- #1
Gregg
- 459
- 0
Homework Statement
Find the distance of the point (2,4,7) from the plane determined by the points (3,4,4) , (4,5,7) and (6,8,9).
2 common vectors for the plane
[itex]\left(
\begin{array}{c}
1 \\
1 \\
3
\end{array}
\right)
[/itex] [itex] \left(
\begin{array}{c}
3 \\
4 \\
5
\end{array}
\right) [/itex]
The normal to the plane is then
[itex]\left(
\begin{array}{c}
1 \\
1 \\
3
\end{array}
\right)\times \left(
\begin{array}{c}
3 \\
4 \\
5
\end{array}
\right)=\left(
\begin{array}{c}
-7 \\
4 \\
1
\end{array}
\right)[/itex]
The vector from the point to a known point on the plane dot product with the normal vector divided by |n| gives
[itex]\frac{\left(
\begin{array}{c}
1 \\
0 \\
-3
\end{array}
\right).\left(
\begin{array}{c}
-7 \\
4 \\
1
\end{array}
\right)}{\sqrt{7^2+4^2+1^2}}=\frac{10}{\sqrt{66}} = \frac{5\sqrt{66}}{33}[/itex]
The answer has just 3 in the denominator, not 33. Have I done this correctly?