Inner and Outer Products

This article introduces basic operations on vectors.

Before beginning, a note about terminology is necessary. For historical reasons, mathematics and physics use many different translations for the terms “inner product” and “outer product”.

In physics, they are generally translated as “scalar product” and “vector product”, indicating that the results of the operations are a scalar and a vector. High-school mathematics textbooks likewise use the semantically translated terms “scalar product” and “vector product”.

In mathematics, they can also be translated as “inner product” and “outer product”, which are direct translations of the two nouns. “Dot product” and “cross product” are informal names derived from the operation symbols, and these names are also very common.

In the “dot product” operation, the dot symbol is often omitted. In linear algebra, it is even treated directly as matrix multiplication, without writing the dot symbol.

Inner Product

The concept of an inner product applies to vectors of any dimension.

Definition

There are different but equivalent ways to define an inner product. Some of them are introduced below.

Geometric Definition

In the $n$-dimensional Euclidean space $\mathbf{R}^n$, let two vectors be $\boldsymbol{a}, \boldsymbol{b}$ and let the angle between them be $\theta$. Then

$$\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta$$

is the inner product of the two vectors, also called the dot product or scalar product. The quantity $|\boldsymbol{b}|\cos \theta$ is called the projection of $\boldsymbol{b}$ in the direction of $\boldsymbol{a}$. The geometric meaning of the inner product is that the inner product $\boldsymbol{a} \cdot \boldsymbol{b}$ equals the product of the magnitude of $\boldsymbol{a}$ and the projection of $\boldsymbol{b}$ in the direction of $\boldsymbol{a}$.

Algebraic Definition

In the $n$-dimensional Euclidean space $\mathbf{R}^n$, let $\boldsymbol{a} = (a_1, a_2, \dots, a_n), \boldsymbol{b} = (b_1, b_2, \dots, b_n)$ be two vectors. Then

$$\boldsymbol{a} \cdot \boldsymbol{b} = \sum_{i = 1}^{n} a_i b_i$$

is the inner product of the two vectors, also called the dot product or scalar product. The geometric and algebraic definitions of the inner product are equivalent in Euclidean space, while the latter is more convenient to use.

When there is no risk of confusion, the dot in an inner product may be omitted. If a vector has a superscript $2$ at its upper right, this is shorthand for the inner product of the vector with itself, namely the square of the vector's magnitude, with the magnitude notation omitted. This superscript $2$ cannot be interpreted as the square of a vector, because the result of an inner product is a scalar, and the inner product of a vector with any number other than $2$ of vectors does not exist. Similarly, the square of the square of a vector's magnitude cannot be abbreviated with a superscript $4$; instead, the result represented by the superscript $2$ must be treated as a whole, and so on.

Properties

The result of an inner product is a scalar. Its special feature is that it is a bilinear operation, linear in each of the two vectors separately. Specifically, the inner product satisfies

$$\begin{aligned} (\boldsymbol{a} + \boldsymbol{b}) \cdot \boldsymbol{c} &= \boldsymbol{a} \cdot \boldsymbol{c} + \boldsymbol{b} \cdot \boldsymbol{c} \\ \boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c}) &= \boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c} \\ (\lambda \boldsymbol{a}) \cdot \boldsymbol{b} &= \lambda (\boldsymbol{a} \cdot \boldsymbol{b}) \\ \boldsymbol{a} \cdot (\lambda \boldsymbol{b}) &= \lambda (\boldsymbol{a} \cdot \boldsymbol{b}) \end{aligned}$$

The inner product also satisfies commutativity:

$$\boldsymbol{a} \cdot \boldsymbol{b} = \boldsymbol{b} \cdot \boldsymbol{a}$$

Applications

Some common applications of the inner product are introduced below.

1. Determine whether two vectors are perpendicular:

   $$\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0$$

   The inner product of two perpendicular vectors is $0$; the inner product of a vector and the zero vector is $0$. If zero inner product is used as the definition of perpendicularity, then the zero vector is perpendicular to every vector.

