Question

Let $\vec{u}$, $\vec{v}$ and $\vec{w}$ be vectors in three-dimensional space, where $\vec{u}$ and $\vec{v}$ are unit vectors which are not perpendicular to each other and

$\vec{u} \cdot \vec{w} = 1$, $\vec{v} \cdot \vec{w} = 1$, $\vec{w} \cdot \vec{w} = 4$

$\begin{cases} \vec{u} \cdot \vec{u} = |\vec{u}|^2 = 1 \\ \vec{v} \cdot \vec{v} = |\vec{v}|^2 = 1 \\ \vec{u} \cdot \vec{v} = Cos\theta \end{cases}$

If the volume of the parallelopiped, whose adjacent sides are represented by the vectors $\vec{u}$, $\vec{v}$ and $\vec{w}$, is $\sqrt{2}$, then the value of $|3\vec{u}+5\vec{v}|$ is __.

Answer 1

7

Answer 2

The square of the magnitude of $3\vec{u}+5\vec{v}$ is calculated as: $|3\vec{u}+5\vec{v}|^2 = (3\vec{u}+5\vec{v}) \cdot (3\vec{u}+5\vec{v})$ $|3\vec{u}+5\vec{v}|^2 = 9(\vec{u} \cdot \vec{u}) + 15(\vec{u} \cdot \vec{v}) + 15(\vec{v} \cdot \vec{u}) + 25(\vec{v} \cdot \vec{v})$ Since $\vec{u}$ and $\vec{v}$ are unit vectors, $|\vec{u}|^2 = \vec{u} \cdot \vec{u} = 1$ and $|\vec{v}|^2 = \vec{v} \cdot \vec{v} = 1$. Also, $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$. $|3\vec{u}+5\vec{v}|^2 = 9(1) + 30(\vec{u} \cdot \vec{v}) + 25(1)$ $|3\vec{u}+5\vec{v}|^2 = 34 + 30(\vec{u} \cdot \vec{v})$

The square of the volume ($V^2$) of the parallelepiped formed by $\vec{u}$, $\vec{v}$, and $\vec{w}$ is given by the determinant of the Gram matrix: $V^2 = \begin{vmatrix} \vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w} \end{vmatrix}$ Given $V = \sqrt{2}$, so $V^2 = 2$. We have: $\vec{u} \cdot \vec{u} = 1$ $\vec{v} \cdot \vec{v} = 1$ $\vec{w} \cdot \vec{w} = 4$ $\vec{u} \cdot \vec{w} = 1$ $\vec{v} \cdot \vec{w} = 1$ Let $x = \vec{u} \cdot \vec{v}$. Substituting these values: $2 = \begin{vmatrix} 1 & x & 1 \\ x & 1 & 1 \\ 1 & 1 & 4 \end{vmatrix}$ $2 = 1(4-1) - x(4x-1) + 1(x-1)$ $2 = 3 - 4x^2 + x + x - 1$ $2 = 2 - 4x^2 + 2x$ $0 = -4x^2 + 2x$ $4x^2 - 2x = 0$ $2x(2x - 1) = 0$ The possible values for $x$ are $x=0$ or $x=1/2$. Since $\vec{u}$ and $\vec{v}$ are not perpendicular, $\vec{u} \cdot \vec{v} \neq 0$. Therefore, $\vec{u} \cdot \vec{v} = 1/2$.

Now, substitute this value back into the expression for $|3\vec{u}+5\vec{v}|^2$: $|3\vec{u}+5\vec{v}|^2 = 34 + 30\left(\frac{1}{2}\right)$ $|3\vec{u}+5\vec{v}|^2 = 34 + 15$ $|3\vec{u}+5\vec{v}|^2 = 49$ Taking the square root: $|3\vec{u}+5\vec{v}| = \sqrt{49} = 7$