Vector Triple Product
Definition
The vector triple product refers to the cross product of a vector with the cross product of two other vectors. Given three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\), the vector triple product is expressed as
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}).
\]
This is not a new kind of product but simply a sequential application of the cross product to three vectors. We can evaluate it directly by first computing \(\mathbf{b} \times \mathbf{c}\) and then taking its cross product with \(\mathbf{a}\). However, rather than performing two separate cross products, this expression can be significantly simplified using an important identity, which reduces computational effort.
Additionally, the vector triple product has a crucial geometric interpretation, making it highly useful in various applications.
Evaluating the Vector Triple Product
The vector triple product identity provides a useful simplification for expressions involving the cross product of three vectors. The key identities are:
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}
\]
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}
\]
These identities allow us to evaluate vector triple products without explicitly computing two separate cross products.
Let
\[
\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}, \quad
\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}, \quad
\mathbf{c} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k}.
\]
The cross product \(\mathbf{b} \times \mathbf{c}\) is
\[
\mathbf{b} \times \mathbf{c} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{vmatrix}
= (b_2 c_3 - b_3 c_2) \mathbf{i} + (b_3 c_1 - b_1 c_3) \mathbf{j} + (b_1 c_2 - b_2 c_1) \mathbf{k}.
\]
Writing this as \(\mathbf{b} \times \mathbf{c} = p \mathbf{i} + q \mathbf{j} + r \mathbf{k}\), where
\[
p = b_2 c_3 - b_3 c_2, \quad q = b_3 c_1 - b_1 c_3, \quad r = b_1 c_2 - b_2 c_1.
\]
Now, computing \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) leads to a long expansion. For simplicity, consider only the \( i \)-component:
\[
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
p & q & r
\end{vmatrix}.
\]
Its \( i \)-component is
\[
a_2 r - a_3 q = a_2 (b_1 c_2 - b_2 c_1) - a_3 (b_3 c_1 - b_1 c_3).
\]
Expanding,
\[
(a_2 b_1 c_2 + a_3 b_1 c_3) - (a_2 b_2 c_1 + a_3 b_3 c_1).
\]
Factoring,
\[
b_1 (a_2 c_2 + a_3 c_3) - c_1 (a_2 b_2 + a_3 b_3).
\]
Adding \( a_1 c_1 \) in the first bracket and \( a_1 b_1 \) in the second (which does not change the value),
\[
b_1 (a_1 c_1 + a_2 c_2 + a_3 c_3) - c_1 (a_1 b_1 + a_2 b_2 + a_3 b_3).
\]
Since \( \mathbf{a} \cdot \mathbf{c} = a_1 c_1 + a_2 c_2 + a_3 c_3 \) and \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \),
\[
b_1 (\mathbf{a} \cdot \mathbf{c}) - c_1 (\mathbf{a} \cdot \mathbf{b}).
\]
Similarly, the \( j \)-component is
\[
b_2 (\mathbf{a} \cdot \mathbf{c}) - c_2 (\mathbf{a} \cdot \mathbf{b}),
\]
and the \( k \)-component is
\[
b_3 (\mathbf{a} \cdot \mathbf{c}) - c_3 (\mathbf{a} \cdot \mathbf{b}).
\]
Thus,
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (b_1 (\mathbf{a} \cdot \mathbf{c}) - c_1 (\mathbf{a} \cdot \mathbf{b})) \mathbf{i} + (b_2 (\mathbf{a} \cdot \mathbf{c}) - c_2 (\mathbf{a} \cdot \mathbf{b})) \mathbf{j} + (b_3 (\mathbf{a} \cdot \mathbf{c}) - c_3 (\mathbf{a} \cdot \mathbf{b})) \mathbf{k}.
\]
Rewriting,
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}.
\]
For the second identity,
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = -\mathbf{c} \times (\mathbf{a} \times \mathbf{b}).
\]
Applying the first identity,
\[
-\left( (\mathbf{c} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} \right).
\]
Distributing the negative sign,
\[
(\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a}.
\]
This gives
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a}.
\]
Example
Given the vectors
\[
\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}, \quad
\mathbf{b} = 2\mathbf{i} + \mathbf{k}, \quad
\mathbf{c} = \mathbf{i} - \mathbf{j} - \mathbf{k},
\]
evaluate both \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) and \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\).
