Question

Show that \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})\) need not equal \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) by calculating both when $$ \mathbf{u}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \text { and } \mathbf{w}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] $$

Short answer

The expressions \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})\) and \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) are different; thus, they are not equal.

Step-by-step solution

Calculate \( \mathbf{v} \times \mathbf{w} \)

The cross product \( \mathbf{v} \times \mathbf{w} \) is calculated using the determinant of a 3x3 matrix which includes the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of \( \mathbf{v} \) and \( \mathbf{w} \): \[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 0 & 0 & 1 \end{vmatrix} \]This results in the vector: \[ \mathbf{v} \times \mathbf{w} = \mathbf{i}(1 \cdot 1 - 0 \cdot 0) - \mathbf{j}(1 \cdot 1 - 0 \cdot 0) + \mathbf{k}(1 \cdot 0 - 1 \cdot 0) = \mathbf{i} - \mathbf{j} \] So, \( \mathbf{v} \times \mathbf{w} = \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} \).

Calculate \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \)

Using \( \mathbf{v} \times \mathbf{w} = \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} \) from Step 1, calculate \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \):\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 1 & -1 & 0 \end{vmatrix} \]This evaluates to: \[ = \mathbf{i}(1 \cdot 0 - 1 \cdot (-1)) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot (-1) - 1 \cdot 1) \]\[ = \mathbf{i}(1) - \mathbf{j}(-1) + \mathbf{k}(-2) = \mathbf{i} + \mathbf{j} - 2\mathbf{k} \]Thus, \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} \).

Calculate \( \mathbf{u} \times \mathbf{v} \)

Now, find the cross product \( \mathbf{u} \times \mathbf{v} \): \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} \]This evaluates to: \[ = \mathbf{i}(1 \cdot 0 - 1 \cdot 1) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot 1 - 1 \cdot 1) \]\[ = \mathbf{i}(0 - 1) - \mathbf{j}(0 - 1) + \mathbf{k}(1 - 1) = -\mathbf{i} + \mathbf{j} \]So, \( \mathbf{u} \times \mathbf{v} = \begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix} \).

Calculate \( (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} \)

Using the vector from Step 3, calculate \( (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} \): \[ (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 1 & 0 \ 0 & 0 & 1 \end{vmatrix} \]This results in:\[ = \mathbf{i}(1 \cdot 1 - 0 \cdot 0) - \mathbf{j}(-1 \cdot 1 - 0 \cdot 0) + \mathbf{k}(-1 \cdot 0 - 1 \cdot 0) \]\[ = \mathbf{i}(1) - \mathbf{j}(-1) + \mathbf{k}(0) = \mathbf{i} + \mathbf{j} \]Thus, \( (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \).

Compare the Results

Compare the vectors from Step 2 and Step 4:- \( \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \begin{bmatrix} 1 \ 1 \ -2 \end{bmatrix} \)- \( (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \)These two vectors are not equal, demonstrating that in general, \( \mathbf{u} \times(\mathbf{v} \times \mathbf{w}) eq (\mathbf{u} \times \mathbf{v}) \times \mathbf{w} \).

Key concepts

Cross Product

The cross product is a vector operation mainly used in three-dimensional geometry to find a vector perpendicular to two given vectors. If vectors \( \mathbf{a} \) and \( \mathbf{b} \) are given by their components \([a_1, a_2, a_3]\) and \([b_1, b_2, b_3]\), the cross product \( \mathbf{a} \times \mathbf{b} \) is computed using the determinant of a 3x3 matrix:

• \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) - unit vectors in x, y, and z directions respectively.
• \( \mathbf{a} \) and \( \mathbf{b} \) components form the second and third rows.
• The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).

Using these calculations, the components of the cross product \( \mathbf{a} \times \mathbf{b} \) are:\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}\]This specific operation is essential in physics and engineering to understand vector fields and forces.

Vector Operations

Vector operations involve various mathematical tools to combine or resolve vectors in geometrically meaningful ways. These operations include addition, subtraction, dot products, and our focus here, cross products. Important aspects of vector operations:

• Addition: Simply combine the components of each vector.
• Subtraction: Similar to addition but subtracting the second vector's components from the first.
• Dot Product: Yields a scalar; it's the sum of the products of corresponding components.
• Cross Product: Results in a vector that's perpendicular to the input vectors, as discussed in our previous section.

Understanding these operations underpins more complex concepts like vector calculus, where these become tools for solving multi-dimensional physics and engineering problems. While addition and subtraction are straightforward, cross products provide unique insights, particularly in understanding orientations and forces in physics.

Determinant

The determinant, particularly in the context of a 3x3 matrix, is a scalar value that can give significant information about a matrix and the system it represents. In vector calculus, the determinant is integral to calculating the cross product.
To find the determinant of a 3x3 matrix:

• For a matrix with rows \( [a, b, c] , [d, e, f] , [g, h, i] \), it's calculated as:

\[\text{Determinant} = a(ei - fh) - b(di - fg) + c(dh - eg)\]This scalar value gives insight into:

• The volume of the parallelepiped described by the matrix rows (when vectors are involved).
• The linear independence of vectors, which is crucial in understanding whether vectors can span a space or not.
• The properties of systems, indicating whether solutions exist or how systems behave.

In our vector cross product calculation, the use of the determinant is crucial, as it helps extract the perpendicular vector in 3D space.