Problem 37 - Algebra and Trigonometry

Question

Order of Operations in the Triple Product Given three vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w},\) their scalar triple product can be performed in six different orders: $$ \begin{array}{ll}{\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}),} & {\mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}), \quad \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w})} \\ {\mathbf{v} \cdot(\mathbf{w} \times \mathbf{u}),} & {\mathbf{w} \cdot(\mathbf{u} \times \mathbf{v}), \quad \mathbf{w} \cdot(\mathbf{v} \times \mathbf{u})}\end{array} $$ (a) Calculate each of these six triple products for the vectors: $$ \mathbf{u}=\langle 0,1,1\rangle, \quad \mathbf{v}=\langle 1,0,1\rangle, \quad \mathbf{w}=\langle 1,1,0\rangle $$ (b) On the basis of your observations in part (a), make a conjecture about the relationships between these six triple products. (c) Prove the conjecture you made in part (b).

Step-by-Step Solution

Answer

Each pair of triple products with switched cross product order has opposite signs. This is due to the properties of the scalar triple product.

Step 1: Calculate \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \)

First, find the cross product \( \mathbf{v} \times \mathbf{w} \). The cross product is calculated as follows:\[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 1 \ 1 & 1 & 0 \end{vmatrix} \]Expanding the determinant, we get:\[ \mathbf{v} \times \mathbf{w} = (0\cdot0 - 1\cdot1)\mathbf{i} - (1\cdot0 - 1\cdot1)\mathbf{j} + (1\cdot1 - 0\cdot1)\mathbf{k} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \]Now, calculate the dot product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \):\[ \mathbf{u} = \langle 0, 1, 1 \rangle \cross \langle -1, 1, 1 \rangle = 0\times(-1) + 1\times1 + 1\times1 = 2 \]

Step 2: Calculate \( \mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}) \)

Find \( \mathbf{w} \times \mathbf{v} \):\[ \mathbf{w} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 1 & 0 & 1 \end{vmatrix} \]Expanding the determinant gives:\[ \mathbf{w} \times \mathbf{v} = (1\cdot1 - 0\cdot1)\mathbf{i} - (1\cdot1 - 0\cdot1)\mathbf{j} + (1\cdot0 - 1\cdot1)\mathbf{k} = \mathbf{i} - \mathbf{j} - \mathbf{k}\]Now, calculate the dot product:\[ \mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}) = \langle 0, 1, 1 \rangle \cross \langle 1, -1, -1 \rangle = 0\times1 + 1\times(-1) + 1\times(-1) = -2 \]

Step 3: Calculate \( \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w}) \)

Find \( \mathbf{u} \times \mathbf{w} \):\[ \mathbf{u} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} \]Expanding the determinant gives:\[ \mathbf{u} \times \mathbf{w} = (1\cdot0 - 1\cdot1)\mathbf{i} - (0\cdot0 - 1\cdot1)\mathbf{j} + (0\cdot1 - 1\cdot1)\mathbf{k} = -\mathbf{i} + \mathbf{j} - \mathbf{k} \]Now, calculate the dot product:\[ \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w}) = \langle 1, 0, 1 \rangle \cross \langle -1, 1, -1 \rangle = 1\times(-1) + 0\times1 + 1\times(-1) = -2 \]

Step 4: Calculate \( \mathbf{v} \cdot(\mathbf{w} \times \mathbf{u}) \)

\( \mathbf{w} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{w}) = \mathbf{i} - \mathbf{j} + \mathbf{k} \) (since reversing the cross product changes the sign)Now, calculate the dot product:\[ \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \langle 1, 0, 1 \rangle \cdot \langle 1, -1, 1 \rangle = 1\times1 + 0\times(-1) + 1\times1 = 2 \]

Step 5: Calculate \( \mathbf{w} \cdot(\mathbf{u} \times \mathbf{v}) \)

Find \( \mathbf{u} \times \mathbf{v} \):\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & 1 \ 1 & 0 & 1 \end{vmatrix} \]Expanding the determinant gives:\[ \mathbf{u} \times \mathbf{v} = (1\cdot1 - 0\cdot1)\mathbf{i} - (0\cdot1 - 1\cdot1)\mathbf{j} + (0\cdot0 - 1\cdot1)\mathbf{k} = \mathbf{i} + \mathbf{j} - \mathbf{k} \]Now, calculate the dot product:\[ \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = \langle 1, 1, 0 \rangle \cdot \langle 1, 1, -1 \rangle = 1\times1 + 1\times1 + 0\times(-1) = 2 \]

