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Mathematics LibreTexts

11.4: The Cross Product

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Learning Objectives
  • Calculate the cross product of two given vectors.
  • Use determinants to calculate a cross product.
  • Find a vector orthogonal to two given vectors.
  • Determine areas and volumes by using the cross product.
  • Calculate the torque of a given force and position vector.

Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies a force at the end of the wrench. This creates rotation, or torque, which tightens the bolt. We can use vectors to represent the force applied by the mechanic, and the distance (radius) from the bolt to the end of the wrench. Then, we can represent torque by a vector oriented along the axis of rotation. Note that the torque vector is orthogonal to both the force vector and the radius vector.

In this section, we develop an operation called the cross product, which allows us to find a vector orthogonal to two given vectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in the section.

The Cross Product and Its Properties

The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let u=u1,u2,u3 and v=v1,v2,v3 be nonzero vectors. We want to find a vector w=w1,w2,w3 orthogonal to both u and v—that is, we want to find w such that uw=0 and vw=0. Therefore, w1, w2, and w3 must satisfy

u1w1+u2w2+u3w3=0

v1w1+v2w2+v3w3=0.

If we multiply the top equation by v3 and the bottom equation by u3 and subtract, we can eliminate the variable w3, which gives

(u1v3v1u3)w1+(u2v3v2u3)w2=0.

If we select

w1=u2v3u3v2w2=(u1v3u3v1),

we get a possible solution vector. Substituting these values back into the original equations (Equations ??? and ???) gives

w3=u1v2u2v1.

That is, vector

w=u2v3u3v2,(u1v3u3v1),u1v2u2v1

is orthogonal to both u and v, which leads us to define the following operation, called the cross product.

Definition: Cross Product

Let u=u1,u2,u3 and v=v1,v2,v3. Then, the cross product u×v is vector

u×v=(u2v3u3v2)ˆi(u1v3u3v1)ˆj+(u1v2u2v1)ˆk=u2v3u3v2,(u1v3u3v1),u1v2u2v1.

From the way we have developed u×v, it should be clear that the cross product is orthogonal to both u and v. However, it never hurts to check. To show that u×v is orthogonal to u, we calculate the dot product of u and u×v.

u(u×v)=u1,u2,u3u2v3u3v2,u1v3+u3v1,u1v2u2v1=u1(u2v3u3v2)+u2(u1v3+u3v1)+u3(u1v2u2v1)=u1u2v3u1u3v2u1u2v3+u2u3v1+u1u3v2u2u3v1=(u1u2v3u1u2v3)+(u1u3v2+u1u3v2)+(u2u3v1u2u3v1)=0

In a similar manner, we can show that the cross product is also orthogonal to v.

ezgif-2-1a13a291e4.gif
The cross product a×b (vertical, in pink) changes as the angle between the vectors a (blue) and b (red) changes. The cross product (purple) is always perpendicular to both vectors, and has magnitude zero when the vectors are parallel and maximum magnitude ab when they are perpendicular. (Public Domain; LucasVB).
Example 11.4.1: Finding a Cross Product

Let p=1,2,5 and q=4,0,3 (Figure 11.4.1). Find p×q.

This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled “p = <-1, 2, 5>.” The second vector is labeled “q = <4, 0, -3>.”
Figure 11.4.1: Finding a cross product to two given vectors.
Solution

Substitute the components of the vectors into Equation 11.4.7:

p×q=1,2,5×4,0,3=p2q3p3q2,(p1q3p3q1),p1q2p2q1=2(3)5(0),(1)(3)+5(4),(1)(0)2(4)=6,17,8.

Exercise 11.4.1

Find p×q for p=5,1,2 and q=2,0,1. Express the answer using standard unit vectors.

Hint

Use the formula u×v=(u2v3u3v2)ˆi(u1v3u3v1)ˆj+(u1v2u2v1)ˆk.

Answer

p×q=ˆi9ˆj+2ˆk

Although it may not be obvious from Equation 11.4.7, the direction of u×v is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of u, then curl the fingers toward vector v, the thumb points in the direction of the cross product, as shown in Figure 11.4.2.

This figure has two images. The first image has three vectors with the same initial point. Two of the vectors are labeled “u” and “v.” The angle between u and v is theta. The third vector is perpendicular to u and v. It is labeled “u cross v.” The second image has three vectors. The vectors are labeled “u, v, and u cross v.” “u cross v” is perpendicular to u and v. Also, on the image of these three vectors is a right hand. The fingers are in the direction of u. As the hand is closing, the direction of the closing fingers is the direction of v. The thumb is up and in the direction of “u cross v.”
Figure 11.4.2: The direction of u×v is determined by the right-hand rule.

Notice what this means for the direction of v×u. If we apply the right-hand rule to v×u, we start with our fingers pointed in the direction of v, then curl our fingers toward the vector u. In this case, the thumb points in the opposite direction of u×v. (Try it!)

Example 11.4.2: Anticommutativity of the Cross Product

Let u=0,2,1 and v=3,1,0. Calculate u×v and v×u and graph them.

