2.5 Equations of lines and planes in space  (Page 5/19)

The scalar equations of a plane vary depending on the normal vector and point chosen.

Writing an equation for a plane given a point and a line

Find an equation of the plane that passes through point $\left(1,4,3\right)$ and contains the line given by $x=\frac{y-1}{2}=z+1.$

Symmetric equations describe the line that passes through point $\left(0,1,\text{−}1\right)$ parallel to vector ${\text{v}}_1=⟨1,2,1⟩$ (see the following figure). Use this point and the given point, $\left(1,4,3\right),$ to identify a second vector parallel to the plane:

${\text{v}}_2=⟨1-0,4-1,3-(\mathrm{-1})⟩=⟨1,3,4⟩.$

Use the cross product of these vectors to identify a normal vector for the plane:

$\begin{array}{cc}\hfill \text{n}& ={\text{v}}_1\times {\text{v}}_2\hfill \\ & =\left|\begin{array}{ccc}\hfill \text{i}\hfill & \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right|\hfill \\ & =\left(8-3\right)\text{i}-\left(4-1\right)\text{j}+\left(3-2\right)\text{k}\hfill \\ & =5\text{i}-3\text{j}+\text{k}.\hfill \end{array}$

The scalar equations for the plane are $5x-3\left(y-1\right)+\left(z+1\right)=0$ and $5x-3y+z+4=0.$

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Now that we can write an equation for a plane, we can use the equation to find the distance $d$ between a point $P$ and the plane. It is defined as the shortest possible distance from $P$ to a point on the plane.

Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane. Let $R$ bet the point in the plane such that $\overrightarrow{RP}$ is orthogonal to the plane, and let $Q$ be an arbitrary point in the plane. Then the projection of vector $\overrightarrow{QP}$ onto the normal vector describes vector $\overrightarrow{RP},$ as shown in [link] .

Distance between a point and a plane

Find the distance between point $P=\left(3,1,2\right)$ and the plane given by $x-2y+z=5$ (see the following figure).

The coefficients of the plane’s equation provide a normal vector for the plane: $\text{n}=⟨1,\mathrm{-2},1⟩.$ To find vector $\overrightarrow{QP},$ we need a point in the plane. Any point will work, so set $y=z=0$ to see that point $Q=\left(5,0,0\right)$ lies in the plane. Find the component form of the vector from $Q\text{to}P\text{:}$

$\overrightarrow{QP}=⟨3-5,1-0,2-0⟩=⟨\mathrm{-2},1,2⟩.$

Apply the distance formula from [link] :

$\begin{array}{cc}\hfill d& =\frac{\left|\overrightarrow{QP}·\text{n}\right|}{\Vert \text{n}\Vert }\hfill \\ & =\frac{\left|⟨\mathrm{-2},1,2⟩·⟨1,\mathrm{-2},1⟩\right|}{\sqrt{1^2+{\left(\mathrm{-2}\right)}^2+1^2}}\hfill \\ & =\frac{\left|\mathrm{-2}-2+2\right|}{\sqrt{6}}\hfill \\ & =\frac{2}{\sqrt{6}}.\hfill \end{array}$

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Parallel and intersecting planes

We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line ( [link] ).

We can use the equations of the two planes to find parametric equations for the line of intersection.