2. Determine whether two vectors are collinear:

   $$\exists\lambda \in \mathbf{R} (\boldsymbol{a} = \lambda \boldsymbol{b}) \iff |\boldsymbol{a} \cdot \boldsymbol{b}| = |\boldsymbol{a}| |\boldsymbol{b}|$$

3. Compute the magnitude of a vector:

   $$|\boldsymbol a| = \sqrt{\boldsymbol{a} \cdot \boldsymbol{a}}$$

4. Compute the angle between two vectors:

   $$\theta = \arccos \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol a| |\boldsymbol b|}$$

Second- and Third-Order Determinants

Second- and third-order determinants can be given special definitions as relatively simple cases of determinants. In the final part of calculus, the field theory section, Green's formula uses second-order determinants, Gauss's formula uses the dot product, and Stokes' formula uses third-order determinants.

A second-order determinant can be viewed as a function of four variables, defined by

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$$

A third-order determinant can be viewed as a function of nine variables, defined by

$$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=aei+dhc+gbf-ahf-dbi-gec$$

A special way to remember these is the “diagonal rule”, which applies only to second- and third-order determinants.

Be especially careful: expanding a fourth-order determinant gives 24 terms, and the term on the secondary diagonal has a positive sign. If the “diagonal rule” for third-order determinants is forcibly applied, there are not enough terms and the sign of the secondary-diagonal term is also incorrect. Therefore, the “diagonal rule” for third-order determinants does not apply to higher-order determinants, and higher-order determinants are also unsuitable for computation by direct expansion.

Cross Product

The cross product is an operation specific to three-dimensional vectors.

In physics, three-dimensional vectors are by default vectors related to positions in space and are always written in bold. However, four-dimensional vectors related to relativity in physics are not written in bold; special notation and subscripts are used instead.

In linear algebra, all vectors are written in bold. Since this is inconvenient, and since most operations in linear algebra are between vectors and matrices and are unlikely to be ambiguous, the vector notation may be omitted in handwriting.

Definition

There are different but equivalent ways to define the cross product. Some of them are introduced below.

Geometric Definition

In the three-dimensional Euclidean space $\mathbf{R}^3$, define the cross product of vectors $\boldsymbol{a}, \boldsymbol{b}$ to be a vector, denoted by $\boldsymbol{a} \times \boldsymbol{b}$, whose magnitude and direction are defined as follows:

1. $|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}| |\boldsymbol{b}| \sin \langle \boldsymbol{a}, \boldsymbol{b} \rangle$;
2. $\boldsymbol{a} \times \boldsymbol{b}$ is perpendicular to $\boldsymbol{a}, \boldsymbol{b}$, and the directions of $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{a} \times \boldsymbol{b}$ follow the right-hand rule.

Notice that the magnitude of the cross product is reminiscent of the triangle-area formula $S=\frac{1}{2}ab\sin C$. Thus, the geometric meaning of the cross product is: $|\boldsymbol{a} \times \boldsymbol{b}|$ is the area of the parallelogram whose adjacent sides are $\boldsymbol{a}, \boldsymbol{b}$.

Algebraic Definition

In the three-dimensional Euclidean space $\mathbf{R}^3$, define the cross product of $\boldsymbol{a} = (x_1, y_1, z_1), \boldsymbol{b} = (x_2, y_2, z_2)$ to be a vector $\boldsymbol{c}$, written as $\boldsymbol{c} = \boldsymbol{a} \times \boldsymbol{b}$. It can be represented by the third-order determinant

$$\begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix}$$

where $\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}$ are unit vectors pointing in the directions of the coordinate axes $x, y, z$, respectively, and are placed at the corresponding coordinates. Expanding gives

$$\begin{aligned} \boldsymbol{c} &= \boldsymbol{a} \times \boldsymbol{b} \\ &= (y_1z_2 - y_2z_1)\boldsymbol{i} + (z_1x_2 - z_2x_1)\boldsymbol{j} + (x_1y_2 - x_2y_1)\boldsymbol{k} \\ &= (y_1z_2 - y_2z_1, z_1x_2 - z_2x_1, x_1y_2 - x_2y_1) \end{aligned}$$