Solution:
Using the vector triple product identities,
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c},
\]
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a}.
\]
**Computing \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) **
First, compute the dot products:
\[
\mathbf{a} \cdot \mathbf{c} = (1 \cdot 1) + (1 \cdot (-1)) + (1 \cdot (-1)) = 1 - 1 - 1 = -1.
\]
\[
\mathbf{a} \cdot \mathbf{b} = (1 \cdot 2) + (1 \cdot 0) + (1 \cdot 1) = 2 + 0 + 1 = 3.
\]
Substituting into the identity:
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (-1)(2\mathbf{i} + \mathbf{k}) - (3)(\mathbf{i} - \mathbf{j} - \mathbf{k}).
\]
Expanding,
\[
-2\mathbf{i} - \mathbf{k} - 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}.
\]
\[
= -5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}.
\]
**Computing \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) **
First, compute
\[
\mathbf{c} \cdot \mathbf{a} = (1 \cdot 1) + (-1 \cdot 1) + (-1 \cdot 1) = 1 - 1 - 1 = -1.
\]
\[
\mathbf{c} \cdot \mathbf{b} = (1 \cdot 2) + (-1 \cdot 0) + (-1 \cdot 1) = 2 + 0 - 1 = 1.
\]
Substituting into the identity:
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (-1) \mathbf{b} - (1) \mathbf{a}.
\]
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = - (2\mathbf{i} + \mathbf{k}) - (\mathbf{i} + \mathbf{j} + \mathbf{k}).
\]
Expanding,
\[
-2\mathbf{i} - \mathbf{k} - \mathbf{i} - \mathbf{j} - \mathbf{k}.
\]
\[
= -3\mathbf{i} - \mathbf{j} - 2\mathbf{k}.
\]
Thus,
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = -5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}.
\]
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = -3\mathbf{i} - \mathbf{j} - 2\mathbf{k}.
\]
Geometrical Interpretation of Vector Triple Product
The geometrical interpretation of the vector triple product follows naturally from its algebraic structure. Given three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\), consider the expression
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}).
\]
From the vector triple product identity,
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}.
\]
This result is a linear combination of \(\mathbf{b}\) and \(\mathbf{c}\), which implies that the vector \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) must lie in the plane spanned by \(\mathbf{b}\) and \(\mathbf{c}\). Furthermore, since the cross product \(\mathbf{b} \times \mathbf{c}\) is perpendicular to both \(\mathbf{b}\) and \(\mathbf{c}\), the vector \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) is perpendicular to \(\mathbf{a}\), as illustrated in the figure.
Similarly, for
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c},
\]
the identity
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a}
\]
shows that this vector is a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\), meaning it lies in the plane spanned by \(\mathbf{a}\) and \(\mathbf{b}\). Since \(\mathbf{a} \times \mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\), the vector \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) is necessarily perpendicular to \(\mathbf{c}\).
Thus, the vector triple product identities not only provide computational simplifications but also encode essential geometric information—the resultant vector always lies in a specific plane and is perpendicular to the remaining vector.
Significance of Vector Triple Product
The vector triple product solves the problem of finding a vector that is coplanar with two given, non-collinear vectors \(\mathbf{b}\) and \(\mathbf{c}\) but is also perpendicular to some other vector \(\mathbf{a}\).
Given three vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\), the vector \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) is always in the plane of \(\mathbf{b}\) and \(\mathbf{c}\) and perpendicular to \(\mathbf{a}\). This follows from the identity:
\[
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}.
\]
Since this result is a linear combination of \(\mathbf{b}\) and \(\mathbf{c}\), it must be coplanar with them. Furthermore, as \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\) arises from a cross product involving \(\mathbf{a}\), it is necessarily perpendicular to \(\mathbf{a}\).
Thus, if we need a vector \(\mathbf{v}\) that is coplanar with \(\mathbf{b}, \mathbf{c}\) but perpendicular to \(\mathbf{a}\), we conclude that
\[
\mathbf{v} \parallel \mathbf{a} \times (\mathbf{b} \times \mathbf{c}).
\]
Therefore, the most general form of such a vector can be taken as
\[
\mathbf{v} = \lambda \mathbf{a} \times (\mathbf{b} \times \mathbf{c})
\]
for some scalar \(\lambda\).
