To solve for $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$, we will begin by calculating the cross products $\mathbf{a} \times \mathbf{c}$ and $\mathbf{b} \times \mathbf{d}$ separately.

We have the vectors defined as:

$\mathbf{a} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$

$\mathbf{b} = \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{c} = 2\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - \hat{\mathbf{k}}$

$\mathbf{d} = 2\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$

Step 1: Calculate $\mathbf{a} \times \mathbf{c}$

The cross product $\mathbf{a} \times \mathbf{c}$ is determined using the determinant:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 2 & -3 & -1 \end{vmatrix} $

Calculating the determinant, we have:

$ \hat{\mathbf{i}}[( -1)(-1) - (1)(-3)] - \hat{\mathbf{j}}[(1)(-1) - (1)(2)] + \hat{\mathbf{k}}[(1)(-3) - (-1)(2)] $

This simplifies to:

$ \hat{\mathbf{i}}(1 + 3) - \hat{\mathbf{j}}(-1 - 2) + \hat{\mathbf{k}}(-3 + 2) $

Resulting in:

$ 4\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}} $

Step 2: Calculate $\mathbf{b} \times \mathbf{d}$

Similarly, compute the cross product $\mathbf{b} \times \mathbf{d}$ using the determinant:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & -2 \\ 2 & 1 & 1 \end{vmatrix} $

Calculating this determinant, we find:

$ \hat{\mathbf{i}}[(1)(1) - (-2)(1)] - \hat{\mathbf{j}}[(1)(1) - (-2)(2)] + \hat{\mathbf{k}}[(1)(1) - (1)(2)] $

This gives:

$ \hat{\mathbf{i}}(1 + 2) - \hat{\mathbf{j}}(1 + 4) + \hat{\mathbf{k}}(1 - 2) $

Resulting in:

$ 3\hat{\mathbf{i}} - 5\hat{\mathbf{j}} - \hat{\mathbf{k}} $

Step 3: Calculate $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$

Now use the result from both cross products to find their cross product:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & -1 \\ 3 & -5 & -1 \end{vmatrix} $

This determinant is computed as follows:

$ \hat{\mathbf{i}}[(3)(-1) - (3)(-1)] - \hat{\mathbf{j}}[(4)(-1) - (-1)(-1)] + \hat{\mathbf{k}}[(4)(-5) - (3)(3)] $

Which results in:

$ \hat{\mathbf{i}}(-3 + 5) - \hat{\mathbf{j}}(-4 - 1) + \hat{\mathbf{k}}(-20 - 9) $

Simplifying, we find:

$ -8\hat{\mathbf{i}} + \hat{\mathbf{j}} - 29\hat{\mathbf{k}} $

Thus, the final result of $(\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{d})$ is:

$ -8\hat{\mathbf{i}} + \hat{\mathbf{j}} - 29\hat{\mathbf{k}} $

Vector Algebra

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