Properties

1. The cross product is a bilinear operation, linear in each of the two vectors separately. Specifically, it satisfies

   $$\begin{aligned} (\boldsymbol{a} + \boldsymbol{b}) \times \boldsymbol{c} &= \boldsymbol{a} \times \boldsymbol{c} + \boldsymbol{b} \times \boldsymbol{c} \\ \boldsymbol{a} \times (\boldsymbol{b} + \boldsymbol{c}) &= \boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{a} \times \boldsymbol{c} \\ (\lambda \boldsymbol{a}) \times \boldsymbol{b} &= \lambda (\boldsymbol{a} \times \boldsymbol{b}) \\ \boldsymbol{a} \times (\lambda \boldsymbol{b}) &= \lambda (\boldsymbol{a} \times \boldsymbol{b}) \end{aligned}$$

   The first two properties are also called the distributive law: the cross product is distributive over vector addition.

2. The cross product is anticommutative:

   $$\boldsymbol a \times \boldsymbol b=-\boldsymbol b \times \boldsymbol a$$

3. From the geometric definitions of the inner and cross products above,

   $$\begin{aligned} |\boldsymbol a \times \boldsymbol b| &= |\boldsymbol a| |\boldsymbol b| \sin \langle \boldsymbol a, \boldsymbol b \rangle \\ \boldsymbol a \cdot \boldsymbol b &= |\boldsymbol a| |\boldsymbol b| \cos \theta \\ &= |\boldsymbol a| |\boldsymbol b| \cos \langle \boldsymbol a, \boldsymbol b\rangle \end{aligned}$$

   we can write the identity

   $$(\boldsymbol a\times \boldsymbol b) \cdot (\boldsymbol a\times \boldsymbol b) = |\boldsymbol a|^2 |\boldsymbol b|^2-{(\boldsymbol a \cdot \boldsymbol b)}^2$$

4. The cross product satisfies the Jacobi identity:

   $$\boldsymbol a \times (\boldsymbol b \times \boldsymbol c) + \boldsymbol b \times (\boldsymbol c \times \boldsymbol a) + \boldsymbol c \times (\boldsymbol a \times \boldsymbol b) = \boldsymbol 0$$

Applications

Some common applications of the cross product are introduced below.

1. Determine whether two vectors are collinear:

   $$\exists\lambda \in \mathbf{R} (\boldsymbol{a} = \lambda \boldsymbol{b}) \iff \boldsymbol{a} \times \boldsymbol{b} = \boldsymbol{0}$$

   The cross product of two collinear three-dimensional vectors is $\boldsymbol 0$; the cross product of a three-dimensional vector with itself is $\boldsymbol 0$; and the cross product of a three-dimensional vector with the zero vector is $\boldsymbol 0$. If zero cross product is used as the definition of collinearity, then the zero vector is collinear with every vector.

2. Compute the area of the parallelogram spanned by two vectors:

   $$S \langle \boldsymbol a, \boldsymbol b \rangle = |\boldsymbol a \times \boldsymbol b|$$

The Two-Dimensional Case

For two-dimensional vectors, the cross product cannot be computed, but the area of the parallelogram spanned by two vectors can still be computed.

Let $\boldsymbol{a} = (m, n), \boldsymbol{b} = (p, q)$. Extend the plane rectangular coordinate system to a spatial rectangular coordinate system, with the original plane as the $xOy$ plane of the new system. The original coordinates $(m, n)$ and $(p, q)$ become $(m, n, 0)$ and $(p, q, 0)$.

The cross product of the two vectors is then $(0, 0, mq - np)$, so the area of the parallelogram is $|mq - np|$, which can be viewed as the absolute value of the result of a second-order determinant.

At this point, from the sign of the $z$ coordinate and the right-hand rule, we can infer the direction of $\boldsymbol b$ relative to $\boldsymbol a$: if it is counterclockwise, the $z$ coordinate is positive; otherwise it is negative. This can be remembered as clockwise negative, counterclockwise positive.

Scalar Triple Product

Like the cross product, the scalar triple product of vectors is an operation specific to three-dimensional vectors.

