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Wize University Calculus 3 Textbook > Vectors & Geometry of Space

Equation of Planes

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Equations of Planes

Assume for some plane in R3\mathbb{R}^3,
  • P0=(x0, y0, z0) P_0=(x_0,~y_0,~z_0)~ is on the plane
  • Let n⃗= <a, b, c>\vec{n}=~<a,~b,~c> be some vector orthogonal to the plane called a normal vector
  • Let P=(x,y,z)P=(x,y,z) be any point on the plane

  • Because n⃗ \vec{n}~is orthogonal to the plane, it is also orthogonal to any vector that lies in that plane.
  • Let the vector r⃗−r0⃗\vec{r}-\vec{r_0} be a vector that lies in the plane.
  • Then, n⃗\vec{n} and (r⃗−r0⃗)(\vec{r}-\vec{r_0}) are orthogonal to each other; thus, their dot product is 0.

Therefore,
n⃗⋅(r⃗−r0⃗)=0n⃗⋅r⃗=n⃗⋅r0⃗\begin{array}{rcl} \vec{n}\cdot{}(\vec{r}-\vec{r_0})&=&0\\\\ \vec{n}\cdot{}\vec{r}&=&\vec{n}\cdot{}\vec{r_0}\\\\ \end{array}

  1. This is the vector form for the equation of a plane, where:
  • n⃗= <a, b, c> \vec{n}=~<a,~b,~c>~
  • r⃗= <x, y, z> \vec{r}=~<x,~y,~z>~
  • r0⃗= <x0, y0, z0>\vec{r_0}=~<x_0,~y_0,~z_0>
  1. The scalar form can be found by expanding the vector form.
  • Therefore:
n⃗⋅(r⃗−r0⃗)=0<a,b,c>⋅(<x,y,z>−<x0,y0,z0>)=0a(x−x0) + b(y−y0) + c(z−z0)=0ax+by+cz=dwhere  d=ax0+by0+cz0\begin{array}{rcl} \vec{n}\cdot{}(\vec{r}-\vec{r_0})&=&0 \\\\ <a,b,c>\cdot{}\Big(<x,y,z>-<x_0,y_0,z_0>\Big)&=&0\\\\ a(x-x_0)~+~b(y-y_0)~+~c(z-z_0)&=&0 \\\\ ax+by+cz&=&d&&{\color{magenta}\text{\footnotesize{where}}}~~d=ax_0+by_0+cz_0 \end{array}

Wize Tip
The second form makes it easy to find the normal vector, since n⃗= <a, b, c>\vec{n}=~<a,~b,~c>

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Angle Between Planes

The angle between two planes is equal to the angle between two normal vectors.


  • The normal vector for two planes is n1⃗\vec{n_1} and n2⃗\vec{n_2} and the angle between the two planes is:
cos⁡θ=n1⃗⋅n2⃗∣n1⃗∣∣n2⃗∣\boxed{\cos\theta=\dfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}}

  • The distance of a given point P(x1,y1,z1)P(x_1,y_1,z_1) from plane ax+by+cz+d=0ax+by+cz+d=0 is:
D=∣ax1+by1+cz1+d∣a2+b2+c2\boxed{D=\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}}


Write it Down
1. Two PLANES are parallel if their normal vectors are parallel (cross product of the normal vectors is 0)
2. Two PLANES are orthogonal if their normal vectors are orthogonal (dot product of the normal vectors is 0)

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Example

Find the equation of the plane containing the points P=(1,−2,0), Q=(3,1,4),P=(1,-2,0),~Q=(3,1,4), and R=(0,−1,2)R=(0,-1,2).

In order to find the equation of a plane, we need a point and normal vector (the normal vector will be orthogonal to any two vectors that lie in the plane)


We know how to find two vectors that lie in a plane by turning our coordinate points into point vectors and then finding any two vectors between the three point vectors.
P=(1, −2, 0)→P⃗= <1, −2, 0>Q=(3,1,4)→Q⃗= <3, 1, 4>R=(0, −1, 2)→R⃗= <0, −1, 2>\begin{array}{rcl} P&=&(1,~-2,~0)&{\color{red}{\rightarrow}}&\vec{P}&=&~<1,~-2,~0>\\ Q&=&(3, 1, 4)&{\color{red}{\rightarrow}}&\vec{Q}&=&~<3,~1,~4>\\ R&=&(0,~-1,~2)&{\color{red}{\rightarrow}}&\vec{R}&=&~<0,~-1,~2> \end{array}
Then, any two vectors between these points is:
PQ⃗= <3−1, 1−(−2), 4−0>=<2, 3, 4>PR⃗=<0−1, −1−(−2), 2−0>=<−1, 1, 2>\begin{array}{rcl} \vec{PQ}&=&~<3-1,~1-(-2),~4-0>&=&<2,~3,~4> \\ \vec{PR}&=&<0-1,~-1-(-2),~2-0>&=&<-1,~1,~2> \end{array}
These two vectors lie in the plane as they were created with points that lie in the plane.
We want a vector that is orthogonal to the plane, which means it is also orthogonal to PQ⃗ & PR⃗. \vec{PQ}~\&~\vec{PR}.~

The cross product of PQ⃗ & PR⃗ \vec{PQ}~\&~\vec{PR}~ will give a vector orthogonal to the plane, which is the normal vector to the plane.