Step 6: Calculate \( \mathbf{w} \cdot(\mathbf{v} \times \mathbf{u}) \)

\( \mathbf{v} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{v}) = -\mathbf{i} - \mathbf{j} + \mathbf{k} \) (since reversing the cross product changes the sign)Now, calculate the dot product:\[ \mathbf{w} \cdot (\mathbf{v} \times \mathbf{u}) = \langle 1, 1, 0 \rangle \cdot \langle -1, -1, 1 \rangle = 1\times(-1) + 1\times(-1) + 0\times1 = -2 \]

Step 7: Summarize the Calculated Products

From the above calculations, we observe the following results for triple products:- \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) = 2 \)- \( \mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}) = -2 \)- \( \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w}) = -2 \)- \( \mathbf{v} \cdot(\mathbf{w} \times \mathbf{u}) = 2 \)- \( \mathbf{w} \cdot(\mathbf{u} \times \mathbf{v}) = 2 \)- \( \mathbf{w} \cdot(\mathbf{v} \times \mathbf{u}) = -2 \)

Step 8: Conjecture Relationships of Triple Products

Conjecture that each pair of triple products from the same vector set but with different orders has opposite signs. For each pair, that involves reversing the order of the cross product, the result changes from positive to negative or vice versa. Specifically, these products mirror their signs around a switch in the cross product vectors.

Step 9: Prove the Conjecture

The scalar triple product properties mean changing the order of vectors or reversing the cross product vector arrangement, the result is equivalent to changing the sign. Thus:- \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) = -(\mathbf{u} \cdot(\mathbf{w} \times \mathbf{v})) \)- \( \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w}) = -(\mathbf{v} \cdot(\mathbf{w} \times \mathbf{u})) \)- \( \mathbf{w} \cdot(\mathbf{u} \times \mathbf{v}) = -(\mathbf{w} \cdot(\mathbf{v} \times \mathbf{u})) \)These properties of the cross product give us the basis for interpreting cyclic and anti-cyclic permutations of vectors in scalar triple products.

Key Concepts

Cross Product

The cross product is a vital concept in vector mathematics, particularly in three dimensions. It's a binary operation on two vectors in three-dimensional space, resulting in another vector that is perpendicular to the plane containing the original vectors. The cross product, denoted as \( \mathbf{a} \times \mathbf{b} \), helps compute area, volume, and is used in physics and engineering.
To calculate the cross product of two vectors \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) and \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \), you use the following determinant expansion:

• \( \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ v_1 & v_2 & v_3 \ w_1 & w_2 & w_3 \end{vmatrix} \)

The result is a vector, where each component is obtained from calculating the determinant of submatrices formed by ignoring the respective row and column of each unit vector (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)).
Importantly, the direction of the resulting vector follows the right-hand rule, and changing the order of the vectors changes the sign of the result (\( \mathbf{v} \times \mathbf{w} = - (\mathbf{w} \times \mathbf{v}) \)). This anti-commutative property is fundamental in understanding vector operations.

Dot Product

The dot product, or scalar product, is another fundamental operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. Unlike the cross product, which results in a vector, the dot product gives you a scalar, hence the name. The dot product is calculated as follows for vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \):

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

The dot product measures the extent to which two vectors point in the same direction. A result of zero indicates orthogonal vectors.
This operation is commutative, meaning \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \), and is widely used to compute the angle between vectors, projection of one vector onto another, and in physical contexts, work done by a force.

Order of Operations

The order of operations in mathematical expressions is crucial, particularly in VECTOR operations like the scalar triple product. In the context of vectors, the scalar triple product involves both cross and dot products, necessitating careful attention to sequence in operations.
For vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \), the scalar triple product is defined as \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). Here, the cross product \( \mathbf{v} \times \mathbf{w} \) is computed first, resulting in a vector, which is then dotted with \( \mathbf{u} \) to produce a scalar.

• The order matters because cross products are non-commutative: \( \mathbf{v} \times \mathbf{w} eq \mathbf{w} \times \mathbf{v} \)
• Dot products are commutative: \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (\mathbf{v} \times \mathbf{w}) \cdot \mathbf{u} \)

Failure to follow the correct sequence can yield incorrect results. The properties of vector operations, such as anti-commutativity of the cross product, play a crucial role in evaluating different permutations of operations. Understanding these rules ensures accurate computation in physics, engineering, and applied mathematics.