Definition

Let $\boldsymbol a, \boldsymbol b, \boldsymbol c$ be three vectors in three-dimensional space. Then $(\boldsymbol a \times \boldsymbol b) \cdot \boldsymbol c$ is called the scalar triple product of $\boldsymbol a, \boldsymbol b, \boldsymbol c$, denoted by $[\boldsymbol a \boldsymbol b \boldsymbol c]$, $(\boldsymbol a, \boldsymbol b, \boldsymbol c)$, $(\boldsymbol a \boldsymbol b \boldsymbol c)$, or $\det(\boldsymbol a, \boldsymbol b, \boldsymbol c)$. The geometric meaning of the absolute value $|(\boldsymbol a \times \boldsymbol b) \cdot \boldsymbol c|$ is the volume of the parallelepiped whose edges are $\boldsymbol a, \boldsymbol b, \boldsymbol c$.

The scalar triple product can be represented by a third-order determinant:

$$\begin{aligned} (\boldsymbol a \times \boldsymbol b) \cdot \boldsymbol c &= \det(\boldsymbol a, \boldsymbol b, \boldsymbol c) \\ &= \begin{vmatrix} a_x & b_x & c_x \\ a_y & b_y & c_y \\ a_z & b_z & c_z \end{vmatrix} \\ &= a_x b_y c_z + a_y b_z c_x + a_z b_x c_y - a_z b_y c_x -a _y b_x c_z - a_x b_z c_y \end{aligned}$$

Properties

1. The scalar triple product is linear in each of the three vectors separately. Specifically,

   $$\begin{aligned} \det(\lambda\boldsymbol{u} + \mu\boldsymbol{v}, \boldsymbol{b}, \boldsymbol{c}) &= \lambda\det(\boldsymbol{u}, \boldsymbol{b}, \boldsymbol{c}) + \mu\det(\boldsymbol{v}, \boldsymbol{b}, \boldsymbol{c}) \\ \det(\boldsymbol{a}, \lambda\boldsymbol{u} + \mu\boldsymbol{v}, \boldsymbol{c}) &= \lambda\det(\boldsymbol{a}, \boldsymbol{u}, \boldsymbol{c}) + \mu\det(\boldsymbol{a}, \boldsymbol{v}, \boldsymbol{c}) \\ \det(\boldsymbol{a}, \boldsymbol{b}, \lambda\boldsymbol{u} + \mu\boldsymbol{v}) &= \lambda\det(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{u}) + \mu\det(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{v}) \end{aligned}$$

2. The scalar triple product is antisymmetric: exchanging the positions of two vectors changes it to its opposite, so

   $$\det(\boldsymbol a, \boldsymbol b, \boldsymbol c) = \det(\boldsymbol b, \boldsymbol c, \boldsymbol a) = \det(\boldsymbol c, \boldsymbol a, \boldsymbol b) = -\det(\boldsymbol b, \boldsymbol a, \boldsymbol c) = -\det(\boldsymbol a, \boldsymbol c, \boldsymbol b)= -\det(\boldsymbol c, \boldsymbol b, \boldsymbol a)$$

   This also gives the following relation between the inner and cross products:

   $$(\boldsymbol a \times \boldsymbol b) \cdot \boldsymbol c = \boldsymbol a \cdot (\boldsymbol b \times \boldsymbol c)$$

Applications

Some common applications of the scalar triple product are as follows.

1. Compute the volume of tetrahedron $ABCD$:

   $$V=\frac{1}{6}\left|\det(\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD})\right|$$

2. Determine whether $\boldsymbol a, \boldsymbol b, \boldsymbol c$ are coplanar;

   Three three-dimensional vectors $\boldsymbol a, \boldsymbol b, \boldsymbol c$ are coplanar if and only if $\det(\boldsymbol a, \boldsymbol b, \boldsymbol c)=0$.

