# Points On A Plane

1. Jul 3, 2008

### danago

The three points (1,1,-1), (3,3,2) and (3,-1,-2) determine a plane. Find:
1. A vector normal to the plane
2. The equation of the plane
3. The distance of the plane from the origin

For part 1 i just said let:
$$\begin{array}{l} \overrightarrow {OA} = < 1,1, - 1 > \\ \overrightarrow {OB} = < 3,3,2 > \\ \overrightarrow {OC} = < 3, - 1, - 2 > \\ \end{array}$$

then:

$$\begin{array}{l} \overrightarrow {AB} = < 2,2,3 > \\ \overrightarrow {AC} = < 2, - 2, - 1 > \\ \end{array}$$

Since the vectors AB and AC both lie on the plane, i just need to find a vector normal to both.

$$\overrightarrow {AB} \times \overrightarrow {AC} = < 4,8, - 8 >$$

For the equation of the plane i just said

$$\left( {\vec{p} - \left( {\begin{array}{*{20}c} 1 \\ 1 \\ { - 1} \\ \end{array}} \right)} \right) \cdot \left( {\begin{array}{*{20}c} 4 \\ 8 \\ { - 8} \\ \end{array}} \right) = 0$$

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.

$$\begin{array}{l} {\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\overrightarrow n .\overrightarrow n }}\overrightarrow n = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}}\widehat{n} \\ \therefore \left| {{\mathop{\rm proj}\nolimits} _{\overrightarrow n } \overrightarrow {OA} } \right| = \frac{{\overrightarrow {OA} \cdot \overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = 5/3 \\ \end{array}$$

Does that look ok? Its mainly the last part i was a bit iffy about; i think ive done parts 1 and 2 correctly, unless ive made errors in my calculations.

Dan.

2. Jul 3, 2008

### Defennder

I got the same answer for your first two parts. For the last part you have to find the distance of the perpendicular line from O to the plane. I don't think you need to use projection here. Since the normal is already perpendicular to the plane, you can let $k \vec{n}$ be the vector from the origin to the plane. Clearly the individual directional scalar components of $k\vec{n}[/tex] satisfy the equation of the plane, since [itex]k\vec{n}$ is itself on the plane. You can make use of that to find the length of kn.

3. Jul 4, 2008

### danago

Oh so pretty much form the equation:

$$\left( {\begin{array}{*{20}c} {4k - 1} \\ {8k - 1} \\ { - 8k + 1} \\ \end{array}} \right) \cdot \left( {\begin{array}{*{20}c} 4 \\ 8 \\ { - 8} \\ \end{array}} \right) = 0$$

And then solve for k? And then find the magnitude of $$k\vec{n}$$? I did all that and got the same answer so atleast i know my reasoning in the first method i did was correct