This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled “u = <0, 2, 1>.” The second vector is labeled “v = <3, -1, 0>.”
Figure 11.4.3: Are the cross products u×v and v×u in the same direction?
Solution

We have

u×v=(0+1),(03),(06)=1,3,6

v×u=(10),(30),(60)=1,3,6.

We see that, in this case, u×v=(v×u) (Figure 11.4.4). We prove this in general later in this section.

Three-dimensional coordinate system and 4 vectors. Two of the vectors are labeled V and U; the other two vectors are the cross products V cross U and U cross V. Both are perpendicular to U and V, but they point in opposite directions from each other.

Figure 11.4.4: The cross products u×v and v×u are both orthogonal to u and v, but in opposite directions.

Exercise 11.4.2

Suppose vectors u and v lie in the xy-plane (the z-component of each vector is zero). Now suppose the x- and y-components of u and the y-component of v are all positive, whereas the x-component of v is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does u×v point?

Hint

Remember the right-hand rule (Figure 11.4.2).

Answer

Up (the positive z-direction)

The cross products of the standard unit vectors ˆi, ˆj, and ˆk can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that

ˆi׈i=ˆj׈j=ˆk׈k=0.

(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar 0.) It’s up to you to verify the calculations on your own.

Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of ˆi and ˆj is parallel to ˆk. Similarly, the vector product of ˆi and ˆk is parallel to ˆj, and the vector product of ˆj and ˆk is parallel to ˆi.

We can use the right-hand rule to determine the direction of each product. Then we have

ˆi׈j=ˆkˆj׈i=ˆkˆj׈k=ˆiˆk׈j=ˆiˆk׈i=ˆjˆi׈k=ˆj.

These formulas come in handy later.

Example 11.4.3: Cross Product of Standard Unit Vectors

Find ˆi×(ˆj׈k).

Solution

We know that ˆj׈k=ˆi. Therefore, ˆi×(ˆj׈k)=ˆi׈i=0.

Exercise 11.4.3

Find (ˆi׈j)×(ˆk׈i).

Hint

Remember the right-hand rule (Figure 11.4.2).

Answer

ˆi

As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product. These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.

Properties of the Cross Product

Let u,v, and w be vectors in space, and let c be a scalar.

  1. Anticommutative property: u×v=(v×u)
  2. Distributive property: u×(v+w)=u×v+u×w
  3. Multiplication by a constant: c(u×v)=(cu)×v=u×(cv)
  4. Cross product of the zero vector: u×0=0×u=0
  5. Cross product of a vector with itself: v×v=0
  6. Triple scalar product: u(v×w)=(u×v)w
  7. Triple cross product: u×(v×w)=(uw)v(uv)w
Proof

For property i, we want to show u×v=(v×u). We have

u×v=u1,u2,u3×v1,v2,v3=u2v3u3v2,u1v3+u3v1,u1v2u2v1=u3v2u2v3,u3v1+u1v3,u2v1u1v2=v1,v2,v3×u1,u2,u3=(v×u).

Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.

For property iv., this follows directly from the definition of the cross product. We have

u×0=u2(0)u3(0),(u1(0)u3(0)),u1(0)u2(0)=0,0,0=0.

Then, by property i., 0×u=0 as well. Remember that the dot product of a vector and the zero vector is the scalar 0, whereas the cross product of a vector with the zero vector is the vector 0.

Property vi. looks like the associative property, but note the change in operations:

u(v×w)=uv2w3v3w2,v1w3+v3w1,v1w2v2w1=u1(v2w3v3w2)+u2(v1w3+v3w1)+u3(v1w2v2w1)=u1v2w3u1v3w2u2v1w3+u2v3w1+u3v1w2u3v2w1=(u2v3u3v2)w1+(u3v1u1v3)w2+(u1v2u2v1)w3=u2v3u3v2,u3v1u1v3,u1v2u2v1w1,w2,w3=(u×v)w.

Example 11.4.4: Using the Properties of the Cross Product

Use the cross product properties to calculate (2\mathbf{\hat i}×3\mathbf{\hat j})×\mathbf{\hat j}.

Solution

\begin{align*} (2\mathbf{\hat i}×3 \mathbf{\hat j})×\mathbf{\hat j} &=2(\mathbf{\hat i}×3\mathbf{\hat j})×\mathbf{\hat j} \\[4pt] &=2(3)(\mathbf{\hat i}×\mathbf{\hat j})×\mathbf{\hat j} \\[4pt] &=(6\mathbf{\hat k})×\mathbf{\hat j} \\[4pt] &=6(\mathbf{\hat k}×\mathbf{\hat j}) \\[4pt] &=6(−\mathbf{\hat i})=−6\mathbf{\hat i}. \end{align*}

Exercise \PageIndex{4}

Use the properties of the cross product to calculate (\mathbf{\hat i}×\mathbf{\hat k})×(\mathbf{\hat k}×\mathbf{\hat j}).

Hint

\vecs u×\vecs v=−(\vecs v×\vecs u)

Answer

−\mathbf{\hat k}

So far in this section, we have been concerned with the direction of the vector \vecs u×\vecs v, but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of \vecs u×\vecs v involving the magnitudes of \vecs u and \vecs v, and the sine of the angle between them.