So, the cross product of PQ⃗ & PR⃗ \vec{PQ}~\&~\vec{PR}~is:
PQ⃗×PR⃗=i⃗j⃗k⃗234−112=i⃗(6−4)−j⃗(4+4)+k⃗(2+3)                =2i⃗−8j⃗+5k⃗\vec{PQ}\times\vec{PR}=\begin{array}{|ccc|} \vec{i}&\vec{j}&\vec{k}\\ 2&3&4\\ -1&1&2\end{array}=\vec{i}(6-4)-\vec{j}(4+4)+\vec{k}(2+3) \newline\newline{} ~~~~~~~~~~~~~~~~=2\vec{i}-8\vec{j}+5\vec{k}
The normal vector will be:

n⃗= <a, b, c> = <2, −8, 5> \vec{n}=~<a,~b,~c>~=~<2,~-8,~5>~

and we can use any point for the equation of the plane.
The equation of the plane is:
a(x−x0) + b(y−y0) + c(z−z0)=02(x−1)−8(y+2)+5(z−0)=0Put in Scalar Form2x−8y+5z=18a(x-x_0)~+~b(y-y_0)~+~c(z-z_0)=0\newline 2(x-1)-8(y+2)+5(z-0)=0\newline \text{\color{magenta}\scriptsize{Put in Scalar Form}}\newline2x-8y+5z=18

Watch Out!
If we take the dot product, we are testing to see if these two vectors are orthogonal to EACH OTHER and we do not want that!!!

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Example

What is the angle and line of intersection of the planes x+y+z=1 x+y+z=1~ and 2x+y−z=32x+y-z=3? (Leave all answers in exact form)

Let's take a look at what two planes intersecting looks like:

The angle of intersection can be found using cos⁡(θ)=n1⋅n2∣n1∣∣n2∣\cos(\theta)=\dfrac{n_1\cdot n_2}{|n_1||n_2|}but we need two normal vectors, n1⃗ & n2⃗, \vec{n_1}~\&~\vec{n_2},~to both planes.
They can be identified easily since both planes are in scalar form. Thus,
n1⃗= <1, 1, 1>n2⃗= <2, 1, −1>\vec{n_1}=~<1,~1,~1>\newline \vec{n_2}=~<2,~1,~-1>
Then,
n1⃗⋅n2⃗=<2, 1, −1>=2∣n1⃗∣=12+12+12=3∣n2⃗∣=22+12+(−1)2=6Substitute Valuescos⁡(θ)=n1⋅n2∣n1∣∣n2∣=236=23θ=cos⁡−1(23)\begin{array}{rcl} \vec{n_1}\cdot\vec{n_2}&=&<2,~1,~-1>&=&2\\\\ |\vec{n_1}|&=&\sqrt{1^2+1^2+1^2}&=&\sqrt{3}\\\\ |\vec{n_2}|&=&\sqrt{2^2+1^2+(-1)^2}&=&\sqrt{6}\\\\ \text{\scriptsize{\color{magenta}Substitute Values}}\\\\ \cos(\theta)&=&\dfrac{n_1\cdot n_2}{|n_1||n_2|} \\\\ &=&\dfrac{2}{\sqrt{3}\sqrt{6}}\\\\ &=&\dfrac{\sqrt{2}}{3}\\\\ \theta&=&\cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \end{array}
Now, v⃗=n1×n2 \vec{v}=n_1\times n_2\ will be parallel to the line of intersection:
v⃗=∣i⃗j⃗k⃗11121−1∣=i⃗(−1−1)−j⃗(−1−2)+k⃗(1−2)=−2i⃗+3j⃗−k⃗      v⃗=<−2, 3, −1>\begin{array}{rcl} \vec{v}&=&\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\ 1&1&1\\ 2&1&-1 \end{vmatrix}\\\\ &=&\vec{i}(-1-1)-\vec{j}(-1-2)+\vec{k}(1-2) \\\\ &=&-2\vec{i}+3\vec{j}-\vec{k}~~~~~~\\\\ \color{red}\vec{v}&\color{red}=&\color{red}<-2,~3,~-1> \end{array}
Now, to find a point on the line, let's set x = 0:
y+z=1y−z=3add2y=4y=2∴  2+z=1z=−1\begin{array}{rcl} y+z&=&1\\ y-z&=&3&&\text{\color{magenta}{\scriptsize{add}}}\\\\ 2y&=&4\\ \color{red}y&\color{red}=&\color{red}2\\\\ \therefore~~2+z&=&1\\ \color{red}z&\color{red}=&\color{red}-1 \end{array}
So, our point will be P = (0, 2, -1)

The equation of the line, in vector form, is:
r⃗=<0, 2, −1>+t<−2, 3, −1>\vec{r}=<0,~2,~-1>+t<-2,~3,~-1>


Practice

Let L1 L_1~be a line that passes through the points (1, −2, 13) & (2, 0, −5). (1,~-2,~13)~\&~(2,~0,~-5).~
Let L2 L_2~be the line that is defined as r2(t)⃗=<2+4t, −1−t, 3>. \vec{r_2(t)}=<2+4t,~-1-t,~3>.~
Find the equation of the plane that contains L1 & L2.L_1~\&~L_2.

Practice

Consider the planes 4x−3y+z=104x-3y+z=10 and 10x−4y+7z=−1. 10x-4y+7z=-1.~

What is the line of intersection, L?