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Wize High School Grade 12 Calculus Textbook > Equations of Lines & Planes

Equations of Planes in R3R^3

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Intro to Equations of Planes in R3

The equation of a plane in R3R^3 can be defined by a point on the plane and two direction vectors parallel to the plane, or a point and one normal vector perpendicular to the plane.

Vector Equationr⃗=r0⃗+tm1⃗+sm2⃗Parametric Equationsx=x0+m1xt+m2xsy=y0+m1yt+m2ysz=z0+m1zt+m2zsNormal Equation(r⃗−r0⃗)⋅n⃗=0Cartesian (a.k.a. Scalar) EquationAx+By+Cz+D=0\displaystyle \begin{array}{|c|c|} \hline\\ \text{\textcolor{orange}{Vector Equation}}&\vec r=\vec {r_0}+t\vec {m_1}+s\vec{m_2}\\ \\\hline\\ \text{\textcolor{orange}{Parametric Equations}}&\begin{array}{c} x=x_0+m_{1x}t+m_{2x}s\\y=y_0+m_{1y}t+m_{2y}s\\z=z_0+m_{1z}t+m_{2z}s\end{array}\\ \\\hline\\ \text{\textcolor{orange}{Normal Equation}}&(\vec r-\vec{r_0})\cdot\vec n=0\\ \\\hline\\ \text{\textcolor{orange}{Cartesian (a.k.a. Scalar) Equation}}&Ax+By+Cz+D=0\\ \\\hline \end{array}

What Information is Commonly Provided?

  1. A line on the plane, and a point on the plane that is not on the line
  2. Three non-collinear points on the plane (three points that don't form a line)
  3. Two intersecting lines
  4. Two parallel and non-coincidient lines
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Vector and Parametric Equations of Planes in R3



For any plane Π\Pi in R3R^3, as long as we know one point (x0, y0,z0)(x_0,\ y_0, z_0) on the plane and two non-collinear vectors m⃗1=[m1x, m1y, m1z]\vec m_1=[m_{1x},\ m_{1y},\ m_{1z}] and m2⃗=[m2x, m2y, m2z]\vec {m_2}=[m_{2x},\ m_{2y},\ m_{2z}] that are parallel to the plane, we can define the equation of the plane.

Vector Equation
r→=r→0+tm1→+sm2→\boxed{\overrightarrow r=\overrightarrow r_0+t\overrightarrow {m_1}+s\overrightarrow {m_2}} or [x, y, z]=[x0, y0, z0]+t[m1x, m1y, m1z]+s[m2x, m2y, m2z]\boxed{\left[x,\ y,\ z\right]=\left[x_0,\ y_0,\ z_0\right]+t\left[m_{1x},\ m_{1y},\ m_{1z}\right]+s[m_{2x},\ m_{2y},\ m_{2z}]}
  • r→=[x, y, z]\overrightarrow{r}=[x,\ y,\ z] represents the position vector from the original to any point (x, y, z)(x,\ y,\ z) on the plane Π\Pi
  • r0→=[x0, y0, z0]\overrightarrow{r_0}=[x_0,\ y_0,\ z_0] represents the position vector of a known point on the plane Π\Pi
  • tt and ss are parameters that can take on any real numbers (t, s∈Rt,\ s\in R)
We can split this vector equation up into its xx, yy, and zz components.

Parametric Equations
x=x0+m1xt+m2xsy=y0+m1yt+m2ysz=z0+m1zt+m2zs\boxed{\begin{array}{c} x=x_0+m_{1x}t+m_{2x}s\\ y=y_0+m_{1y}t+m_{2y}s\\ z=z_0+m_{1z}t+m_{2z}s \end{array}}
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Example: Equations of Planes in R3

Find the vector and parametric equations of the following planes:
a) Given a line on the plane and a point on the plane that's not on the line
The plane contains the point (1, 2, 0)\left(1,\ 2,\ 0\right) and the line [0, −1, 0]+t[1, 2, 3]\left[0,\ -1,\ 0\right]+t\left[1,\ 2,\ 3\right].