Magnitude of the Cross Product

Let \vecs u and \vecs v be vectors, and let θ be the angle between them. Then, ‖\vecs u×\vecs v‖=‖\vecs u‖⋅‖\vecs v‖⋅\sin θ.

Proof

Let \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩ be vectors, and let θ denote the angle between them. Then

\begin{align*} ‖\vecs u×\vecs v‖^2 &=(u_2v_3−u_3v_2)^2+(u_3v_1−u_1v_3)^2+(u_1v_2−u_2v_1)^2 \\[4pt] &=u^2_2v^2_3−2u_2u_3v_2v_3+u^2_3v^2_2+u^2_3v^2_1−2u_1u_3v_1v_3+u^2_1v^2_3+u^2_1v^2_2−2u_1u_2v_1v_2+u^2_2v^2_1 \\[4pt] &=u^2_1v^2_1+u^2_1v^2_2+u^2_1v^2_3+u^2_2v^2_1+u^2_2v^2_2+u^2_2v^2_3+u^2_3v^2_1+u^2_3v^2_2+u^2_3v^2_3−(u^2_1v^2_1+u^2_2v^2_2+u^2_3v^2_3+2u_1u_2v_1v_2+2u_1u_3v_1v_3+2u_2u_3v_2v_3) \\[4pt] &=(u^2_1+u^2_2+u^2_3)(v^2_1+v^2_2+v^2_3)−(u_1v_1+u_2v_2+u_3v_3)^2 \\[4pt] &=‖\vecs u‖^2‖\vecs v‖^2−(\vecs u⋅\vecs v)^2 \\[4pt] &=‖\vecs u‖^2‖\vecs v‖^2−‖\vecs u‖^2‖\vecs v‖^2 \cos^2θ \\[4pt] &=‖\vecs u‖^2‖\vecs v‖^2(1−\cos^2θ) \\[4pt] &=‖\vecs u‖^2‖\vecs v‖^2(\sin^2θ). \end{align*} \nonumber

Taking square roots and noting that \sqrt{\sin^2θ}=\sinθ for 0≤θ≤180°, we have the desired result:

‖\vecs u×\vecs v‖=‖\vecs u‖‖\vecs v‖ \sin θ. \nonumber

This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.

Example \PageIndex{5}: Calculating the Cross Product

Use "Magnitude of the Cross Product" to find the magnitude of the cross product of \vecs u=⟨0,4,0⟩ and \vecs v=⟨0,0,−3⟩.

Solution

We have

\begin{align*} ‖\vecs u×\vecs v‖ &= ‖\vecs u‖⋅‖\vecs v‖⋅\sinθ \\[4pt] &=\sqrt{0^2+4^2+0^2}⋅\sqrt{0^2+0^2+(−3)^2}⋅\sin{\dfrac{π}{2}} \\[4pt] &=4(3)(1)=12 \end{align*}

Exercise \PageIndex{5}

Use "Magnitude of the Cross Product" to find the magnitude of \vecs u×\vecs v, where \vecs u=⟨−8,0,0⟩ and \vecs v=⟨0,2,0⟩.

Hint

Vectors \vecs u and \vecs v are orthogonal.

Answer

16

Determinants and the Cross Product

Using Equation \ref{cross} to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant notation.

A 2×2 determinant is defined by

\begin{vmatrix}a_1 & b_1\\a_2 & b_2\end{vmatrix} =a_1b_2−b_1a_2. \nonumber

For example,

\begin{vmatrix}3 & −2\\5 & 1\end{vmatrix} =3(1)−5(−2)=3+10=13. \nonumber

A 3×3 determinant is defined in terms of 2×2 determinants as follows:

\begin{vmatrix}a_1 & a_2 & a_3\\b_1 & b_2 & b_3\\c_1 & c_2 & c_3\end{vmatrix}=a_1\begin{vmatrix}b_2 & b_3\\c_2 & c_3\end{vmatrix}−a_2\begin{vmatrix}b_1 & b_3\\c_1 & c_3\end{vmatrix}+a_3\begin{vmatrix}b_1 & b_2\\c_1 & c_2\end{vmatrix}.\label{expandEqn}

Equation \ref{expandEqn} is referred to as the expansion of the determinant along the first row. Notice that the multipliers of each of the 2×2 determinants on the right side of this expression are the entries in the first row of the 3×3 determinant. Furthermore, each of the 2×2 determinants contains the entries from the 3×3 determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, a_1 is the multiplier, and the 2×2 determinant contains the entries that remain if you cross out the first row and first column of the 3×3 determinant. Similarly, for the second term, the multiplier is a_2, and the 2×2 determinant contains the entries that remain if you cross out the first row and second column of the 3×3 determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.

Example \PageIndex{6}: Using Expansion Along the First Row to Compute a 3×3 Determinant

Evaluate the determinant \begin{vmatrix}2 & 5 &−1\\−1 & 1 & 3\\−2 & 3 & 4\end{vmatrix}.

