Magnitude of Combination of Unit Vectors from Parallelopiped Volume

To find the value of $|3 \vec{u} + 5 \vec{v}|$, we first determine the dot product $\vec{u} \cdot \vec{v}$.

Given data:

• $\vec{u}$ and $\vec{v}$ are unit vectors, so $|\vec{u}| = 1$ and $|\vec{v}| = 1$, which implies: $\vec{u} \cdot \vec{u} = 1 \quad \text{and} \quad \vec{v} \cdot \vec{v} = 1$
• $\vec{u}$ and $\vec{v}$ are not orthogonal, so $\vec{u} \cdot \vec{v} \neq 0$. Let $\vec{u} \cdot \vec{v} = x$.
• Given scalar products involving $\vec{w}$: $\vec{u} \cdot \vec{w} = 1, \quad \vec{v} \cdot \vec{w} = 1, \quad \vec{w} \cdot \vec{w} = 4$
• The volume $V$ of the parallelopiped formed by $\vec{u}$, $\vec{v}$, and $\vec{w}$ is given as $\sqrt{2}$.

The square of the volume of the parallelopiped $V^2$ is equal to the Gram determinant of the vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$:

$$V^2 = [\vec{u} \ \vec{v} \ \vec{w}]^2 = \begin{vmatrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{vmatrix}$$

Substituting $V = \sqrt{2}$ and the given dot products into the matrix:

$$(\sqrt{2})^2 = \begin{vmatrix} 1 & x & 1 \\ x & 1 & 1 \\ 1 & 1 & 4 \end{vmatrix}$$

Now, expand the determinant along the first row: $2 = 1 \cdot (1 \cdot 4 - 1 \cdot 1) - x \cdot (x \cdot 4 - 1 \cdot 1) + 1 \cdot (x \cdot 1 - 1 \cdot 1)$ $2 = 1(4 - 1) - x(4x - 1) + 1(x - 1)$ $2 = 3 - 4x^2 + x + x - 1$ $2 = 2 - 4x^2 + 2x$

Simplifying the equation: $4x^2 - 2x = 0$ $2x(2x - 1) = 0$

Since $\vec{u}$ and $\vec{v}$ are not orthogonal, $x = \vec{u} \cdot \vec{v} \neq 0$. Therefore: $2x - 1 = 0 \implies x = \frac{1}{2}$

Thus, $\vec{u} \cdot \vec{v} = \frac{1}{2}$.

Now, we evaluate the magnitude $|3 \vec{u} + 5 \vec{v}|$: $|3 \vec{u} + 5 \vec{v}|^2 = (3 \vec{u} + 5 \vec{v}) \cdot (3 \vec{u} + 5 \vec{v})$ $|3 \vec{u} + 5 \vec{v}|^2 = 9(\vec{u} \cdot \vec{u}) + 25(\vec{v} \cdot \vec{v}) + 30(\vec{u} \cdot \vec{v})$

Substitute the known values: $|3 \vec{u} + 5 \vec{v}|^2 = 9(1) + 25(1) + 30\left(\frac{1}{2}\right)$ $|3 \vec{u} + 5 \vec{v}|^2 = 9 + 25 + 15 = 49$

Taking the positive square root on both sides: $|3 \vec{u} + 5 \vec{v}| = \sqrt{49} = 7$