  • Position vector of a point on the plane: [1, 2, 0]\left[1,\ 2,\ 0\right]
  • Direction vector 1: m1⃗=[1, 2, 3]\vec{m_1}=\left[1,\ 2,\ 3\right]
  • Direction vector 2: m2⃗=[1, 2, 0]−[0, −1, 0]=[1, 3, 0]\vec{m_2}=\left[1,\ 2,\ 0\right]-\left[0,\ -1,\ 0\right]=\left[1,\ 3,\ 0\right]
Therefore, the vector equation of the plane is r⃗=[1, 2, 0]+t[1, 2, 3]+s[1, 3, 0]\vec{r}=\left[1,\ 2,\ 0\right]+t\left[1,\ 2,\ 3\right]+s\left[1,\ 3,\ 0\right]

The parametric equations are x=1+t+sy=2+2t+3sz=3t\begin{array}{l} x=1+t+s\\ y=2+2t+3s\\ z=3t \end{array}.


b) Given three non-collinear points on the plane
The plane contains the points (1, 2, 3)\left(1,\ 2,\ 3\right), (2, 0, 1)\left(2,\ 0,\ 1\right), and (0, −1, −2)\left(0,\ -1,\ -2\right).

  • Position vector of a point on the plane: [1,2,3][1, 2, 3]
  • Direction vector 1: m1⃗=[2, 0, 1]−[1, 2, 3]=[1, −2, −2]\vec{m_1}=\left[2,\ 0,\ 1\right]-\left[1,\ 2,\ 3\right]=\left[1,\ -2,\ -2\right]
  • Direction vector 2: m2⃗=[2, 0 ,1]−[0, −1, −2]=[2, 1, 3]\vec{m_2}=\left[2,\ 0\ ,1\right]-\left[0,\ -1,\ -2\right]=\left[2,\ 1,\ 3\right]
Therefore, the vector equation of the plane is r⃗=[1, 2, 3]+t[1, −2, −2]+s[2, 1, 3]\vec{r}=\left[1,\ 2,\ 3\right]+t\left[1,\ -2,\ -2\right]+s\left[2,\ 1,\ 3\right]

The parametric equations are x=1+t+2sy=2−2t+sz=3−2t+3s\begin{array}{l} x=1+t+2s\\ y=2-2t+s\\ z=3-2t+3s \end{array}

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c) Given two intersecting lines
The plane contains the two lines [0, −1, 0]+t[1, 2, 3]\left[0,\ -1,\ 0\right]+t\left[1,\ 2,\ 3\right] and [2, −1, 0]+s[1, 0, 0]\left[2,\ -1,\ 0\right]+s\left[1,\ 0,\ 0\right]

  • Position vector of a point on the plane: [0,−1,0][0, -1, 0]
  • Direction vector 1: m1⃗=[1, 2, 3]\vec{m_1}=\left[1,\ 2,\ 3\right]
  • Direction vector 2: m2⃗=[1, 0, 0]\vec{m_2}=\left[1,\ 0,\ 0\right]
Therefore, the vector equation of the plane is r⃗=[0, −1, 0]+t[1, 2, 3]+s[1, 0, 0]\vec{r}=\left[0,\ -1,\ 0\right]+t\left[1,\ 2,\ 3\right]+s\left[1,\ 0,\ 0\right]

The parametric equations are x=t+sy=−1+2tz=3t\begin{array}{l} x=t+s\\ y=-1+2t\\ z=3t \end{array}

d) Given two parallel and non-coincident lines
The plane contains the two lines [1, 0, −1]+t[−1, 2, 0]\left[1,\ 0,\ -1\right]+t\left[-1,\ 2,\ 0\right] and [0, 1, 1]+s[2, −4, 0]\left[0,\ 1,\ 1\right]+s\left[2,\ -4,\ 0\right]

  • Position vector of a point on the plane: [1,0,−1][1, 0, -1]
  • Direction vector 1: m1⃗=[−1, 2, 0]\vec{m_1}=[-1,\ 2,\ 0]
  • Direction vector 2: m2⃗=[1, 0, −1]−[0, 1, 1]=[1, −1, −2]\vec{m_2}=\left[1,\ 0,\ -1\right]-\left[0,\ 1,\ 1\right]=\left[1,\ -1,\ -2\right]
Therefore, the vector equation of the plane is r⃗=[1, 0, −1]+t[−1, 2, 0]+s[1, −1, −2]\vec{r}=\left[1,\ 0,\ -1\right]+t\left[-1,\ 2,\ 0\right]+s\left[1,\ -1,\ -2\right]

The parametric equations are x=1−t+sy=2t−sz=−1−2s\begin{array}{l} x=1-t+s\\ y=2t-s\\ z=-1-2s \end{array}

Practice: Equations of Planes in R3

Find the vector equation of the plane that contains the lines [1, 0, 0]+t[−4, 0, 2]\left[1,\ 0,\ 0\right]+t\left[-4,\ 0,\ 2\right] and [2, −3, 5]+s[6, 0, −3]\left[2,\ -3,\ 5\right]+s\left[6,\ 0,\ -3\right].
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Example: Equations of Planes in R3

A plane Π\Pi is given by the vector equation r⃗=[1, 0, 3]+t[1,−1, 1]+s[3, 1, 1]\vec{r}=\left[1,\ 0,\ 3\right]+t\left[1,-1,\ 1\right]+s\left[3,\ 1,\ 1\right]

a) Find 2 different points on the plane.