Solution

We have

\begin{align*} \begin{vmatrix}2 & 5 & −1\\−1 & 1 & 3\\−2 & 3 & 4\end{vmatrix} &=2\begin{vmatrix}1 & 3\\3 & 4\end{vmatrix}−5\begin{vmatrix}−1 & 3\\−2 & 4\end{vmatrix}−1\begin{vmatrix}−1 & 1\\−2 & 3\end{vmatrix} \\[4pt] &=2(4−9)−5(−4+6)−1(−3+2) \\[4pt] &= 2(−5)−5(2)−1(−1)=−10−10+1 \\[4pt] &=−19 \end{align*}

Exercise \PageIndex{6}

Evaluate the determinant \begin{vmatrix}1 & −2 & −1\\3 & 2 & −3\\1 & 5 & 4\end{vmatrix}.

Hint

Expand along the first row. Don’t forget the second term is negative!

Answer

40

Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

Rule: Cross Product Calculated by a Determinant

Let \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩ be vectors. Then the cross product \vecs u×\vecs v is given by

\vecs u×\vecs v=\begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3\end{vmatrix}=\begin{vmatrix}u_2 & u_3\\v_2 & v_3\end{vmatrix}\mathbf{\hat i}−\begin{vmatrix}u_1 & u_3\\v_1 & v_3\end{vmatrix}\mathbf{\hat j}+\begin{vmatrix}u_1 & u_2\\v_1 & v_2\end{vmatrix}\mathbf{\hat k}. \nonumber

Example \PageIndex{7}: Using Determinant Notation to find \vecs p×\vecs q

Let \vecs p=⟨−1,2,5⟩ and \vecs q=⟨4,0,−3⟩. Find \vecs p×\vecs q.

Solution

We set up our determinant by putting the standard unit vectors across the first row, the components of \vecs u in the second row, and the components of \vecs v in the third row. Then, we have

\begin{align*} \vecs p×\vecs q &=\begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\−1 & 2 & 5\\4 & 0 & −3\end{vmatrix}=\begin{vmatrix}2 & 5\\0 & −3\end{vmatrix}\mathbf{\hat i}−\begin{vmatrix}−1 & 5\\4 & −3\end{vmatrix}\mathbf{\hat j}+\begin{vmatrix}−1 & 2\\4 & 0\end{vmatrix}\mathbf{\hat k} \\[4pt] &= (−6−0)\mathbf{\hat i}−(3−20)\mathbf{\hat j}+(0−8)\mathbf{\hat k} \\[4pt] &=−6\mathbf{\hat i}+17\mathbf{\hat j}−8\mathbf{\hat k}.\end{align*}

Notice that this answer confirms the calculation of the cross product in Example \PageIndex{1}.

Exercise \PageIndex{7}

Use determinant notation to find \vecs a×\vecs b, where \vecs a=⟨8,2,3⟩ and \vecs b=⟨−1,0,4⟩.

Hint

Calculate the determinant \begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\8 & 2 & 3\\−1 & 0 & 4\end{vmatrix}.

Answer

\vecs a×\vecs b = 8\mathbf{\hat i}−35\mathbf{\hat j}+2\mathbf{\hat k}

Using the Cross Product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped. The following examples illustrate these calculations.

Example \PageIndex{8}: Finding a Unit Vector Orthogonal to Two Given Vectors

Let \vecs a=⟨5,2,−1⟩ and \vecs b=⟨0,−1,4⟩. Find a unit vector orthogonal to both \vecs a and \vecs b.

Solution

The cross product \vecs a×\vecs b is orthogonal to both vectors \vecs a and \vecs b. We can calculate it with a determinant:

\begin{align*} \vecs a×\vecs b &=\begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\5 & 2 & −1\\0 & −1 & 4\end{vmatrix}=\begin{vmatrix}2 & −1\\−1 & 4\end{vmatrix}\mathbf{\hat i}−\begin{vmatrix}5 & −1\\0 & 4\end{vmatrix}\mathbf{\hat j}+\begin{vmatrix}5 & 2\\0 & −1\end{vmatrix}\mathbf{\hat k} \\[4pt] &=(8−1)\mathbf{\hat i}−(20−0)\mathbf{\hat j}+(−5−0)\mathbf{\hat k} \\[4pt] &=7\mathbf{\hat i}−20\mathbf{\hat j}−5\mathbf{\hat k}.\end{align*} \nonumber

Normalize this vector to find a unit vector in the same direction:

\|\vecs a×\vecs b\|=\sqrt{(7)^2+(−20)^2+(−5)^2}=\sqrt{474}.

Thus, \left\langle\dfrac{7}{\sqrt{474}},\dfrac{−20}{\sqrt{474}},\dfrac{−5}{\sqrt{474}}\right\rangle is a unit vector orthogonal to \vecs a and \vecs b.

Simplified, this vector becomes \left\langle\dfrac{7\sqrt{474}}{474},\dfrac{−10\sqrt{474}}{237},\dfrac{−5\sqrt{474}}{474}\right\rangle.

Exercise \PageIndex{8}

Find a unit vector orthogonal to both \vecs a and \vecs b, where \vecs a=⟨4,0,3⟩ and \vecs b=⟨1,1,4⟩.