3. Determine the handedness of the coordinate system formed by $\boldsymbol a, \boldsymbol b, \boldsymbol c$;

   Whether the scalar triple product $\det(\boldsymbol a, \boldsymbol b, \boldsymbol c)$ is positive or negative depends on whether the angle formed by $\boldsymbol a \times \boldsymbol b$ and $\boldsymbol c$ is acute or obtuse, that is, whether they point to the same or opposite sides of the plane spanned by $\boldsymbol a$ and $\boldsymbol b$. Equivalently, the vectors $\boldsymbol a, \boldsymbol b, \boldsymbol c$ form a right-handed or left-handed system in that order. Specifically:

   • $\det(\boldsymbol a, \boldsymbol b, \boldsymbol c) < 0$ is equivalent to $\boldsymbol a, \boldsymbol b, \boldsymbol c$ forming a left-handed system in that order;
   • $\det(\boldsymbol a, \boldsymbol b, \boldsymbol c) > 0$ is equivalent to $\boldsymbol a, \boldsymbol b, \boldsymbol c$ forming a right-handed system in that order.

Double Cross Product

The scalar triple product of three-dimensional vectors is a combination of the inner and cross products and has cyclic symmetry. The cross product of two three-dimensional vectors is still a three-dimensional vector, so does the cross product of cross products have a related result?

We first prove a lemma.

$$(\boldsymbol a \times \boldsymbol b)\times \boldsymbol a = (\boldsymbol a \cdot \boldsymbol a) \boldsymbol b - (\boldsymbol a \cdot \boldsymbol b) \boldsymbol a$$

Proof: by the right-hand rule, $\boldsymbol a \times \boldsymbol b$ is perpendicular to both $\boldsymbol a$ and $\boldsymbol b$. The left-hand side of the desired equality is perpendicular to $\boldsymbol a \times \boldsymbol b$, so it is coplanar with $\boldsymbol a$ and $\boldsymbol b$.

Therefore, we may assume

$$(\boldsymbol a \times \boldsymbol b)\times \boldsymbol a = \lambda \boldsymbol a + \mu \boldsymbol b$$

Taking inner products of both sides with $\boldsymbol a$ and $\boldsymbol b$, respectively, and using the relevant result for the scalar triple product, gives

$$\begin{aligned} \lambda (\boldsymbol a \cdot \boldsymbol a)+\mu (\boldsymbol a \cdot \boldsymbol b) &= 0 \\ \lambda (\boldsymbol a \cdot \boldsymbol b) + \mu (\boldsymbol b \cdot \boldsymbol b) &= \det(\boldsymbol b, \boldsymbol a \times \boldsymbol b, \boldsymbol a) \\ &= (\boldsymbol a \times \boldsymbol b) \cdot (\boldsymbol a \times \boldsymbol b) \end{aligned}$$

Using the identity derived above,

$$(\boldsymbol a \times \boldsymbol b) \cdot (\boldsymbol a \times \boldsymbol b) = |\boldsymbol a|^2|\boldsymbol b|^2-(\boldsymbol a \cdot \boldsymbol b)^2$$

we obtain

$$\begin{aligned} \lambda &= -\boldsymbol a \cdot \boldsymbol b \\ \mu &= \boldsymbol a \cdot \boldsymbol a \end{aligned}$$

This completes the proof.

The proof above noted that taking the cross product of $\boldsymbol a \times \boldsymbol b$ with any vector produces a vector coplanar with $\boldsymbol a$ and $\boldsymbol b$. We now prove the result for the double cross product:

$$(\boldsymbol a\times \boldsymbol b)\times \boldsymbol c=(\boldsymbol a \cdot \boldsymbol c)\boldsymbol b - (\boldsymbol b \cdot \boldsymbol c)\boldsymbol a$$

The coplanarity above helps us remember the double-cross-product result. The lemma above is a special case of the double cross product.

Proof: it suffices to consider the case where the three vectors are all nonzero and not collinear; the other special cases are immediate.

The three-dimensional vectors $\boldsymbol a$, $\boldsymbol b$, and $\boldsymbol a \times \boldsymbol b$ are not coplanar, so we may assume

$$\boldsymbol c = \alpha \boldsymbol a + \beta \boldsymbol b + \gamma(\boldsymbol a \times \boldsymbol b)$$

Therefore,

$$\begin{aligned} (\boldsymbol a \times \boldsymbol b) \times \boldsymbol c &= (\boldsymbol a \times \boldsymbol b) \times (\alpha \boldsymbol a + \beta \boldsymbol b + \gamma(\boldsymbol a \times \boldsymbol b)) \\ &= \alpha(\boldsymbol a \times \boldsymbol b) \times \boldsymbol a + \beta(\boldsymbol a \times \boldsymbol b) \times \boldsymbol b \end{aligned}$$

