The vectors u,v and w span a parallelepiped of volume V 0 in the three-dimensionalspace. Consider the vectors
u′=u+v−w
v′=u−v−2w
w′= 2u+v+w
i)Determine the volume of the parallelepiped spanned by the vectors u′,v′and w′
ii)Do the vectors u′,v′,w′ have the same orientation as u,v,w?
This is my working for part (i)
Volume of (u',v'w') = (u' x v' | w')
u' x v' =
= (u+v-w) x (u-v-2w)
= (u x u) + (u x-v) + (u x-2w) + (v x u) + (v x-v) + (v x -2w) + (-w x u) + (-w x -v) + (-w x -2w)
=-(u x v) -2(u x w) + (v x u) -2(v x w) -(w x u) + (w x v)
=2(v x u) – (u x w) – 3(v x w)
Now that we have (u' x v') we just have to take the dot product with w'.
So, [ 2(v x u) – (u x w) – 3(v x w) ] \cdot (2u + v + w) = -3u(v x w) *** I didnt write the multiplication working for this one.
So I got -3u(v x w) for the volume of the parallelepiped of u', v' and w', but is this correct? Also I need some help with part (ii) if possible. Thank you so much! 🙂
Best Answer
The volume of the parallelepiped spanned by three vectors $\mathbf{u,v, w} $ is given by the absolute value of the determinant
$V = | \begin{vmatrix}\mathbf{ u }, \mathbf{v} , \mathbf{w} \end{vmatrix} | $
Now the vectors $\mathbf{u',v', w'} $ are related to $\mathbf{u,v,w}$ as follows:
$\begin{bmatrix} \mathbf{u'}, \mathbf{v'}, \mathbf{w'} \end{bmatrix} = \begin{bmatrix} \mathbf{u}, \mathbf{v}, \mathbf{w} \end{bmatrix} \begin{bmatrix} 1 &&1&&2 \\1 && -1 && 1 \\-1&&-2&&1 \end{bmatrix}$
Hence, the volume of the parallelepiped spanned by $\mathbf{u',v',w'}$ is
$V' = | \begin{vmatrix} \mathbf{u', v', w'} \end{vmatrix} | = | \begin{vmatrix} \mathbf{u}, \mathbf{v}, \mathbf{w} \end{vmatrix} |\hspace{6pt} |\begin{vmatrix} 1 &&1&&2 \\1 && -1 && 1 \\-1&&-2&&1 \end{vmatrix} | $
The first term on the right hand side is just $V_0$, and the second term evaluates to $|-7| = 7 $
Hence,
$V' = 7 V_0 $
For (ii), obviously the vectors $\mathbf{u',v',w'} $ do not have the same orientation as $\mathbf{u,v,w}$.