Hint

Normalize the cross product.

Answer

\left\langle\dfrac{−3}{\sqrt{194}},\dfrac{−13}{\sqrt{194}},\dfrac{4}{\sqrt{194}}\right\rangle or, simplified as \left\langle\dfrac{−3\sqrt{194}}{194},\dfrac{−13\sqrt{194}}{194},\dfrac{2\sqrt{194}}{97}\right\rangle

To use the cross product for calculating areas, we state and prove the following theorem.

Area of a Parallelogram

If we locate vectors \vecs u and \vecs v such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by ‖\vecs u×\vecs v‖ (Figure \PageIndex{5}).

This figure is a parallelogram. One side is represented with a vector labeled “v.” The second side, the base, has the same initial point as vector v and is labeled “u.” The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled “|v|sin(theta).”
Figure \PageIndex{5}: The parallelogram with adjacent sides \vecs u and \vecs v has base ‖\vecs u‖ and height ‖\vecs v‖\sin θ.
Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

\begin{align*} \text{Area of a parallelogram} &= \text{base} × \text{height} \\[4pt] &=‖\vecs u‖(‖\vecs v‖\sin θ) \\[4pt] &=‖\vecs u×\vecs v‖ \end{align*}

Example \PageIndex{9}: Finding the Area of a Triangle

Let P=(1,0,0),Q=(0,1,0), and R=(0,0,1) be the vertices of a triangle (Figure \PageIndex{6}). Find its area.

This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1).
Figure \PageIndex{6}: Finding the area of a triangle by using the cross product.
Solution

We have \vecd{PQ}=⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩ and \vecd{PR}=⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩. The area of the parallelogram with adjacent sides \vecd{PQ} and \vecd{PR} is given by ∥\vecd{PQ}×\vecd{PR}∥:

\begin{align*} \vecd{PQ} \times \vecd{PR} &= \begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\−1 & 1 & 0\\−1 & 0 & 1\end{vmatrix} \\[4pt] &=(1−0)\mathbf{\hat i}−(−1−0)\mathbf{\hat j}+(0−(−1))\mathbf{\hat k} \\[4pt] &=\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k} \\[10pt] ∥\vecd{PQ}×\vecd{PR}∥ &=∥⟨1,1,1⟩∥ \\[4pt] &=\sqrt{1^2+1^2+1^2} \\[4pt] &=\sqrt{3}. \end{align*} \nonumber

The area of ΔPQR is half the area of the parallelogram or \sqrt{3}/2 \, \text{units}^2.

Exercise \PageIndex{9}

Find the area of the parallelogram PQRS with vertices P(1,1,0), Q(7,1,0), R(9,4,2), and S(3,4,2).

Hint

Sketch the parallelogram and identify two vectors that form adjacent sides of the parallelogram.

Answer

6\sqrt{13}\, \text{units}^2

The Triple Scalar Product

Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.

Definition: Triple Scalar Product

The triple scalar product of vectors \vecs u, \vecs v, and \vecs w is

\vecs u⋅( \vecs v× \vecs w). \nonumber

Calculating a Triple Scalar Product

The triple scalar product of vectors

\vecs u=u_1 \mathbf{\hat i}+u_2 \mathbf{\hat j}+u_3\mathbf{\hat k} \nonumber

\vecs v=v_1\mathbf{\hat i}+v_2\mathbf{\hat j}+v_3\mathbf{\hat k} \nonumber

and

\vecs w=w_1 \mathbf{\hat i}+w_2\mathbf{\hat j}+w_3\mathbf{\hat k} \nonumber

is the determinant of the 3×3 matrix formed by the components of the vectors:

\vecs u⋅( \vecs v× \vecs w)=\begin{vmatrix}u_1 & u_2 & u_3\\v_1 & v_2 & v_3\\w_1 & w_2 & w_3\end{vmatrix}. \label{triple2}

Proof

The calculation is straightforward.

\begin{align*} \vecs u⋅( \vecs v× \vecs w) &=⟨u_1,u_2,u_3⟩⋅⟨v_2w_3−v_3w_2,−v_1w_3+v_3w_1,v_1w_2−v_2w_1⟩\\[4pt] &=u_1(v_2w_3−v_3w_2)+u_2(−v_1w_3+v_3w_1)+u_3(v_1w_2−v_2w_1) \\[4pt] &=u_1(v_2w_3−v_3w_2)−u_2(v_1w_3−v_3w_1)+u_3(v_1w_2−v_2w_1) \\[4pt] &=\begin{vmatrix}u_1 & u_2 & u_3\\v_1 & v_2 & v_3\\w_1 & w_2 & w_3\end{vmatrix}.\end{align*} \nonumber

Example \PageIndex{10}: Calculating the Triple Scalar Product

Let \vecs u=⟨1,3,5⟩,\,\vecs v=⟨2,−1,0⟩ and \vecs w=⟨−3,0,−1⟩. Calculate the triple scalar product \vecs u⋅(\vecs v×\vecs w).