To find a point on the plane, we just need to set different values for our parameters.
Let t=1, s=0t=1,\ s=0: (1, 0, 3)+(1, −1, 1)=(2, −1, 4)\left(1,\ 0,\ 3\right)+\left(1,\ -1,\ 1\right)=\left(2,\ -1,\ 4\right)
Let t=0, s=1t=0,\ s=1: (1, 0, 3)+(3, 1, 1)=(4, 1, 4)\left(1,\ 0,\ 3\right)+\left(3,\ 1,\ 1\right)=\left(4,\ 1,\ 4\right)
There are infinitely many possible such points.

b) Find a line on the plane.

For a line, we need a point and a direction vector.
  • Position vector of a point on the line: [1, 0, 3]\left[1,\ 0,\ 3\right]
  • Direction vector of the line: [1, −1, 1]\left[1,\ -1,\ 1\right]
Therefore, one line on the plane is [1, 0, 3]+t[1, −1, 1]\left[1,\ 0,\ 3\right]+t\left[1,\ -1,\ 1\right]
There are infinitely many possible such lines.

c) Check if the point (0, −1, 2)\left(0,\ -1,\ 2\right) is on the plane.

The parametric equations of the plane are:
x=1+t+3sx=1+t+3s
y=−t+sy=-t+s
z=3+t+sz=3+t+s

Substituting the point (0, −1, 2)\left(0,\ -1,\ 2\right) into these equations, we get
0=1+t+3s0=1+t+3s
−1=−t+s-1=-t+s
2=3+t+s2=3+t+s

From equations 1: t=−1−3st=-1-3s
Sub this into equation 2: −1=−(−1−3s)+s →s=−12-1=-\left(-1-3s\right)+s\ \to s=-\frac{1}{2}
Sub this back into equation 1: t=−1−3(−12)=12t=-1-3\left(-\frac{1}{2}\right)=\frac{1}{2}

Substitute t=12t=\frac{1}{2} and s=−12s=-\frac{1}{2} into equation 3:
2=3+(12)+(−12) → 2=32=3+\left(\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\ \to\ 2=3
Since we don't get left side equal right side, the point (0, −1, 2)\left(0,\ -1,\ 2\right) is NOT on the plane.
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Cartesian (Scalar) & Normal Equations of Planes in R3


For any plane Π\Pi in R3R^3, as long as we know one point (x0, y0,z0)(x_0,\ y_0, z_0) on the plane and a normal vector n⃗=[A, B, C]\vec n=[A,\ B,\ C] that is perpendicular to the plane, we can define the equation of the plane.

Normal Equation
(r→−r0→)⋅n→=0\boxed{\left(\overrightarrow r-\overrightarrow{r_0}\right)\cdot\overrightarrow{n}=0}
  • r→=[x, y, z]\overrightarrow{r}=[x,\ y,\ z] represents the position vector from the original to any point (x, y, z)(x,\ y,\ z) on the plane Π\Pi
  • r0→=[x0, y0, z0]\overrightarrow{r_0}=[x_0,\ y_0,\ z_0] represents the position vector of a known point on the plane Π\Pi
  • n→=[A, B, C]\overrightarrow{ n}=[A,\ B,\ C] represents the normal vector that is perpendicular to the plane Π\Pi

Cartesian (a.k.a. Scalar) Equation
Ax+By+Cz+D=0\boxed{Ax+By+Cz+D=0}
  • n→=[A,B,C]\overrightarrow{n}=[A,B,C] represents the normal vector that is perpendicular to the plane Π\Pi
  • D=−[x0, y0, z0]⋅[A, B, C]D=-[x_0,\ y_0,\ z_0]\cdot[A,\ B,\ C] where [x0, y0, z0][x_0,\ y_0,\ z_0] represents the position vector of a known point on the plane Π\Pi
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Example: Equations of Planes in R3

Find the Normal and Cartesian equations of the following planes:

a) Given a line on the plane and a point on the plane that's not on the line
The plane contains the point (1, 2, 0)\left(1,\ 2,\ 0\right) and the line [0, −1, 0]+t[1, 2, 3]\left[0,\ -1,\ 0\right]+t\left[1,\ 2,\ 3\right].

