MHB -z.60 Find the vector perpendicular to the plane PQR

[karush]

$\textsf{Find the vector perpendicular to the plane PQR
determined by the points}$
\begin{align*}\displaystyle
&P(2,1,3), \, Q(1,1,2), \, R(2,2,1)\\ \end{align*}
$$\textbf{ solution:}$$
\begin{align*}\displaystyle
\vec{PQ}&=(1-2)i+(1-1)J+(2-3)k=&-i-k\\
\vec{PR}&=(2-2)i+(2-1)j+(1-3)k=&j-2k
\end{align*}

\begin{align*}\displaystyle
\vec{PQ} \times \vec{PR}&=
\begin{vmatrix}
i&j&k\\-1&0&-1\\0&1&-2
\end{vmatrix}\\
&=
\begin{vmatrix}
0&-1\\1&-2
\end{vmatrix}i-
\begin{vmatrix}
-1&-1\\0&-2
\end{vmatrix}j-
\begin{vmatrix}
-1&0\\0&1
\end{vmatrix}k\\
&=(0+1)i-(2+0)j+(-1+0)k\\
&=i-2j-k\\
%&=\\
\text{book answer}&=\frac{1}{\sqrt{6}}(i-2j-k)
\end{align*}

ok I don't know where the $\displaystyle \frac{1}{\sqrt{6}}$ comes from

just that $\displaystyle\sqrt{1^2+2^2+1^1} = \sqrt{6}$

 

[skeeter]

ok I don't know where the $\displaystyle \frac{1}{\sqrt{6}}$ comes from

just that $\displaystyle\sqrt{1^2+2^2+1^1} = \sqrt{6}$

the book answer is just the unit vector perpendicular to the plane

 

[HallsofIvy]

Your problem says "find the vector perpendicular to the plane". But there are an infinite number of vectors perpendicular to a given plane. There are, in fact, two unit vectors perpendicular to the given plane, one the negative of the other.

 

[karush]

Your problem says "find the vector perpendicular to the plane". But there are an infinite number of vectors perpendicular to a given plane. There are, in fact, two unit vectors perpendicular to the given plane, one the negative of the other.

ok its

Find a unit vector perpendicular to the plane [FONT=MathJax_SansSerif]PQR determined by the points
$\begin{align*}\displaystyle &P(2,1,3), \, Q(1,1,2), \, R(2,2,1)
\end{align*}$

$\text{book answer}=\frac{1}{\sqrt{6}}(i-2j-k)$

 

[karush]

https://dl.orangedox.com/SNILuKcDnZqpqsUsH1