Scalar Triple Product

Scalar Triple Product: Definition and Calculation

In vector algebra, we have learned about the scalar product (dot product) and the vector product (cross product). The scalar triple product (also known as the mixed product or box product) is an operation that combines both of these products, involving three vectors in three-dimensional space. It results in a scalar quantity.

Definition

Given three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, their scalar triple product is defined as the dot product of one vector with the cross product of the other two vectors, taken in a specific order.

The standard definition is the scalar product of $\vec{a}$ with the vector product of $\vec{b}$ and $\vec{c}$:

Result

Let's analyze the order of operations and the type of result:

• The cross product $\vec{b} \times \vec{c}$ is calculated first, and it results in a vector that is perpendicular to both $\vec{b}$ and $\vec{c}$.
• Then, the dot product of vector $\vec{a}$ with this resulting vector $(\vec{b} \times \vec{c})$ is computed. The dot product of two vectors yields a scalar quantity.

Therefore, the result of the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is a scalar quantity (a real number). This is why it is called the 'scalar' triple product.

Notation

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is often denoted using a compact bracket notation, which is read as "box a b c":

So, $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$. As we will see in the properties, the position of the dot and cross can be interchanged, so $[\vec{a} \vec{b} \vec{c}]$ also equals $(\vec{a} \times \vec{b}) \cdot \vec{c}$.

Calculation using Components

When the three vectors are given in component form with respect to the standard orthonormal basis $\{\hat{i}, \hat{j}, \hat{k}\}$, the scalar triple product can be calculated using a convenient determinant formula.

Let the vectors be:

First, calculate the cross product $\vec{b} \times \vec{c}$ using the determinant formula for the cross product:

Now, calculate the dot product of $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ with this resulting vector $(\vec{b} \times \vec{c})$ using the component formula for the dot product ($\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$):

This expression is precisely the expansion of the $3 \times 3$ determinant whose rows (or columns) are the scalar components of the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ in that order:

$$ \mathbf{[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}} $$

This determinant formula is the standard and most efficient way to calculate the scalar triple product when vector components are known.

Answer:

Properties of Scalar Triple Product

The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ has several key properties that are direct consequences of its definition and the properties of dot products, cross products, and determinants. These properties are very useful in simplifying calculations and understanding the geometric significance of the scalar triple product.

Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors, and $k$ be any scalar.

1. Interchange of Dot and Cross Product:

   The position of the dot product and the cross product can be interchanged in the scalar triple product without changing the value.

   $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$

   This property can be proven using the component formula (both expressions expand to the same determinant) or through geometric arguments. It justifies the ambiguity-free bracket notation $[\vec{a} \vec{b} \vec{c}]$.

2. Cyclic Permutation of Vectors:

   The value of the scalar triple product remains unchanged if the three vectors are permuted cyclically (keeping their relative order in the cycle).

   $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$

   Reason (Determinant): This property follows from the properties of determinants. Cyclically shifting the rows of a $3 \times 3$ determinant (Row 1 to Row 2, Row 2 to Row 3, Row 3 to Row 1) requires two row swaps. Each row swap multiplies the determinant by -1. Two swaps result in a multiplication by $(-1)^2 = 1$, leaving the determinant value unchanged.

3. Effect of Interchanging Adjacent Vectors:

   Interchanging any two adjacent vectors in the scalar triple product (a non-cyclic permutation) negates the value of the product.

   $[\vec{a} \vec{b} \vec{c}] = -[\vec{a} \vec{c} \vec{b}]$

   $[\vec{a} \vec{b} \vec{c}] = -[\vec{b} \vec{a} \vec{c}]$

   (and also $[\vec{b} \vec{c} \vec{a}] = -[\vec{b} \vec{a} \vec{c}]$, which is consistent with the other properties).

   Reason (Determinant): This property follows directly from the determinant property that swapping any two rows multiplies the determinant by -1.

4. Zero Value Conditions:

   The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to zero if and only if the three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar. This is a crucial geometric condition derived from the algebraic property that the determinant of a matrix is zero if its rows (or columns) are linearly dependent.

   The scalar triple product is zero if any of the following conditions hold (all imply coplanarity):

   • At least one of the vectors is the zero vector ($\vec{0}$). (If $\vec{a}=\vec{0}$, the first row of the determinant is zero, so the determinant is 0).
   • Any two of the vectors are equal ($\vec{a} = \vec{b}$ or $\vec{b} = \vec{c}$ or $\vec{c} = \vec{a}$). (If $\vec{a}=\vec{b}$, two rows of the determinant are identical, so the determinant is 0).
   • Any two of the vectors are collinear ($\vec{a} = k\vec{b}$ or $\vec{b} = k\vec{c}$ or $\vec{c} = k\vec{a}$ for some scalar $k$). (If $\vec{a}=k\vec{b}$, the first row is a scalar multiple of the second row, so the determinant is 0).
   • The three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar. (If they are coplanar when originating from the same point, then one vector can be expressed as a linear combination of the other two - assuming those two are non-collinear. This linear dependence of rows/columns makes the determinant zero. If two are collinear, the STP is zero as per the previous point). This condition is discussed in detail in the next subheading.
5. Linearity Property:

   The scalar triple product is linear with respect to each of its vector arguments. This means that scalar multiplication and vector addition distribute over the STP for each vector position.

