that βvΓβu=β(βuΓβv) in the preceding example; one can show using Definition 10.4.1 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.Theorem 10.4.4. Properties of the Cross Product.
Let βu, βv and βw be vectors in R3 and let c be a scalar. The following identities hold:
βuΓβv=β(βvΓβu) Anticommutative Property
(βu+βv)Γβw=βuΓβw+βvΓβw Distributive Properties
βuΓ(βv+βw)=βuΓβv+βuΓβw
c(βuΓβv)=(cβu)Γβv=βuΓ(cβv)
(βuΓβv)β βu=0 Orthogonality Properties
(βuΓβv)β βv=0
βuΓβu=β0
βuΓβ0=β0
βuβ (βvΓβw)=(βuΓβv)β βw Triple Scalar Product
Theorem 10.4.5. The Cross Product and Angles.
Let βu and βv be nonzero vectors in R3. Then
where \theta\text{,} 0\leq \theta \leq \pi\text{,} is the angle between \vecu and \vecv\text{.}
Example 10.4.6. The cross product and angles.
Let \vec u = \la 1,3,6\ra and \vec v = \la -1,2,1\ra as in Example 10.4.3. Verify Theorem 10.4.5 by finding \theta\text{,} the angle between \vecu and \vecv\text{,} and the magnitude of \crossp uv\text{.}
We use Theorem 10.3.5 to find the angle between \(\vecu\) and \(\vecv\text{.}\)
Our work in Example 10.4.3 showed that \(\crossp uv = \la -9,-7,5\ra\text{,}\) hence \(\norm{\crossp uv} = \sqrt{155}\text{.}\) Is \(\norm{\crossp uv} = \vnorm u\, \vnorm v\sin(\theta)\text{?}\) Using numerical approximations, we find:
Numerically, they seem equal. Using a right triangle, one can show that
which allows us to verify the theorem exactly.
Right Hand Rule.
The anticommutative property of the cross product demonstrates that \crossp uv and \crossp vu differ only by a sign β these vectors have the same magnitude but point in the opposite direction. When seeking a vector perpendicular to \vec u and \vec v\text{,} we essentially have two directions to choose from, one in the direction of \crossp uv and one in the direction of \crossp vu\text{.} Does it matter which we choose? How can we tell which one we will get without graphing, etc.? Another wonderful property of the cross product, as defined, is that it follows the right hand rule. Given \vec u and \vec v in \mathbb{R}^3 with the same initial point, point the index finger of your right hand in the direction of \vecu and let your middle finger point in the direction of \vecv (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 \crossp uv\text{.} One can βpracticeβ this using Figure 10.4.7. If you switch, and point the index finder in the direction of \vecv and the middle finger in the direction of \vecu\text{,} your thumb will now point in the opposite direction, allowing you to βvisualizeβ the anticommutative property of the cross product.Subsection 10.4.2 Applications of the Cross Product
There are a number of ways in which the cross product is useful in mathematics, physics and other areas of science beyond βjustβ finding a vector perpendicular to two others. We highlight a few here.Area of a Parallelogram.
It is a standard geometry fact that the area of a parallelogram is A = bh\text{,} where b is the length of the base and h is the height of the parallelogram, as illustrated in Figure 10.4.8.(a). As shown when defining the Parallelogram Law of vector addition, two vectors \vecu and \vecv define a parallelogram when drawn from the same initial point, as illustrated in Figure 10.4.8.(b). Trigonometry tells us that h = \vnorm u \sin(\theta)\text{,} hence the area of the parallelogram isExample 10.4.9. Finding the area of a parallelogram.
Find the area of the parallelogram defined by the vectors \vecu = \la 2,1\ra and \vecv = \la 1,3\ra\text{.}
Verify that the points A = (1,1,1)\text{,} B = (2,3,2)\text{,} C = (4,5,3) and D = (3,3,2) are the vertices of a parallelogram. Find the area of the parallelogram.
