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 $\vec{ u } \cdot \vec{ w }=1, \vec{ v } \cdot \vec{ w }=1, \vec{ w } \cdot \vec{ w }=4$.
If the volume of the parallelepiped, 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

$[\vec{ u } \,\,\vec{ v }\,\, \vec{ w }]^{2}=\begin{vmatrix}\vec{ u } \cdot \vec{ u } & \vec{ u } \cdot \vec{ v } & \vec{ u } \cdot \vec{ w } \\ \vec{ u } \cdot \vec{ v } \cdot \vec{ v } & \vec{ v } \cdot \vec{ v } & \vec{ w } \cdot \vec{ w }\end{vmatrix}$
$=\begin{vmatrix}1 & \vec{ u } \cdot \vec{ v } & 1 \\ \vec{ u } \cdot \vec{ u } & 1 & 1 \\ 1 & 4\end{vmatrix}=2$
$=1(3)-\vec{ u } \cdot \vec{ v }(4 \vec{ u } \cdot \vec{ v }-1)+1(\vec{ u } \cdot \vec{ v }-1)=2$
$=3-4(\vec{ u } \cdot \vec{ v })^{2}+\vec{ u } \cdot \vec{ v }+\vec{ u } \cdot \vec{ v }-1=2$
$=-4(\vec{ u } \cdot \vec{ v })^{2}+2 \vec{ u } \cdot \vec{ v }=0$
$(\vec{ u } \cdot \vec{ v })(2-4(\vec{ u } \cdot \vec{ v }))=0 ; \vec{ u } \cdot \vec{ v }=0$,
$ \vec{ u } \cdot \vec{ v }=\frac{1}{2}$
$|3 \vec{ u }+5 \vec{ v }|^{2}=9|\vec{ u }|^{2}+30 \vec{ u } \cdot \vec{ v }+25|\vec{ v }|^{2}$
$=9+15+25=49$
$\therefore |3 \vec{ u }+5 \vec{ v }|=7$