   • Scalar Multiplication:

     $[(k\vec{a}) \vec{b} \vec{c}] = k[\vec{a} \vec{b} \vec{c}]$

     (Similarly for $k$ multiplying $\vec{b}$ or $\vec{c}$).
   • Vector Addition:

     $[(\vec{a} + \vec{d}) \vec{b} \vec{c}] = [\vec{a} \vec{b} \vec{c}] + [\vec{d} \vec{b} \vec{c}]$

     (Similarly for $(\vec{b}+\vec{d})$ or $(\vec{c}+\vec{d})$ in the respective positions).

   Reason (Determinant): These properties follow from the linearity properties of determinants with respect to their rows (or columns).

6. Independence of Coordinate System:

   The value of the scalar triple product depends only on the vectors themselves, not on the specific Cartesian coordinate system used to represent them in component form.

   Reason: The geometric meaning (volume of a parallelepiped, discussed next) is independent of the coordinate system. The determinant value is also invariant under rotation and translation of the coordinate system (though this is a deeper result from linear algebra).

Geometric Interpretation (Volume of Parallelepiped/Tetrahedron)

The scalar triple product, while an algebraic operation, possesses a very significant and intuitive geometric meaning. Its absolute value is directly related to the volume of a parallelepiped and a tetrahedron formed by the three vectors.

Geometric Interpretation: Volume of Parallelepiped

The absolute value of the scalar triple product $|[\vec{a} \vec{b} \vec{c}]| = |\vec{a} \cdot (\vec{b} \times \vec{c})|$ represents the volume of the parallelepiped formed by the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ when they are taken as coterminous edges (edges originating from the same vertex).

Derivation of Volume Formula

Consider a parallelepiped with coterminous edges represented by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

The volume of a parallelepiped can be calculated as the area of its base multiplied by its perpendicular height.

1. Choose the parallelogram formed by vectors $\vec{b}$ and $\vec{c}$ as the base of the parallelepiped.
2. The area of this base parallelogram is given by the magnitude of the cross product of its adjacent sides $\vec{b}$ and $\vec{c}$.

   Area of Base $= |\vec{b} \times \vec{c}|$

3. The vector $\vec{n} = \vec{b} \times \vec{c}$ is perpendicular (normal) to the plane containing the base parallelogram (the plane of $\vec{b}$ and $\vec{c}$). The magnitude of $\vec{n}$ is the area of the base.
4. The perpendicular height $h$ of the parallelepiped is the component of the third vector $\vec{a}$ along the direction perpendicular to the base. This height is the magnitude of the scalar projection of $\vec{a}$ onto the normal vector $\vec{n} = \vec{b} \times \vec{c}$.

   Scalar Projection of $\vec{a}$ onto $\vec{n} = \frac{\vec{a} \cdot \vec{n}}{|\vec{n}|} = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}$

   The perpendicular height $h$ must be a non-negative value, so we take the absolute value of this scalar projection:

   Height $h = \left| \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} \right| = \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|}$

   (Note: This is valid provided $\vec{b} \times \vec{c} \ne \vec{0}$, i.e., $\vec{b}$ and $\vec{c}$ are not collinear. If they are collinear, the base area is 0 and the volume is 0, which is also given by the STP being 0).
5. Now, calculate the volume of the parallelepiped:

   Volume $=$ Area of Base $\times$ Height

   Volume $= |\vec{b} \times \vec{c}| \times \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|}$

   Assuming $|\vec{b} \times \vec{c}| \ne 0$, we can cancel the terms:

   Volume $= |\vec{a} \cdot (\vec{b} \times \vec{c})|$

Thus, the volume of the parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is the absolute value of their scalar triple product:

$$ \mathbf{\text{Volume of Parallelepiped} = |[\vec{a} \vec{b} \vec{c}]|} $$

The sign of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ itself indicates whether the vectors $\vec{a}, \vec{b}, \vec{c}$ form a right-handed or left-handed system in that specific order. However, volume is a non-negative quantity, hence the absolute value.

Geometric Interpretation: Volume of Tetrahedron

A tetrahedron is a polyhedron with four triangular faces. If three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ represent the coterminous edges of a tetrahedron originating from one vertex, its volume is related to the volume of the parallelepiped formed by the same vectors.

The volume of a tetrahedron (a type of pyramid) is given by the formula: Volume $= \frac{1}{3} \times (\text{Area of Base Triangle}) \times (\text{Perpendicular Height})$.