Solution

Apply Equation \ref{triple2} directly:

\begin{align*} \vecs u⋅(\vecs v×\vecs w) &=\begin{vmatrix}1 & 3 & 5\\2 & −1 & 0\\−3 & 0 & −1\end{vmatrix} \\[4pt] &=1\begin{vmatrix}−1 & 0\\0 & −1\end{vmatrix}−3\begin{vmatrix}2 & 0\\−3 & −1\end{vmatrix}+5\begin{vmatrix}2 & −1\\−3 & 0\end{vmatrix} \\[4pt] &=(1−0)−3(−2−0)+5(0−3) \\[4pt] &=1+6−15=−8. \end{align*} \nonumber

Exercise \PageIndex{10}

Calculate the triple scalar product \vecs a⋅(\vecs b×\vecs c), where \vecs a=⟨2,−4,1⟩, \vecs b=⟨0,3,−1⟩, and \vecs c=⟨5,−3,3⟩.

Hint

Place the vectors as the rows of a 3×3 matrix, then calculate the determinant.

Answer

17

When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

\begin{vmatrix}a_1 & a_2 & a_3\\b_1 & b_2 & b_3\\c_1 & c_2 & c_3\end{vmatrix}=d \quad\quad \begin{vmatrix}b_1 & b_2 & b_3\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix}=−d \quad\quad \begin{vmatrix}b_1 & b_2 & b_3\\c_1 & c_2 & c_3\\a_1 & a_2 & a_3\end{vmatrix}=d \quad\quad \begin{vmatrix}c_1 & c_2 & c_3\\b_1 & b_2 & b_3\\a_1 & a_2 & a_3\end{vmatrix}=−d

Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:

\begin{align*} \begin{vmatrix}1 & 2 & 1\\−2 & 0 & 3\\4 & 1 & −1\end{vmatrix} &=\begin{vmatrix}0 & 3\\1 & −1\end{vmatrix}−2\begin{vmatrix}−2 & 3\\4 & −1\end{vmatrix}+\begin{vmatrix}−2 & 0\\4 & 1\end{vmatrix} \\[4pt] &=(0−3)−2(2−12)+(−2−0) \\[4pt] &=−3+20−2=15. \end{align*} \nonumber

Switching the top two rows we have

\begin{align*} \begin{vmatrix}−2 & 0 & 3\\1 & 2 & 1\\4 & 1 & −1\end{vmatrix} &=-2\begin{vmatrix}2 & 1\\1 & −1\end{vmatrix}+3\begin{vmatrix}1 & 2\\4 & 1\end{vmatrix} \\[4pt] &=−2(−2−1)+3(1−8)\\[4pt] &=6−21=−15. \end{align*} \nonumber

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let \vecs u=u_1\mathbf{\hat i}+u_2\mathbf{\hat j}+u_3\mathbf{\hat k}, \vecs v=v_1\mathbf{\hat i}+v_2\mathbf{\hat j}+v_3\mathbf{\hat k}, and \vecs w=w_1\mathbf{\hat i}+w_2\mathbf{\hat j}+w_3\mathbf{\hat k}. Applying Calculating a Triple Scalar Product, we have

\vecs u⋅(\vecs v×\vecs w)=\begin{vmatrix}u_1 & u_2 & u_3\\v_1 & v_2 & v_3\\w_1 & w_2 & w_3\end{vmatrix} \nonumber

and

\vecs u⋅(\vecs w×\vecs v)=\begin{vmatrix}u_1 & u_2 & u_3\\w_1 & w_2 & w_3\\v_1 & v_2 & v_3\end{vmatrix}. \nonumber

We can obtain the determinant for calculating \vecs u⋅(\vecs w×\vecs v) by switching the bottom two rows of \vecs u⋅(\vecs v×\vecs w). Therefore, \vecs u⋅(\vecs v×\vecs w)=−\vecs u⋅(\vecs w×\vecs v).

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

\begin{align} \vecs u⋅(\vecs v×\vecs w)&=−\vecs u⋅(\vecs w×\vecs v)\\[10pt] \vecs u⋅(\vecs v×\vecs w)&=\vecs v⋅(\vecs w×\vecs u)=\vecs w⋅(\vecs u×\vecs v).\end{align} \nonumber

Let \vecs u and \vecs v be two vectors in standard position. If \vecs u and \vecs v are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in Area of a Parallelogram that the area of this parallelogram is ‖\vecs u×\vecs v‖. Now suppose we add a third vector \vecs w that does not lie in the same plane as \vecs u and \vecs v but still shares the same initial point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are each parallelograms, as shown in Figure \PageIndex{7}. The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of \vecs u,\vecs v, and \vecs w provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

Volume of a Parallelepiped

The volume of a parallelepiped with adjacent edges given by the vectors \vecs u,\vecs v, and \vecs w is the absolute value of the triple scalar product (Figure \PageIndex{7}):

V=|\vecs u⋅(\vecs v×\vecs w)|. \nonumber

Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.

