Given that $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c} = \vec{b} \times \vec{d}$, and also given that $\vec{a} \neq \vec{d}$ and $\vec{b} \neq \vec{c}$, show that $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$.

1. Problem Description

Given that $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c} = \vec{b} \times \vec{d}$, and also given that $\vec{a} \neq \vec{d}$ and $\vec{b} \neq \vec{c}$, show that $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$.

2. Solution Steps

We are given two equations:

$$\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \hspace{1cm} (1)$$

$$\vec{a} \times \vec{c} = \vec{b} \times \vec{d} \hspace{1cm} (2)$$

Rearranging equation (1), we get:

$$\vec{a} \times \vec{b} - \vec{c} \times \vec{d} = \vec{0}$$

$$\vec{a} \times \vec{b} + \vec{d} \times \vec{c} = \vec{0} \hspace{1cm} (3)$$

Rearranging equation (2), we get:

$$\vec{a} \times \vec{c} - \vec{b} \times \vec{d} = \vec{0}$$

$$\vec{a} \times \vec{c} + \vec{d} \times \vec{b} = \vec{0} \hspace{1cm} (4)$$

Subtracting equation (4) from equation (3), we have:

$$(\vec{a} \times \vec{b} + \vec{d} \times \vec{c}) - (\vec{a} \times \vec{c} + \vec{d} \times \vec{b}) = \vec{0} - \vec{0}$$

$$\vec{a} \times \vec{b} + \vec{d} \times \vec{c} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} = \vec{0}$$

$$\vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{d} \times \vec{c} - \vec{d} \times \vec{b} = \vec{0}$$

$$\vec{a} \times (\vec{b} - \vec{c}) + \vec{d} \times (\vec{c} - \vec{b}) = \vec{0}$$

$$\vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = \vec{0}$$

$$(\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0}$$

If the cross product of two vectors is the zero vector, then the vectors are parallel.

Since $(\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0}$, then $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$. Also, since $\vec{a} \neq \vec{d}$ and $\vec{b} \neq \vec{c}$, $\vec{a} - \vec{d} \neq \vec{0}$ and $\vec{b} - \vec{c} \neq \vec{0}$.

3. Final Answer

$\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$.