Find the area of a parallelogram defined by coordinates (1,3,4) and (2,1,3) from the origin (0,0,0).

The two adjacent sides of the parallelogram are given by the vectors originating from the origin:

$$\vec{u} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}$$ $$\vec{v} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$$

The area of the parallelogram is the magnitude of the cross product of its adjacent side vectors.

$$\vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 4 \\ 2 & 1 & 3 \end{vmatrix}$$ $$= \mathbf{i}((3)(3) - (4)(1)) - \mathbf{j}((1)(3) - (4)(2)) + \mathbf{k}((1)(1) - (3)(2))$$ $$= \mathbf{i}(9 - 4) - \mathbf{j}(3 - 8) + \mathbf{k}(1 - 6)$$ $$= 5\mathbf{i} - (-5)\mathbf{j} - 5\mathbf{k}$$ $$= \begin{pmatrix} 5 \\ 5 \\ -5 \end{pmatrix}$$

The magnitude of the vector $\begin{pmatrix} 5 \\ 5 \\ -5 \end{pmatrix}$ is:

$$||\vec{u} \times \vec{v}|| = \sqrt{5^2 + 5^2 + (-5)^2}$$ $$= \sqrt{25 + 25 + 25}$$ $$= \sqrt{75}$$ $$= \sqrt{25 \times 3}$$ $$= 5\sqrt{3}$$

The area of the parallelogram is $\boxed{5\sqrt{3} \text{ square units}}$.

---

: Show that if $\vec{a} = \vec{b} + \lambda \vec{c}$ for some scalar $\lambda$, then $\vec{a} \times \vec{c} = \vec{b} \times \vec{c}$.

We are given the relationship:

$$\vec{a} = \vec{b} + \lambda \vec{c}$$

The cross product is distributive over vector addition:

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

Since $\lambda \vec{c}$ is a scalar multiple of $\vec{c}$, the cross product $\lambda \vec{c} \times \vec{c}$ is zero because the vectors are parallel.

$$\lambda \vec{c} \times \vec{c} = \lambda (\vec{c} \times \vec{c}) = \lambda \vec{0} = \vec{0}$$

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

This proves the statement.