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Figure 10.4.10.(a) sketches the parallelogram defined by the vectors \(\vec u\) and \(\vec v\text{.}\) We have a slight problem in that our vectors exist in \(\mathbb{R}^2\text{,}\) not \(\mathbb{R}^3\text{,}\) and the cross product is only defined on vectors in \(\mathbb{R}^3\text{.}\) We skirt this issue by viewing \(\vec u\) and \(\vecv\) as vectors in the \(x-y\) plane of \(\mathbb{R}^3\text{,}\) and rewrite them as \(\vec u = \la 2,1,0\ra\) and \(\vecv =\la 1,3,0\ra\text{.}\) We can now compute the cross product. It is easy to show that \(\crossp uv = \la 0,0,5\ra\text{;}\) therefore the area of the parallelogram is \(A = \norm{\crossp uv} = 5\text{.}\)
(a) (b) Figure 10.4.10. Sketching the parallelograms in Example 10.4.9 -
To show that the quadrilateral \(ABCD\) is a parallelogram (shown in Figure 10.4.10.(b)), we need to show that the opposite sides are parallel. We can quickly show that \(\overrightarrow{AB} =\overrightarrow{DC} = \la 1,2,1\ra\) and \(\overrightarrow{BC} = \overrightarrow{AD} = \la 2,2,1\ra\text{.}\) We find the area by computing the magnitude of the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\text{:}\)
\begin{equation*} \overrightarrow{AB} \times \overrightarrow{BC} = \la 0,1,-2\ra \Rightarrow \norm{\overrightarrow{AB}\times\overrightarrow{BC}} = \sqrt{5} \approx 2.236\text{.} \end{equation*}
Example 10.4.11. Area of a triangle.
Find the area of the triangle with vertices A=(1,2)\text{,} B=(2,3) and C=(3,1)\text{,} as pictured in Figure 10.4.12.
We found the area of this triangle in Example 7.1.11 to be \(1.5\) using integration. There we discussed the fact that finding the area of a triangle can be inconvenient using the β\(\frac12bh\)β 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 \(\overrightarrow{AB} = \la 1,1\ra\) and \(\overrightarrow{AC}=\la 2,-1\ra\text{.}\) 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
We arrive at the same answer as before with less work.
Volume of a Parallelepiped.
The three dimensional analogue to the parallelogram is the parallelepiped. Each face is parallel to the opposite face, as illustrated in Figure 10.4.13. By crossing \vec v and \vec w\text{,} one gets a vector whose magnitude is the area of the base. Dotting this vector with \vecu computes the volume of parallelepiped! (Up to a sign; take the absolute value.)Example 10.4.14. Finding the volume of parallelepiped.
Find the volume of the parallelepiped defined by the vectors \vecu = \la 1,1,0\ra\text{,} \vecv = \la -1,1,0\ra and \vecw = \la 0,1,1\ra\text{.}
We apply Equation (10.4.2). We first find \(\crossp vw =\la 1,1,-1\ra\text{.}\) Then
So the volume of the parallelepiped is 2 cubic units.
Torque.
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 \vec F and \vec \ell\text{,} we see that the bolt moves (because of the threads) in a direction orthogonal to \vec F and \vec \ell\text{.} Torque is usually represented by the Greek letter \tau\text{,} or tau, and has units of N\cdotm, a Newtonβmeter, or ft\cdotlb, a footβpound. While a full understanding of torque is beyond the purposes of this book, when a force \vec F is applied to a lever arm \vec \ell\text{,} the resulting torque isExample 10.4.16. Computing torque.
A lever of length 2ft makes an angle with the horizontal of 45^\circ\text{.} Find the resulting torque when a force of 10lb is applied to the end of the level where:
the force is perpendicular to the lever, and
the force makes an angle of 60^\circ with the lever, as shown in Figure 10.4.17.