This figure is a parallelepided, a three dimensional parallelogram. Three of the sides are represented with vectors. The base has vectors v and w. The vertical side has vector u. All three vectors have the same initial point. A perpendicular vector is drawn from this common point. It is labeled “proj sub (v x w) u.”
Figure \PageIndex{7}: The height of the parallelepiped is given by \|\text{proj}_{\vecs v×\vecs w}\vecs u\|.
Proof

The area of the base of the parallelepiped is given by ‖\vecs v×\vecs w‖. The height of the figure is given by \|\text{proj}_{\vecs v×\vecs w}\vecs u\|. The volume of the parallelepiped is the product of the height and the area of the base, so we have

\begin{align*} V &=∥\text{proj}_{\vecs v×\vecs w}\vecs u∥‖\vecs v×\vecs w‖ \\[4pt] &=\left|\dfrac{\vecs u⋅(\vecs v×\vecs w)}{‖\vecs v×\vecs w‖}\right|‖\vecs v×\vecs w‖ \\[4pt] &=|\vecs u⋅(\vecs v×\vecs w)|. \end{align*}

Example \PageIndex{11}: Calculating the Volume of a Parallelepiped

Let \vecs u=⟨−1,−2,1⟩,\vecs v=⟨4,3,2⟩, and \vecs w=⟨0,−5,−2⟩. Find the volume of the parallelepiped with adjacent edges \vecs u,\vecs v, and \vecs w (Figure \PageIndex{8}).

This figure is the 3-dimensional coordinate system. It has three vectors in standard position. The vectors are u = <-1, -2, 1>; v = <4, 3, 2>; and w = <0, -5, -2>.
Figure \PageIndex{8}
Solution

We have

\begin{align*} \vecs u⋅(\vecs v×\vecs w) &=\begin{vmatrix}−1 & −2 & 1\\4 & 3 & 2\\0 & −5 & −2\end{vmatrix} \\[4pt] &= (−1)\begin{vmatrix}3 & 2\\−5 & −2\end{vmatrix}+2\begin{vmatrix}4 & 2\\0 & −2\end{vmatrix}+\begin{vmatrix}4 & 3\\0 & −5\end{vmatrix} \\[4pt] &=(−1)(−6+10)+2(−8−0)+(−20−0) \\[4pt] &=−4−16−20 \\[4pt] &=−40.\end{align*}

Thus, the volume of the parallelepiped is |−40|=40 units3

Exercise \PageIndex{11}

Find the volume of the parallelepiped formed by the vectors \vecs a=3\mathbf{\hat i}+4\mathbf{\hat j}−\mathbf{\hat k}, \vecs b=2\mathbf{\hat i}−\mathbf{\hat j}−\mathbf{\hat k}, and \vecs c=3\mathbf{\hat j}+\mathbf{\hat k}.

Hint

Calculate the triple scalar product by finding a determinant.

Answer

8 units3

Applications of the Cross Product

The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (Introduction to Vector Calculus).

Example \PageIndex{12}: Using the Triple Scalar Product

Use the triple scalar product to show that vectors \vecs u=⟨2,0,5⟩,\vecs v=⟨2,2,4⟩, and \vecs w=⟨1,−1,3⟩ are coplanar—that is, show that these vectors lie in the same plane.

Solution

Start by calculating the triple scalar product to find the volume of the parallelepiped defined by \vecs u,\vecs v, and \vecs w:

\begin{align*} \vecs u⋅(\vecs v×\vecs w)&=\begin{vmatrix}2 & 0 & 5\\2 & 2 & 4\\1 & −1 & 3\end{vmatrix} \\[4pt] &=[2(2)(3)+(0)(4)(1)+5(2)(−1)]−[5(2)(1)+(2)(4)(−1)+(0)(2)(3)] \\[4pt] &=2−2 =0. \end{align*}

The volume of the parallelepiped is 0 units3, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.

Exercise \PageIndex{12}

Are the vectors \vecs a=\mathbf{\hat i}+\mathbf{\hat j}−\mathbf{\hat k}, \vecs b=\mathbf{\hat i}−\mathbf{\hat j}+\mathbf{\hat k}, and \vecs c=\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k} coplanar?

Hint

Calculate the triple scalar product.

Answer

No, the triple scalar product is −4≠0, so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.

Example \PageIndex{13}: Finding an Orthogonal Vector

Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points P=(9,−3,−2),Q=(1,3,0), and R=(−2,5,0).

Solution

The plane must contain vectors \vecd{PQ} and \vecd{QR}:

\vecd{PQ}=⟨1−9,3−(−3),0−(−2)⟩=⟨−8,6,2⟩

\vecd{QR}=⟨−2−1,5−3,0−0⟩=⟨−3,2,0⟩.

The cross product \vecd{PQ}×\vecd{QR} produces a vector orthogonal to both \vecd{PQ} and \vecd{QR}. Therefore, the cross product is orthogonal to the plane that contains these two vectors:

\begin{align*} \vecd{PQ}×\vecd{QR} &= \begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\−8 & 6 & 2\\−3 & 2 & 0\end{vmatrix}\\[4pt] &=0\mathbf{\hat i}−6\mathbf{\hat j}−16\mathbf{\hat k}−(−18\mathbf{\hat k}+4\mathbf{\hat i}+0\mathbf{\hat j})\\[4pt] &=−4\mathbf{\hat i}−6\mathbf{\hat j}+2\mathbf{\hat k}. \end{align*}

We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.

Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.

Definition: Torque

Torque, \vecs \tau (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let \vecs r be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector \vecs F represent the force. Then torque is equal to the cross product of \vecs r and \vecs F:

\vecs \tau=\vecs r×\vecs F. \nonumber

See Figure \PageIndex{9}.

This figure has a vector r from an “axis of rotation”. At the terminal point of r there is a vector labeled “F”. The angle between r and F is theta.
Figure \PageIndex{9}: Torque measures how a force causes an object to rotate.

Think about using a wrench to tighten a bolt. The torque τ applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.

Example \PageIndex{14}: Evaluating Torque

A bolt is tightened by applying a force of 6 N to a 0.15-m wrench (Figure \PageIndex{10}). The angle between the wrench and the force vector is 40°. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.

This figure is the image of an open-end wrench. The length of the wrench is labeled “0.15 m.” The angle the wrench makes with a vertical vector is 40 degrees. The vector is labeled with “6 N.”
Figure \PageIndex{10}: Torque describes the twisting action of the wrench.

Solution:

Substitute the given information into the equation defining torque:

\begin{align*} ‖\vecs τ‖ &=\|\vecs r×\vecs F\| \\[4pt] &=‖\vecs r‖∥\vecs F∥\sinθ \\[4pt] &=(0.15\,\text{m})(6\,\text{N})\sin 40° \\[4pt] &≈0.58\,\text{N⋅m.} \end{align*}

Exercise \PageIndex{14}

Calculate the force required to produce 15 N⋅m torque at an angle of 30º from a 150-cm rod.

Hint

‖\vecs τ‖=15 N⋅m and ‖\vecs r‖=1.5 m

Answer

20 N

Key Concepts

  • The cross product \vecs u×\vecs v of two vectors \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩ is a vector orthogonal to both \vecs u and \vecs v. Its length is given by ‖\vecs u×\vecs v‖=‖\vecs u‖⋅‖\vecs v‖⋅\sin θ, where θ is the angle between \vecs u and \vecs v. Its direction is given by the right-hand rule.
  • The algebraic formula for calculating the cross product of two vectors,

\vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩, is

\vecs u×\vecs v=(u_2v_3−u_3v_2)\mathbf{\hat i}−(u_1v_3−u_3v_1)\mathbf{\hat j}+(u_1v_2−u_2v_1)\mathbf{\hat k}.

  • The cross product satisfies the following properties for vectors \vecs u,\vecs v, and \vecs w, and scalar c:

\vecs u×\vecs v=−(\vecs v×\vecs u)

\vecs u×(\vecs v+\vecs w)=\vecs u×\vecs v+\vecs u×\vecs w

c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)

\vecs u×\vecs 0=\vecs 0×\vecs u=\vecs 0

\vecs v×\vecs v=\vecs 0

\vecs u⋅(\vecs v×\vecs w)=(\vecs u×\vecs v)⋅\vecs w

  • The cross product of vectors \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩ is the determinant \begin{vmatrix}\mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3\end{vmatrix}
  • If vectors \vecs u and \vecs v form adjacent sides of a parallelogram, then the area of the parallelogram is given by \|\vecs u×\vecs v\|.
  • The triple scalar product of vectors \vecs u, \vecs v, and \vecs w is \vecs u⋅(\vecs v×\vecs w).
  • The volume of a parallelepiped with adjacent edges given by vectors \vecs u,\vecs v, and \vecs w is V=|\vecs u⋅(\vecs v×\vecs w)|.
  • If the triple scalar product of vectors \vecs u,\vecs v, and \vecs w is zero, then the vectors are coplanar. The converse is also true: If the vectors are coplanar, then their triple scalar product is zero.
  • The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
  • Torque \vecs τ measures the tendency of a force to produce rotation about an axis of rotation. If force \vecs F is acting at a distance (displacement) \vecs r from the axis, then torque is equal to the cross product of \vecs r and \vecs F: \vecs τ=\vecs r×\vecs F.

Key Equations

  • The cross product of two vectors in terms of the unit vectors

\vecs u×\vecs v=(u_2v_3−u_3v_2)\mathbf{\hat i}−(u_1v_3−u_3v_1)\mathbf{\hat j}+(u_1v_2−u_2v_1)\mathbf{\hat k} \nonumber

Glossary

cross product

\vecs u×\vecs v=(u_2v_3−u_3v_2)\mathbf{\hat i}−(u_1v_3−u_3v_1)\mathbf{\hat j}+(u_1v_2−u_2v_1)\mathbf{\hat k}, where \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩

determinant

a real number associated with a square matrix

parallelepiped

a three-dimensional prism with six faces that are parallelograms

torque

the effect of a force that causes an object to rotate

triple scalar product

the dot product of a vector with the cross product of two other vectors: \vecs u⋅(\vecs v×\vecs w)

vector product

the cross product of two vectors


11.4: The Cross Product is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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