Another very common situation is finding a vector \(\mathbf{v}\) that is coplanar with two non-collinear vectors \(\mathbf{a}\) and \(\mathbf{b}\), but also perpendicular to \(\mathbf{a}\).
Since \(\mathbf{a} \times \mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\), taking the cross product of \(\mathbf{a}\) with \(\mathbf{a} \times \mathbf{b}\) ensures that the resulting vector lies in the plane spanned by \(\mathbf{a}\) and \(\mathbf{b}\), while still being perpendicular to \(\mathbf{a}\). This follows from the identity:
\[
\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{b}.
\]
Since this vector is both coplanar with \(\mathbf{a}\) and \(\mathbf{b}\) and also perpendicular to \(\mathbf{a}\), any vector \(\mathbf{v}\) satisfying these conditions must be collinear with \(\mathbf{a} \times (\mathbf{a} \times \mathbf{b})\), meaning
\[
\mathbf{v} \parallel \mathbf{a} \times (\mathbf{a} \times \mathbf{b}).
\]
Thus, the most general form of such a vector is
\[
\mathbf{v} = \lambda \mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = \lambda \left( (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{b} \right).
\]
Lagrange's Identity
The expression
\[
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d})
\]
is not immediately suitable for algebraic manipulation. However, we can transform it using the cyclic property of the scalar triple product. Define \(\mathbf{u} = \mathbf{a} \times \mathbf{b}\), so that
\[
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = \mathbf{u} \cdot (\mathbf{c} \times \mathbf{d}).
\]
This is now in the form of a scalar triple product, which satisfies the identity
\[
\mathbf{u} \cdot (\mathbf{c} \times \mathbf{d}) = \mathbf{c} \cdot (\mathbf{d} \times \mathbf{u}).
\]
Substituting back \(\mathbf{u} = \mathbf{a} \times \mathbf{b}\), we obtain
\[
\mathbf{c} \cdot (\mathbf{d} \times (\mathbf{a} \times \mathbf{b})).
\]
Now applying the vector triple product identity,
\[
\mathbf{d} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{d} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{d} \cdot \mathbf{a}) \mathbf{b},
\]
we substitute this into the expression:
\[
\mathbf{c} \cdot \left( (\mathbf{d} \cdot \mathbf{b}) \mathbf{a} - (\mathbf{d} \cdot \mathbf{a}) \mathbf{b} \right).
\]
Using the distributive property of the dot product,
\[
(\mathbf{d} \cdot \mathbf{b}) (\mathbf{c} \cdot \mathbf{a}) - (\mathbf{d} \cdot \mathbf{a}) (\mathbf{c} \cdot \mathbf{b}).
\]
Since \(\mathbf{d} \cdot \mathbf{b}\) and \(\mathbf{d} \cdot \mathbf{a}\) are scalars, this expression is equivalent to the determinant
\[
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) =
\begin{vmatrix}
\mathbf{a} \cdot \mathbf{c} & \mathbf{a} \cdot \mathbf{d} \\
\mathbf{b} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{d}
\end{vmatrix}.
\]
This result is known as Lagrange’s identity, which expresses the dot product of two cross products in terms of a determinant of inner products.
As a special case, setting \(\mathbf{c} = \mathbf{a}\) and \(\mathbf{d} = \mathbf{b}\) in Lagrange’s identity,
\[
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) =
\begin{vmatrix}
\mathbf{a} \cdot \mathbf{c} & \mathbf{a} \cdot \mathbf{d} \\
\mathbf{b} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{d}
\end{vmatrix},
\]
yields
\[
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) =
\begin{vmatrix}
\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} \\
\mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b}
\end{vmatrix}.
\]
Since \((\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = |\mathbf{a} \times \mathbf{b}|^2\), we obtain
\[
|\mathbf{a} \times \mathbf{b}|^2 =
\begin{vmatrix}
|\mathbf{a}|^2 & \mathbf{a} \cdot \mathbf{b} \\
\mathbf{a} \cdot \mathbf{b} & |\mathbf{b}|^2
\end{vmatrix}.
\]
Expanding the determinant,
\[
|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2.
\]
Rearranging,
\[
|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2 |\mathbf{b}|^2.
