Find the unit vector normal to vec a=\-2,-6,-1\ and vec…

Find the unit vector normal to $\vec a=\{-2,-6,-1\}$ and $\vec b=\{1,2,0\}$ that makes an acute angle with $\vec c=\{2,-1,0\}$. In its direction find $\vec d$ such that $\vec a,\vec b,\vec d$ span a parallelepiped of volume $18$.

Explanation of Method

A unit normal making an acute angle with c, and d for a given volume.

Step 1: Normal

\[ \vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\-2&-6&-1\\1&2&0\end{vmatrix}=\{2,-1,2\},\qquad |\vec a\times\vec b|=3. \]

Step 2: Choosing the sign

\[ \tfrac13\{2,-1,2\}\cdot\vec c=\tfrac13(4+1+0)=\tfrac53\gt0. \]

\[\vec n_0=\tfrac13\{2,-1,2\}\]

Step 3: The vector $\vec d$

Put $\vec d=t\,\vec n_0$ with $t\gt0$. The base area is $3$ and the height is $t$, so

\[ [\vec a,\vec b,\vec d]=(\vec a\times\vec b)\cdot t\vec n_0=3t=18\ \Longrightarrow\ t=6. \]

\[\vec d=6\vec n_0=\{4,-2,4\}\]

Check:

\[ \begin{vmatrix}-2&-6&-1\\1&2&0\\4&-2&4\end{vmatrix}=-16+24+10=18\ \checkmark \]

$O$ $\vec a$ $\vec b$ $\vec c$ $\vec d$

$\vec d=6\vec n_0=\{4,-2,4\}$

$\vec d$ is perpendicular to the base parallelogram of $\vec a,\vec b$ (area $3$) and has length $6$, so $V=3\cdot6=18$; it points to the side of $\vec c$ (acute angle).

Result

\[\boxed{\,\vec n_0=\tfrac13\{2,-1,2\}\,}\]

\[\boxed{\,\vec d=6\vec n_0=\{4,-2,4\}\,}\]