  • Position vector of a point on the plane: [1, 2, 0]\left[1,\ 2,\ 0\right]
  • Direction vector 1: m1⃗=[1, 2, 3]\vec{m_1}=\left[1,\ 2,\ 3\right]
  • Direction vector 2: m2⃗=[1, 2, 0]−[0, −1, 0]=[1, 3, 0]\vec{m_2}=\left[1,\ 2,\ 0\right]-\left[0,\ -1,\ 0\right]=\left[1,\ 3,\ 0\right]
  • Normal vector: n⃗=[1, 2, 3]×[1, 3, 0]=[−9, 3, 1]\vec{n}=\left[1,\ 2,\ 3\right]\times\left[1,\ 3,\ 0\right]=\left[-9,\ 3,\ 1\right] → use [9, −3, −1]\left[9,\ -3,\ -1\right] (or any scalar multiple)
Therefore, the normal equation of the plane is ([x, ,y, z]−[1, 2, 0])⋅[9, −3, −1]=0\left(\left[x,\ ,y,\ z\right]-\left[1,\ 2,\ 0\right]\right)\cdot\left[9,\ -3,\ -1\right]=0

The cartesian equation is 9x−3y−z−[1, 2, 0]⋅[9, −3, −1]=09x-3y-z-\left[1,\ 2,\ 0\right]\cdot\left[9,\ -3,\ -1\right]=0 or 9x−3y−z−3=09x-3y-z-3=0


b) Given three non-collinear points on the plane
The plane contains the points (1, 2, 3)\left(1,\ 2,\ 3\right), (2, 0, 1)\left(2,\ 0,\ 1\right), and (0, −1, −2)\left(0,\ -1,\ -2\right).

  • Position vector of a point on the plane: [1,2,3][1, 2, 3]
  • Direction vector 1: m1⃗=[2, 0, 1]−[1, 2, 3]=[1, −2, −2]\vec{m_1}=\left[2,\ 0,\ 1\right]-\left[1,\ 2,\ 3\right]=\left[1,\ -2,\ -2\right]
  • Direction vector 2: m2⃗=[2, 0 ,1]−[0, −1, −2]=[2, 1, 3]\vec{m_2}=\left[2,\ 0\ ,1\right]-\left[0,\ -1,\ -2\right]=\left[2,\ 1,\ 3\right]
  • Normal vector: n⃗=[1, −2, −2]×[2, 1, 3]=[−4, −7, 5]\vec{n}=\left[1,\ -2,\ -2\right]\times\left[2,\ 1,\ 3\right]=\left[-4,\ -7,\ 5\right] → use [4, 7, −5]\left[4,\ 7,\ -5\right] (or any scalar multiple)
Therefore, the normal equation of the plane is ([x, y, z]−[1, 2, 3])⋅[4, 7, −5]=0\left(\left[x,\ y,\ z\right]-\left[1,\ 2,\ 3\right]\right)\cdot\left[4,\ 7,\ -5\right]=0

The cartesian equation is 4x+7y−5z−[1,2,3]⋅[4,7,−5]=04x+7y-5z-\left[1,2,3\right]\cdot\left[4,7,-5\right]=0 or 4x+7y−5z−3=04x+7y-5z-3=0

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c) Given two intersecting lines
The plane contains the two lines [0, −1, 0]+t[1, 2, 3]\left[0,\ -1,\ 0\right]+t\left[1,\ 2,\ 3\right] and [2, −1, 0]+s[1, 0, 0]\left[2,\ -1,\ 0\right]+s\left[1,\ 0,\ 0\right]

  • Position vector of a point on the plane: [0,−1,0][0, -1, 0]
  • Direction vector 1: m1⃗=[1, 2, 3]\vec{m_1}=\left[1,\ 2,\ 3\right]
  • Direction vector 2: m2⃗=[1, 0, 0]\vec{m_2}=\left[1,\ 0,\ 0\right]
  • Normal vector: n⃗=[1,2,3]×[1,0,0]=[0, 3, −2]\vec{n}=\left[1,2,3\right]\times\left[1,0,0\right]=\left[0,\ 3,\ -2\right]
Therefore, the normal equation is ([x,y,z]−[0,−1,0])⋅[0, 3, −2]=0\left(\left[x,y,z\right]-\left[0,-1,0\right]\right)\cdot\left[0,\ 3,\ -2\right]=0

