Cartesian and vector equation of a plane

A plane can be completely illustrated by denoting two intersecting lines which can be translated into a fixed point A and two nonparallel direction vectors. The position vector $\overrightarrow{r}$ of any general point P on the plane passing through point A and having direction vectors  $\overrightarrow{b}$ and $\overrightarrow{c}$ is given by the equation

Vector equation of a plane

$\overrightarrow{r}=\overrightarrow{a}+\lambda  \overrightarrow{b}+µ\overrightarrow{c}                \lambda, µ∈R                        (\overrightarrow{AP}=\lambda  \overrightarrow{b}+µ\overrightarrow{c})$

Parametric equation of a plane: λ , μ are called a parameters λ,μ ∈ R
$ (\begin{matrix}
x & \\
y & \\
z & \\
\end{matrix}
)=(\begin{matrix}
a_{1} & \\
a_{2} & \\
a_{3} & \\
\end{matrix}
)+\lambda (\begin{matrix}
b_{1} & \\
b_{2} & \\
b_{3} & \\
\end{matrix}
)+\mu (\begin{matrix}
c_{1} & \\
c_{2} & \\
c_{3} & \\
\end{matrix}
)  =>    \begin{matrix}
x=a_{1}+\lambda b_{1}+\mu c_{1} & \\
y=a_{2}+\lambda b_{2}+\mu c_{2} & \\
z=a_{3}+\lambda b_{3}+\mu c_{3} & \\
\end{matrix}
$

If N is considered to be normal to a given plane, then all other normals to that plane are considered parallel to which are resultantly scalar multiples of N., In particular,we can say that there are two normals of length 1:

Normal/Scalar product form of vector equation of a plane

Consider a vector n passing through a point A. Only one plane through A can be is perpendicular to the vector. Now consider R being any point on the plane other than A as shown above. Then we can say that

$\overrightarrow{n}.\overrightarrow{AR}=0$

$\overrightarrow{n}.\overrightarrow{(r}-\overrightarrow{a})=0$

$\overrightarrow{n}.\overrightarrow{r}=  \overrightarrow{n}.(\overrightarrow{a}+\lambda \overrightarrow{b}+µ\overrightarrow{c})$\textit{⇒}$
     \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}   or  \overrightarrow{n}.(\overrightarrow{r}-\overrightarrow{a})= 0
$

Cartesian equation of a plane

$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}    ⤇     n_{1}x+n_{2}y+n_{3}z=n_{1}a_{1}+n_{2}a_{2}+n_{3}a_{3 }=d$

$n_{1}x+n_{2}y+n_{3}z=d$

$D=  \vert \overrightarrow{r}.{n}\vert =\vert \overrightarrow{a}.{n}\vert $

{=}$\frac{\vert \overrightarrow{a}.\overrightarrow{n}\vert }{\sqrt{n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}=\frac{\vert n_{1}a_{1}+n_{2}a_{2}+n_{3}a_{3 }\vert }{\sqrt{n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}$

Therefore, the Cartesian form is

where n1, n2 and n3 are the components of n and where is called the normal vector.

Example: Find the equation of the plane passing through the three points P1(1,-1,4), P2(2,7,-1), and P3(5,0,-1).

$\overrightarrow{b}=\overrightarrow{P_{1}P_{2}}=  (\begin{matrix}
1 & \\
8 & \\
-5 & \\
\end{matrix})$

$\overrightarrow{c}=  \overrightarrow{P_{1}P_{3}} =  (\begin{matrix}
4 & \\
1 & \\
-5 & \\
\end{matrix}
)$

$P_{1}=(\begin{matrix}
1 & \\
-1 & \\
4 & \\
\end{matrix}
)$

In vector form:

$ \overrightarrow{r}=(\begin{matrix}
1 & \\
-1 & \\
4 & \\
\end{matrix}
) +\lambda  (\begin{matrix}
1 & \\
8 & \\
-5 & \\
\end{matrix}
)+µ(\begin{matrix}
4 & \\
1 & \\
-5 & \\
\end{matrix}
)$

$\overrightarrow{n}=\vert \begin{matrix}
{i} & {j} & {k} & \\
1 & 8 & -5 & \\
4 & 1 & -5 & \\
\end{matrix}
\vert =(\begin{matrix}
-35 & \\
-15 & \\
-31 & \\
\end{matrix}
)$

Any non-zero scalar multiples of $\overrightarrow{n}$ is also a normal vector of the plane. Therefore, Multiply by -1.

$\overrightarrow{n}=(\begin{matrix}
35 & \\
15 & \\
31 & \\
\end{matrix}
)$

$(\begin{matrix}
35 & \\
15 & \\
31 & \\
\end{matrix}
).(\begin{matrix}
1 & \\
-1 & \\
4 & \\
\end{matrix}
)=144$

$Cartesian form:$

$35x+15y+31z=144$

Example: Find the equation of the plane with normal vector $
(\begin{matrix}
1 & \\
3 & \\
5 & \\
\end{matrix}
)$containing the point (-2, 3, 4).

$(\begin{matrix}
1 & \\
3 & \\
5 & \\
\end{matrix}
).(-2,3,4)=-2+9+20=27$

x+3y+5z=27

Example: Find the distance of the plane $
\overrightarrow{r}.(\begin{matrix}
3 & \\
2 & \\
-4 & \\
\end{matrix}
)$ = 8 from the origin, and the unit vector perpendicular to the plane.

$\vert (\begin{matrix}
3 & \\
2 & \\
-4 & \\
\end{matrix}
) \vert =\sqrt{29}$

$\frac{1}{\sqrt{29}}\lbrack  \overrightarrow{r}.(\begin{matrix}
3 & \\
2 & \\
-4 & \\
\end{matrix}
)\rbrack  = \frac{8}{\sqrt{29}}$

$D=  \frac{8}{\sqrt{29}}        {n}= \frac{1}{\sqrt{29}}(\begin{matrix}
3 & \\
2 & \\
-4 & \\
\end{matrix}
)$

Example: Find the Cartesian equation of the plane through the point A (1, 1, 1) perpendicular to the vector

Solution:

Example: Show that the following vector is perpendicular to the plane containing the points A(1, 0, 2), B(2, 3, -1) and C(2, 2, -1 ).

Solution:

In conclusion, n is a vector that is perpendicular to 2 vectors in the plane so it is perpendicular to the plane.

 

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