Consider the base of the tetrahedron to be the triangle formed by vectors $\vec{b}$ and $\vec{c}$. The area of this base triangle is half the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$.

The perpendicular height $h$ of the tetrahedron from the terminal point of $\vec{a}$ to the base plane is the same as the height of the parallelepiped derived earlier:

Now, substitute these into the volume formula for the tetrahedron:

Assuming $|\vec{b} \times \vec{c}| \ne 0$, we cancel the magnitudes:

Thus, the volume of the tetrahedron formed by vectors $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is one-sixth of the absolute value of their scalar triple product:

$$ \mathbf{\text{Volume of Tetrahedron} = \frac{1}{6} |\vec{a} \cdot (\vec{b} \times \vec{c})| = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|} $$

Answer:

Condition for Coplanarity of Three Vectors using Scalar Triple Product

The scalar triple product provides a simple and powerful algebraic condition to determine whether three vectors lie in the same plane when they share a common initial point. This condition is directly related to the geometric interpretation of the scalar triple product as volume.

Condition for Coplanarity

Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is equal to zero.

$$ \mathbf{\vec{a}, \vec{b}, \vec{c} \text{ are coplanar } \iff [\vec{a} \vec{b} \vec{c}] = 0} $$

Explanation

We can understand this condition from both geometric and algebraic perspectives:

• Geometric Explanation:

  Consider the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as coterminous edges. The volume of this parallelepiped is given by $|[\vec{a} \vec{b} \vec{c}]|$.

  If the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar when placed with a common initial point, they will all lie in the same plane. In this case, they cannot form a parallelepiped that occupies any three-dimensional space. The "parallelepiped" they form will be flattened into a 2D shape within the plane, effectively having zero volume.

  Since the volume is 0, its absolute value is 0: $|[\vec{a} \vec{b} \vec{c}]| = 0$. This directly implies that the scalar triple product itself is zero: $[\vec{a} \vec{b} \vec{c}] = 0$.

  Conversely, if $[\vec{a} \vec{b} \vec{c}] = 0$, then the volume of the parallelepiped formed by these vectors is 0. A non-degenerate parallelepiped can only have zero volume if its defining edges are coplanar. Therefore, if the scalar triple product is zero, the vectors must be coplanar.

• Algebraic Explanation (using Determinants):

  The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is calculated as the determinant of the matrix formed by the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

  $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

  From linear algebra, a determinant is zero if and only if its row vectors (or column vectors) are linearly dependent. The row vectors of this determinant are the component representations of $\vec{a}, \vec{b}, \vec{c}$.

  The vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if one of them can be expressed as a linear combination of the other two (assuming the other two are non-collinear). For example, if $\vec{a}, \vec{b}, \vec{c}$ are coplanar and $\vec{b}, \vec{c}$ are non-collinear, then there exist scalars $x$ and $y$ such that $\vec{a} = x\vec{b} + y\vec{c}$. This means the vector $\vec{a}$ is a linear combination of $\vec{b}$ and $\vec{c}$. If the components $a_i = x b_i + y c_i$ for $i=1, 2, 3$, then the first row of the determinant is a linear combination of the second and third rows. A determinant with linearly dependent rows is always zero. Thus, coplanarity implies $[\vec{a} \vec{b} \vec{c}] = 0$.

  Conversely, if the determinant is zero, its rows are linearly dependent. This means there exist scalars $x, y, z$, not all zero, such that $x(a_1, a_2, a_3) + y(b_1, b_2, b_3) + z(c_1, c_2, c_3) = (0, 0, 0)$. This is equivalent to $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$. This vector equation implies that $\vec{a}, \vec{b}, \vec{c}$ are coplanar. (If, say, $x \neq 0$, then $\vec{a} = -\frac{y}{x}\vec{b} - \frac{z}{x}\vec{c}$, showing $\vec{a}$ is a linear combination of $\vec{b}$ and $\vec{c}$, which means they lie in the same plane). Thus, $[\vec{a} \vec{b} \vec{c}] = 0$ implies coplanarity.

• Vector Product View:

  Consider the vector $\vec{n} = \vec{b} \times \vec{c}$. This vector is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. If $\vec{a}$ lies in the same plane as $\vec{b}$ and $\vec{c}$ (or is coplanar with them), then $\vec{a}$ must also be perpendicular to the normal vector $\vec{n}$. The condition for $\vec{a}$ to be perpendicular to $\vec{n}$ is that their dot product is zero: $\vec{a} \cdot \vec{n} = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. Conversely, if $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$, then $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$. Since $\vec{b} \times \vec{c}$ is normal to the plane of $\vec{b}$ and $\vec{c}$, any vector perpendicular to $\vec{b} \times \vec{c}$ must lie in that plane (or be the zero vector). Thus, $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$.

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