The cartesian equation is 3y−2z−[0,−1,0]⋅[0,3,−2]=03y-2z-\left[0,-1,0\right]\cdot\left[0,3,-2\right]=0 or 3y−2z+3=03y-2z+3=0


d) Given two parallel and non-coincident lines
The plane contains the two lines [1, 0, −1]+t[−1, 2, 0]\left[1,\ 0,\ -1\right]+t\left[-1,\ 2,\ 0\right] and [0, 1, 1]+s[2, −4, 0]\left[0,\ 1,\ 1\right]+s\left[2,\ -4,\ 0\right]

  • Position vector of a point on the plane: [1,0,−1][1, 0, -1]
  • Direction vector 1: m1⃗=[−1, 2, 0]\vec{m_1}=[-1,\ 2,\ 0]
  • Direction vector 2: m2⃗=[1, 0, −1]−[0, 1, 1]=[1, −1, −2]\vec{m_2}=\left[1,\ 0,\ -1\right]-\left[0,\ 1,\ 1\right]=\left[1,\ -1,\ -2\right]
  • Normal vector: n⃗=[−1, 2, 0]×[1, −1, −2]=[−4, −2, −1]\vec{n}=\left[-1,\ 2,\ 0\right]\times\left[1,\ -1,\ -2\right]=\left[-4,\ -2,\ -1\right] → use [4, 2, 1]\left[4,\ 2,\ 1\right] (or any scalar multiple)
Therefore, the normal equation is ([x, y, z]−[1, 0, −1])⋅[4, 2, 1]=0\left(\left[x,\ y,\ z\right]-\left[1,\ 0,\ -1\right]\right)\cdot\left[4,\ 2,\ 1\right]=0

The cartesian equation is 4x+2y+z−[1, 0, −1]⋅[4, 2, 1]=04x+2y+z-\left[1,\ 0,\ -1\right]\cdot\left[4,\ 2,\ 1\right]=0 or 4x+2y+z−3=04x+2y+z-3=0
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Example: Equations of Planes in R3

The Cartesian equation of a plane is 3x−2y+6z−12=03x-2y+6z-12=0.
a) Find three different points on this plane.

b) Find a normal vector to this plane.

c) Determine if the point (1, 2, −3)\left(1,\ 2,\ -3\right) is on this plane.

Practice: Equations of Planes in R3

The Cartesian equation of a plane is 2x+y+3z−6=02x+y+3z-6=0.
Find the coordinates of the points where the plane intersects the xx-axis.

Practice: Converting Equations of Planes in R3

Which of the following are the parametric equations of the plane x+y−3z=3x+y-3z=3?
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Angle Between Planes

The angle between any two planes in R3R^3 is the angle between the two normal vectors of the planes.

Write it Down
The angle between any two planes in R3R^3 is calculated by:
θ=cos⁡−1(n1⃗⋅n2⃗∣∣n1⃗∣∣∣∣n2⃗∣∣)\displaystyle \theta=\cos^{-1}\left(\frac{\vec{n_1}\cdot\vec{n_2}}{\left|\left|\vec{n_1}\right|\right|\left|\left|\vec{n_2}\right|\right|}\right)

Parallel Planes

n1⃗\vec{n_1} and n2⃗\vec{n_2} are parallel (scalar multiples)

Perpendicular Planes

n1⃗\vec{n_1} and n2⃗\vec{n_2} are perpendicular (dot product is 0)



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Example
Find the angle between the planes x+y+z=0x+y+z=0 and x−2y−5=0x-2y-5=0.

Plane 1: n1⃗=[1, 1, 1]\vec{n_1}=\left[1,\ 1,\ 1\right]
Plane 2: n2⃗=[1, −2, 0]\vec{n_2}=\left[1,\ -2,\ 0\right]

Therefore, the angle between the two planes can be calculated with the formula
cos⁡−1([1, 1, 1]⋅[1, −2, 0]∣∣[1, ,1, 1]∣∣∣∣[1, −2, 0]∣∣)=cos⁡−1(−135)=cos⁡−1(−115)≈104.96°\cos^{-1}\left(\frac{\left[1,\ 1,\ 1\right]\cdot\left[1,\ -2,\ 0\right]}{\left|\left|\left[1,\ ,1,\ 1\right]\right|\right|\left|\left|\left[1,\ -2,\ 0\right]\right|\right|}\right)=\cos^{-1}\left(\frac{-1}{\sqrt{3}\sqrt{5}}\right)=\cos^{-1}\left(-\frac{1}{\sqrt{15}}\right)\approx104.96\degree