Question

1. find the area of the parallelogram defined by vectors \\(\mathbf{u}\\) and \\(\mathbf{v}\\) given that the absolute value of the box product of \\(\mathbf{u}\\), \\(\mathbf{v}\\) and \\(2\mathbf{u} \times 5\mathbf{v}\\) is 360. that is,\\|(\mathbf{u} \times \mathbf{v}) \cdot (2\mathbf{u} \times 5\mathbf{v})| = 360.\\

Explanation:

Step1: Simplify the scalar triple product

First, use the linearity property of the cross product and scalar triple product. For scalars $a,b$, $a\mathbf{u} \times b\mathbf{v} = ab(\mathbf{u} \times \mathbf{v})$, so:
$$|(\mathbf{u} \times \mathbf{v}) \cdot (2\mathbf{u} \times 5\mathbf{v})| = |(\mathbf{u} \times \mathbf{v}) \cdot [10(\mathbf{u} \times \mathbf{v})]|$$

Step2: Rewrite as a dot product of same vector

Let $\mathbf{w} = \mathbf{u} \times \mathbf{v}$. The expression becomes:
$$|10(\mathbf{w} \cdot \mathbf{w})| = 10|\mathbf{w}|^2$$

Step3: Set equal to given value

We know this equals 360, so:
$$10|\mathbf{w}|^2 = 360$$

Step4: Solve for $|\mathbf{w}|$

First divide both sides by 10:
$$|\mathbf{w}|^2 = 36$$
Take the positive square root (since magnitude is non-negative):
$$|\mathbf{w}| = \sqrt{36} = 6$$

Answer:

6