If u, v and w are three non-coplanar vectors, then $(u + v − w) \cdot (u − v) \times (v − w) $is equal to

Question

The answer given was $u\cdot (v \times w)$.

Mine was $1$?

I found this by solving the determinant with $u,v$ and $w$ as three different vectors like -

$a = (1)u + 1(v) + (-1)w$ and so on $b$ and $c$ then $a.(b \times c)$

Where did I go wrong?

Answer 1

You correctly worked out that (treating all of the vectors as row vectors) $$\begin{bmatrix} u+v-w \\ u-v \\ v-w \end{bmatrix} = \begin{bmatrix}1&1&-1\\1&-1&0\\0&1&-1\end{bmatrix} \begin{bmatrix}u\\v\\w\end{bmatrix},$$ but then appear to have forgotten that $\det(AB)=\det(A)\det(B)$ and neglected to multiply by the determinant of the rightmost matrix, which is equal to $u\cdot v\times w$.

A different way to get this result: Using elementary row operations, we have $$\begin{vmatrix} u+v-w \\ u-v \\ v-w \end{vmatrix} = \begin{vmatrix} u \\ u-v \\ v-w \end{vmatrix} = \begin{vmatrix} u \\ -v \\ v-w \end{vmatrix} = \begin{vmatrix} u \\ -v \\ -w \end{vmatrix} = (-1)^2\begin{vmatrix} u \\ v \\ w \end{vmatrix},$$ therefore $(u+v-w)\cdot(u-v)\times(v-w) = u\cdot v\times w$.

Answer 2

Your given question is like the Scalar triple product.

$(u+v-w)\cdot\big((u-v)\times (v-w)\big)$

$= (u+v-w)\cdot\big(u\times v-u\times w- v\times v +v\times w\big) $

$= (u+v-w)\cdot\big(u\times v-u\times w+v\times w\big)$

$ = u\cdot( u\times v)-u\cdot(u\times w)+u\cdot(v\times w)+v\cdot(u\times v)-v\cdot(u\times w)-v\cdot(v\times w)-w\cdot(u\times v)-w\cdot(u\times w)-w\cdot(v\times w)$

$= u\cdot(v\times w)-v\cdot(u\times w)-w\cdot(u\times v)$

$= u\cdot(v\times w)-v\cdot(u\times w)+v\cdot(u\times w)$

$= u\cdot(v\times w)$

Answer 3

Using the distributive laws, together with the identities $$a{\times}a=0$$ $$a{\,\cdot\,}(a{\times}b)=0$$ $$b{\,\cdot\,}(a{\times}b)=0$$ we get \begin{align*} &(u + v - w){\,\cdot\,}\bigl((u - v){\times}(v - w)\bigr)\\[4pt] =\;&\bigl(u + (v - w)\bigr){\,\cdot\,}\bigl((u - v){\times}(v - w)\bigr)\\[4pt] =\;& \Bigl(u{\,\cdot\,}\bigl((u - v){\times}(v - w)\bigr)\Bigr) + \Bigl((v-w){\,\cdot\,}\bigl((u - v){\times}(v - w)\bigr)\Bigr) \\[4pt] =\;& u{\,\cdot\,}\bigl((u - v){\times}(v - w)\bigr) \\[4pt] =\;& u{\,\cdot\,}\bigl(u{\times}v-u{\times}w+v{\times}w\bigr)\\[4pt] =\;& u{\,\cdot\,}(u{\times}v)-u{\,\cdot\,}(u{\times}w)+u{\,\cdot\,}(v{\times}w)\\[4pt] =\;&u{\,\cdot\,}(v{\times}w)\\[4pt] \end{align*}