vec a=\α,1,1\ , vec b=\1,α,1\ , vec c=\1,1,α\ are edges of a…
$\vec a=\{\alpha,1,1\}$, $\vec b=\{1,\alpha,1\}$, $\vec c=\{1,1,\alpha\}$ are edges of a parallelepiped. $1^\circ$ Find its volume. $2^\circ$ Find the height vector $\vec h$ perpendicular to $\vec a$ and $\vec b$. $3^\circ$ For $\alpha\in[-2,1]$, find $\alpha$ giving the maximum and the minimum volume, those volumes, and the vectors when $V$ is minimal.
Explanation of Method
Parallelepiped volume, height vector and extrema over the parameter α.
Step 1: $1^\circ$ Volume
\[ [\vec a,\vec b,\vec c]=\begin{vmatrix}\alpha&1&1\\1&\alpha&1\\1&1&\alpha\end{vmatrix}=\alpha(\alpha^2-1)-(\alpha-1)+(1-\alpha)=\alpha^3-3\alpha+2=(\alpha-1)^2(\alpha+2). \]
\[V=\big|(\alpha-1)^2(\alpha+2)\big|\]
Step 2: $2^\circ$ Height vector
$\vec h$ is the component of $\vec c$ along the normal $\vec a\times\vec b$ of the base: \[\vec h=\frac{[\vec a,\vec b,\vec c]}{|\vec a\times\vec b|^2}\,(\vec a\times\vec b),\qquad |\vec h|=\frac{V}{|\vec a\times\vec b|}.\]
\[ \vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\\alpha&1&1\\1&\alpha&1\end{vmatrix}=\{1-\alpha,\ 1-\alpha,\ \alpha^2-1\}=(1-\alpha)\{1,\ 1,\ -(\alpha+1)\}, \]
\[ |\vec a\times\vec b|^2=(1-\alpha)^2(\alpha^2+2\alpha+3). \]
For $\alpha\neq1$: $\vec h=\frac{(\alpha+2)(1-\alpha)}{\alpha^2+2\alpha+3}\,\{1,\ 1,\ -(\alpha+1)\}$.
Check: $\{1,1,-(\alpha+1)\}\cdot\vec a=\alpha+1-(\alpha+1)=0$ and $\{1,1,-(\alpha+1)\}\cdot\vec b=1+\alpha-(\alpha+1)=0$ $\checkmark$
Step 3: $3^\circ$ Extrema on $[-2,1]$
On this interval $V=(\alpha-1)^2(\alpha+2)\ge0$, and
\[ V'(\alpha)=2(\alpha-1)(\alpha+2)+(\alpha-1)^2=3(\alpha-1)(\alpha+1). \]
| $\alpha$ | $-2$ | $-1$ | $1$ |
|---|---|---|---|
| $V$ | $0$ | $4$ | $0$ |
\[V_{\max}=4\ \text{at}\ \alpha=-1;\qquad V_{\min}=0\ \text{at}\ \alpha=-2\ \text{and}\ \alpha=1\]
- $\alpha=-1$: $\vec a=\{-1,1,1\}$, $\vec b=\{1,-1,1\}$, $\vec c=\{1,1,-1\}$.
- $\alpha=1$: $\vec a=\vec b=\vec c=\{1,1,1\}$. The vectors are equal (collinear), so there is no parallelepiped at all.
- $\alpha=-2$: $\vec a=\{-2,1,1\}$, $\vec b=\{1,-2,1\}$, $\vec c=\{1,1,-2\}$ with $\vec a+\vec b+\vec c=\vec 0$, so they are coplanar (all lie in the plane $x+y+z=0$).
Result
\[\boxed{\,V=\big|(\alpha-1)^2(\alpha+2)\big|\,}\]
\[\boxed{\,\vec h=\frac{(\alpha+2)(1-\alpha)}{\alpha^2+2\alpha+3}\,\{1,\ 1,\ -(\alpha+1)\}\,}\]
\[\boxed{\,V_{\max}=4\ \text{at}\ \alpha=-1;\qquad V_{\min}=0\ \text{at}\ \alpha=-2\ \text{and}\ \alpha=1\,}\]
Remark
Correction. The $\vec j$-component of $\vec a\times\vec b$ is $-(\alpha\cdot1-1\cdot1)=1-\alpha$, not $\alpha-1$. With the source's $\{1,-1,-(\alpha+1)\}$ the "height" would not be perpendicular to $\vec a$: the product gives $-2$. The source also mixes up the pyramid formula $h=3V/S$ with the parallelepiped formula $h=V/S$; for a parallelepiped $h=V/|\vec a\times\vec b|$.