\]
This is a fundamental identity in vector algebra, showing that the sum of the squared magnitudes of the cross product and the dot product equals the product of the squared magnitudes of the vectors. It expresses the Pythagorean relation in terms of vector products, capturing the decomposition of \(\mathbf{a}\) and \(\mathbf{b}\) into perpendicular and parallel components.
Double Vector Triple Product
The expression
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d})
\]
is a double vector triple product, a structure that can be simplified using the vector triple product identity.
To evaluate it, define \(\mathbf{u} = \mathbf{a} \times \mathbf{b}\), so that
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \mathbf{u} \times (\mathbf{c} \times \mathbf{d}).
\]
Applying the vector triple product identity
\[
\mathbf{u} \times (\mathbf{c} \times \mathbf{d}) = (\mathbf{u} \cdot \mathbf{d}) \mathbf{c} - (\mathbf{u} \cdot \mathbf{c}) \mathbf{d},
\]
substituting back \(\mathbf{u} = \mathbf{a} \times \mathbf{b}\),
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = ((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}) \mathbf{c} - ((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}) \mathbf{d}.
\]
Using the scalar triple product identity,
\[
(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = [\mathbf{a}, \mathbf{b}, \mathbf{d}] \quad \text{and} \quad (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = [\mathbf{a}, \mathbf{b}, \mathbf{c}],
\]
this simplifies to
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{b}, \mathbf{d}] \mathbf{c} - [\mathbf{a}, \mathbf{b}, \mathbf{c}] \mathbf{d}.
\]
Alternatively, defining \(\mathbf{v} = \mathbf{c} \times \mathbf{d}\), the expression rewrites as
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{v}.
\]
Applying the vector triple product identity again,
\[
(\mathbf{a} \times \mathbf{b}) \times \mathbf{v} = (\mathbf{a} \cdot \mathbf{v}) \mathbf{b} - (\mathbf{b} \cdot \mathbf{v}) \mathbf{a}.
\]
Using the scalar triple product,
\[
\mathbf{a} \cdot (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{c}, \mathbf{d}] \quad \text{and} \quad \mathbf{b} \cdot (\mathbf{c} \times \mathbf{d}) = [\mathbf{b}, \mathbf{c}, \mathbf{d}],
\]
this gives
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{c}, \mathbf{d}] \mathbf{b} - [\mathbf{b}, \mathbf{c}, \mathbf{d}] \mathbf{a}.
\]
Thus, the double vector triple product satisfies both representations:
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{b}, \mathbf{d}] \mathbf{c} - [\mathbf{a}, \mathbf{b}, \mathbf{c}] \mathbf{d}
\]
\[
= [\mathbf{a}, \mathbf{c}, \mathbf{d}] \mathbf{b} - [\mathbf{b}, \mathbf{c}, \mathbf{d}] \mathbf{a}.
\]
The expression
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d})
\]
has both an algebraic simplification and a geometric interpretation. Using the vector triple product identity, it expands as
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{b}, \mathbf{d}] \mathbf{c} - [\mathbf{a}, \mathbf{b}, \mathbf{c}] \mathbf{d}.
\]
Alternatively, by treating \(\mathbf{c} \times \mathbf{d}\) as a single vector, we obtain
\[
(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = [\mathbf{a}, \mathbf{c}, \mathbf{d}] \mathbf{b} - [\mathbf{b}, \mathbf{c}, \mathbf{d}] \mathbf{a}.
\]
This means the result is a linear combination of \(\mathbf{c}\) and \(\mathbf{d}\) in the first form, and a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\) in the second.
Geometric Interpretation
Since the result is a combination of two vectors, it must lie in the plane formed by those two vectors.
- The first form shows that \((\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d})\) lies in the plane spanned by \(\mathbf{c}\) and \(\mathbf{d}\).
- The second form shows that it also lies in the plane spanned by \(\mathbf{a}\) and \(\mathbf{b}\).
This means that the resulting vector is in both planes at the same time. The only vectors that satisfy this condition lie in the intersection of the two planes.
Thus, \((\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d})\) is a vector that belongs to the common plane of both \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}, \mathbf{d}\). This is clearly reflected in its algebraic form, which expresses it in terms of vectors from both planes.