By the lemma above,

$$\begin{aligned} (\boldsymbol a \times \boldsymbol b) \times \boldsymbol a &=(\boldsymbol a \cdot \boldsymbol a) \boldsymbol b - (\boldsymbol a \cdot \boldsymbol b) \boldsymbol a \\ (\boldsymbol a \times \boldsymbol b) \times \boldsymbol b &= -(\boldsymbol b \times \boldsymbol a) \times \boldsymbol b \\ &= -(\boldsymbol b \cdot \boldsymbol b)\boldsymbol a+(\boldsymbol a \cdot \boldsymbol b) \boldsymbol b \end{aligned}$$

Thus,

$$\begin{aligned} (\boldsymbol a \times \boldsymbol b) \times \boldsymbol c &= \alpha((\boldsymbol a \cdot \boldsymbol a)\boldsymbol b - (\boldsymbol a \cdot \boldsymbol b)\boldsymbol a) + \beta((\boldsymbol a \cdot \boldsymbol b)\boldsymbol b - (\boldsymbol b \cdot \boldsymbol b)\boldsymbol a) \\ &=(\alpha(-\boldsymbol a \cdot \boldsymbol b) + \beta(-\boldsymbol b \cdot \boldsymbol b))\boldsymbol a + (\alpha \boldsymbol a \cdot \boldsymbol a + \beta \boldsymbol a \cdot \boldsymbol b)\boldsymbol b \\ &= (\boldsymbol a \cdot \boldsymbol c) \boldsymbol b - (\boldsymbol b \cdot \boldsymbol c) \boldsymbol a \end{aligned}$$

This completes the proof.

By the anticommutativity of the cross product, we obtain two formulas for the double cross product:

$$\begin{aligned} (\boldsymbol a\times \boldsymbol b)\times \boldsymbol c &=(\boldsymbol a \cdot \boldsymbol c)\boldsymbol b - (\boldsymbol b \cdot \boldsymbol c)\boldsymbol a \\ \boldsymbol a \times(\boldsymbol b \times \boldsymbol c) &= (\boldsymbol a \cdot \boldsymbol c)\boldsymbol b - (\boldsymbol a \cdot \boldsymbol b)\boldsymbol c \end{aligned}$$

Thus, the order of operations in a double cross product is strictly significant.

With the scalar triple product and the double cross product, we can also prove Lagrange's identity.

$$(\boldsymbol a \times \boldsymbol b) \cdot (\boldsymbol c \times \boldsymbol d)=(\boldsymbol a \cdot \boldsymbol c)(\boldsymbol b \cdot \boldsymbol d)-(\boldsymbol a \cdot \boldsymbol d)(\boldsymbol b \cdot \boldsymbol c)$$

Proof:

$$\begin{aligned} (\boldsymbol a \times \boldsymbol b) \cdot (\boldsymbol c \times \boldsymbol d) &= \det(\boldsymbol c, \boldsymbol d, \boldsymbol a \times \boldsymbol b) \\ &= \det(\boldsymbol a \times \boldsymbol b, \boldsymbol c, \boldsymbol d) \\ &= ((\boldsymbol a \times \boldsymbol b)\times \boldsymbol c)\cdot \boldsymbol d \\ &= (\boldsymbol b(\boldsymbol a \cdot \boldsymbol c)- \boldsymbol a(\boldsymbol b \cdot \boldsymbol c))\cdot \boldsymbol d \\ &= (\boldsymbol a \cdot \boldsymbol c)(\boldsymbol b \cdot \boldsymbol d) - (\boldsymbol a \cdot \boldsymbol d)(\boldsymbol b \cdot \boldsymbol c) \end{aligned}$$

The identity from earlier,

$$(\boldsymbol a \times \boldsymbol b) \cdot (\boldsymbol a \times \boldsymbol b) = |\boldsymbol a|^2|\boldsymbol b|^2 - (\boldsymbol a \cdot \boldsymbol b)^2$$

is a special case of Lagrange's identity.