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We start by determining vectors for the force and lever arm. Since the lever arm makes a \(45^\circ\) angle with the horizontal and is 2ft long, we can state that \(\vec \ell = 2\la \cos(45^\circ) ,\sin(45^\circ) \ra = \la \sqrt2,\sqrt2\ra\text{.}\) Since the force vector is perpendicular to the lever arm (as seen in the left hand side of Figure 10.4.17), we can conclude it is making an angle of \(-45^\circ\) with the horizontal. As it has a magnitude of 10lb, we can state \(\vec F = 10\la \cos(-45^\circ), \sin(-45^\circ)\ra = \la 5\sqrt2,-5\sqrt2\ra\text{.}\) Using Equation (10.4.3) to find the torque requires a cross product. We again let the third component of each vector be 0 and compute the cross product:
\begin{align*} \vec\tau \amp = \crossp \ell F\\ \amp = \la \sqrt2,\sqrt2,0\ra \times \la 5\sqrt2,-5\sqrt2,0\ra\\ \amp = \la 0,0,-20\ra \end{align*}This clearly has a magnitude of 20 ft-lb. We can view the force and lever arm vectors as lying βon the pageβ; our computation of \(\vec\tau\) shows that the torque goes βinto the page.β 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.
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Our lever arm can still be represented by \(\vec \ell = \la \sqrt2,\sqrt2\ra\text{.}\) As our force vector makes a \(60^\circ\) angle with \(\vec \ell\text{,}\) we can see (referencing the right hand side of the figure) that \(\vec F\) makes a \(-15^\circ\) angle with the horizontal. Thus
\begin{align*} \vec F = 10\la \cos-15^\circ,\sin-15^\circ\ra \amp = \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2}\ra\\ \amp \approx \la 9.659,-2.588\ra\text{.} \end{align*}We again make the third component 0 and take the cross product to find the torque:
\begin{align*} \vec\tau \amp = \crossp \ell F\\ \amp = \la \sqrt2,\sqrt2,0\ra \times \la \frac{5(1+\sqrt3)}{\sqrt2},\frac{5(-1+\sqrt3)}{\sqrt2},0\ra\\ \amp = \la 0,0,-10\sqrt3\ra\\ \amp \approx \la 0,0,-17.321\ra\text{.} \end{align*}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.
Exercises 10.4.3 Exercises
Terms and Concepts
1.
The cross product of two vectors is a , not a scalar.
2.
One can visualize the direction of \vec u\times\vec v using the
.
3.
Give a synonym for βorthogonal.β
4.
True or False? A fundamental principle of the cross product is that \vec u\times\vec v is orthogonal to \vec u and \vec v\text{.}
True
False
5.
6.
T/F: If \vec u and \vec v are parallel, then \vec u\times\vec v=\vec 0\text{.}
In the following exercises, vectors \vec u and \vec v are given. Compute \vec u\times\vec v and check this is orthogonal to both \vec u and \vec v\text{.}
7.
Let \vec u = \la 3,2,-2\ra\text{,} \vec v = \la 0,1,5\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
8.
Let \vec u = \la 5, -4, 3\ra\text{,} \vec v = \la 2, -5, 1\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
9.
Let \vec u = \la 4, -5, -5\ra\text{,} \vec v = \la 3, 3, 4\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
10.
Let \vec u = \la -4, 7, -10\ra\text{,} \vec v = \la 4, 4, 1\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
11.
Let \vec u = \la 1, 0, 1\ra\text{,} \vec v = \la 5, 0, 7\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
12.
Let \vec u = \la 1, 5, -4\ra\text{,} \vec v = \la -2, -10, 8\ra\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
13.
\vec u = \langle a,b,0\rangle\text{,} \vec v = \langle c,d,0\rangle
14.
Let \vec u = \hat\imath\text{,} \vec v = \hat\jmath\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
15.
Let \vec u = \hat\imath\text{,} \vec v = \hat{k}\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
16.
Let \vec u = \hat\jmath\text{,} \vec v = \hat{k}\text{.}
\vec u\times\vec v=
Check this is orthogonal to both \vec u and \vec v\text{.}
17.
Pick any vectors \vec u\text{,} \vec v and \vec w in \mathbb{R}^3 and show that \vec u \times (\vec v+\vec w) = \vec u\times \vec v+\vec u\times \vec w\text{.}
18.
