Find Vector Normal, Plane Equation & Distance from Origin for Points On A Plane

[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:
<br /> \begin{array}{l}<br /> \overrightarrow {OA} = &lt; 1,1, - 1 &gt; \\ <br /> \overrightarrow {OB} = &lt; 3,3,2 &gt; \\ <br /> \overrightarrow {OC} = &lt; 3, - 1, - 2 &gt; \\ <br /> \end{array}<br />

then:

<br /> \begin{array}{l}<br /> \overrightarrow {AB} = &lt; 2,2,3 &gt; \\ <br /> \overrightarrow {AC} = &lt; 2, - 2, - 1 &gt; \\ <br /> \end{array}<br />

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

<br /> \overrightarrow {AB} \times \overrightarrow {AC} = &lt; 4,8, - 8 &gt; <br />


For the equation of the plane i just said

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

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.

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

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.

 

[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 k\vec{n} is itself on the plane. You can make use of that to find the length of kn.

 

[danago]

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 k\vec{n} is itself on the plane. You can make use of that to find the length of kn.

<br /> <br /> Oh so pretty much form the equation:<br /> <br /> &lt;br /&gt; \left( {\begin{array}{*{20}c}&lt;br /&gt; {4k - 1} \\&lt;br /&gt; {8k - 1} \\&lt;br /&gt; { - 8k + 1} \\&lt;br /&gt; \end{array}} \right) \cdot \left( {\begin{array}{*{20}c}&lt;br /&gt; 4 \\&lt;br /&gt; 8 \\&lt;br /&gt; { - 8} \\&lt;br /&gt; \end{array}} \right) = 0&lt;br /&gt;<br /> <br /> 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 <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> Thanks again for your help.