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

Answer: 7

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