Pick any vectors \vec u\text{,} \vec v and \vec w in \mathbb{R}^3 and show that \vec u \cdot (\vec v\times\vec w) = (\vec u\times \vec v)\cdot \vec w\text{.}
In the following exercises, the magnitudes of vectors \vec u and \vec v in \mathbb{R}^3 are given, along with the angle \theta between them. Use this information to find the magnitude of \vec u\times\vec v\text{.}
19.
If \vnorm{u} = 2\text{,} \vnorm{v} = 5\text{,} and \theta = 30^{\circ} is the angle between \vec u and \vec v\text{,} then \norm{\vec u\times\vec v}=
20.
If \vnorm{u} = 3\text{,} \vnorm{v} = 7\text{,} and \theta = \pi/2 is the angle between \vec u and \vec v\text{,} then \norm{\vec u\times\vec v}=
21.
If \vnorm{u} = 3\text{,} \vnorm{v} = 4\text{,} and \theta = \pi is the angle between \vec u and \vec v\text{,} then \norm{\vec u\times\vec v}=
22.
If \vnorm{u} = 2\text{,} \vnorm{v} = 5\text{,} and \theta = 5\pi/6 is the angle between \vec u and \vec v\text{,} then \norm{\vec u\times\vec v}=
In the following exercises, find the area of the parallelogram defined by the given vectors.
23.
Find the area of the parallelogram defined by \vec u = \la 1,1,2\ra\text{,} and \vec v = \la 2,0,3\ra\text{.}
24.
Find the area of the parallelogram defined by \vec u = \la -2,1,5\ra\text{,} and \vec v = \la -1,3,1\ra\text{.}
25.
Find the area of the parallelogram defined by \vec u = \la 1,2\ra\text{,} and \vec v = \la 2,1\ra\text{.}
26.
Find the area of the parallelogram defined by \vec u = \la 2,0\ra\text{,} and \vec v = \la 0,3\ra\text{.}
In the following exercises, find the area of the triangle with the given vertices.
27.
Find the area of the triangle with vertices (0,0,0)\text{,} (1,3,-1) and (2,1,1)\text{.}
28.
Find the area of the triangle with vertices (5,2,-1)\text{,} (3,6,2) and (1,0,4)\text{.}
29.
Find the area of the triangle with vertices (1,1)\text{,} (1,3) and (2,2)\text{.}
30.
Find the area of the triangle with vertices (3,1)\text{,} (1,2) and (4,3)\text{.}
In the following exercises, find the area of the quadrilateral with the given vertices. (Hint: break the quadrilateral into two triangles.)
In the following exercises, find the volume of the parallelepiped defined by the given vectors.
In the following exercises, find a unit vector orthogonal to both \vec u and \vec v\text{.}
35.
Find a unit vector orthogonal to both \vec u = \la 1,1,1\ra\text{,} and \vec v=\la 2,0,1\ra\text{.}
36.
Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra\text{,} and \vec v=\la 3,2,1\ra\text{.}
37.
Find a unit vector orthogonal to both \vec u = \la 5,0,2\ra\text{,} and \vec v=\la -3,0,7\ra\text{.}
38.
Find a unit vector orthogonal to both \vec u = \la 1,-2,1\ra\text{,} and \vec v=\la -2,4,-2\ra\text{.}
39.
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.
40.
A bicycle rider applies 150lb of force, straight down, onto a pedal that extends 7in from the crankshaft, making a 30^\circ angle with the horizontal. Find the magnitude of the torque applied to the crankshaft.
41.
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?
42.
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^\circ angle with the wrench. How much torque is subsequently applied to the wrench?
43.
Show, using the definition of the Cross Product, that \vec u\cdot(\vec u\times\vec v)=0\text{;} that is, that \vec u is orthogonal to the cross product of \vec u and \vec v\text{.}
44.
Show, using the definition of the Cross Product, that \vec u\times\vec u=\vec 0\text{.}