- #1
danago
Gold Member
- 1,123
- 4
The three points (1,1,-1), (3,3,2) and (3,-1,-2) determine a plane. Find:
For part 1 i just said let:
[tex]
\begin{array}{l}
\overrightarrow {OA} = < 1,1, - 1 > \\
\overrightarrow {OB} = < 3,3,2 > \\
\overrightarrow {OC} = < 3, - 1, - 2 > \\
\end{array}
[/tex]
then:
[tex]
\begin{array}{l}
\overrightarrow {AB} = < 2,2,3 > \\
\overrightarrow {AC} = < 2, - 2, - 1 > \\
\end{array}
[/tex]
Since the vectors AB and AC both lie on the plane, i just need to find a vector normal to both.
[tex]
\overrightarrow {AB} \times \overrightarrow {AC} = < 4,8, - 8 >
[/tex]
For the equation of the plane i just said
[tex]
\left( {\vec{p} - \left( {\begin{array}{*{20}c}
1 \\
1 \\
{ - 1} \\
\end{array}} \right)} \right) \cdot \left( {\begin{array}{*{20}c}
4 \\
8 \\
{ - 8} \\
\end{array}} \right) = 0
[/tex]
Where vector p is some point on the plane.
For the final part, i think i did it correctly, but not sure if it is the most efficient way of doing it. I assumed that the question was supposed to say the perpendicular distance.
What i did was find the projection of OA (since A lies on the plane) onto a vector which is normal to the plane, n, and then calculate its magnitude.
[tex]
\begin{array}{l}
{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\overrightarrow n .\overrightarrow n }}\overrightarrow n = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}}\widehat{n} \\
\therefore \left| {{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} } \right| = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = 5/3 \\
\end{array}
[/tex]
Does that look ok? Its mainly the last part i was a bit iffy about; i think I've done parts 1 and 2 correctly, unless I've made errors in my calculations.
Thanks in advance,
Dan.
- A vector normal to the plane
- The equation of the plane
- The distance of the plane from the origin
For part 1 i just said let:
[tex]
\begin{array}{l}
\overrightarrow {OA} = < 1,1, - 1 > \\
\overrightarrow {OB} = < 3,3,2 > \\
\overrightarrow {OC} = < 3, - 1, - 2 > \\
\end{array}
[/tex]
then:
[tex]
\begin{array}{l}
\overrightarrow {AB} = < 2,2,3 > \\
\overrightarrow {AC} = < 2, - 2, - 1 > \\
\end{array}
[/tex]
Since the vectors AB and AC both lie on the plane, i just need to find a vector normal to both.
[tex]
\overrightarrow {AB} \times \overrightarrow {AC} = < 4,8, - 8 >
[/tex]
For the equation of the plane i just said
[tex]
\left( {\vec{p} - \left( {\begin{array}{*{20}c}
1 \\
1 \\
{ - 1} \\
\end{array}} \right)} \right) \cdot \left( {\begin{array}{*{20}c}
4 \\
8 \\
{ - 8} \\
\end{array}} \right) = 0
[/tex]
Where vector p is some point on the plane.
For the final part, i think i did it correctly, but not sure if it is the most efficient way of doing it. I assumed that the question was supposed to say the perpendicular distance.
What i did was find the projection of OA (since A lies on the plane) onto a vector which is normal to the plane, n, and then calculate its magnitude.
[tex]
\begin{array}{l}
{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\overrightarrow n .\overrightarrow n }}\overrightarrow n = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}}\widehat{n} \\
\therefore \left| {{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} } \right| = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = 5/3 \\
\end{array}
[/tex]
Does that look ok? Its mainly the last part i was a bit iffy about; i think I've done parts 1 and 2 correctly, unless I've made errors in my calculations.
Thanks in advance,
Dan.