Equations of Planes and Lines in 3D

No pressure, no diamonds, Thomas Carlyle.

For every problem there is always, at least, a solution which seems quite plausible. It is simple and clean, direct, neat and nice, and yet very wrong, #Anawim, justtothepoint.com.

Recall: Vector Basics

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.

Basic Vector Operations

• Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.

• Vector Subtraction: The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.

• The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$
  Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$

• The projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.

• The cross product, denoted by $\vec{A} \times \vec{B}$, is a binary operation on two vectors in three-dimensional space. It yields a new vector that is perpendicular (orthogonal, at a right angle) to both original vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
  The cross product $\vec{A}x\vec{B}$ can be computed using the determinant of a $3 \times 3$ matrix formed by the standard basis vectors ($\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$) and the components of the two vectors: $\vec{A} \times \vec{B} = det(\begin{smallmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$.
  Example: If $\vec{A} = \langle 1, 2, 3 \rangle$ and $\vec{B} = \langle 4, 5, 6 \rangle$: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = (2\cdot 6 - 3\cdot 5)\hat{\mathbf{i}} - (1\cdot 6 - 3\cdot 4)\hat{\mathbf{j}} + (1\cdot 5 - 2\cdot 4)\hat{\mathbf{k}} = -3\hat{\mathbf{i}} + 6\hat{\mathbf{j}} - 3\hat{\mathbf{k}} = \langle -3, 6, -3 \rangle$

Defining a Plane with a Normal Vector

A plane is entirely flat and extends infinitely in two dimensions. To pin down a specific plane’s equation, you need two things:

1. A known point on the plane.
2. A normal vector ($\vec{N}$), which is a vector that is perfectly perpendicular (orthogonal) to every single vector lying within the plane.

Because the normal vector $\vec{N}$ is perpendicular to the plane, if you take any random point $P$ on the plane and connect it to a known point $P_1$, the vector $\vec{P_1P}$ must be perpendicular to $\vec{N}$. Mathematically, this means their dot product must be exactly zero: $\boxed{\vec{P_1P} \cdot \vec{N} = 0}$

Find the Equation of a Plane Given Three Points

1. To find the equation of a plane given three points $P_1, P_2$, and $P_3$ in the plane, we need to determine a condition that ensures a fourth point P lies in the same plane.
2. Consider the vectors $\vec{P_1P_2} = \vec{P_2}-\vec{P_1}$ and $\vec{P_1P_3} = \vec{P_3}-\vec{P_1}$. These two vectors lie in the plane and (assuming they’re not collinear) span it.
3. Now take a fourth point 𝑃. The vector $\vec{P_1P} = \vec{P}-\vec{P_1}$ lies in the plane if and only if it can be written as a linear combination of $\vec{P_1P_2}$ and $\vec{P_1P_3}$. Geometrically, the three vectors $\vec{P_1P}, \vec{P_1P_2}, \vec{P_1P_3}$ must be coplanar.
4. The condition that the parallelepiped spanned by $\vec{P_1P}, \vec{P_1P_2}, \vec{P_1P_3}$ is flat means its volume is zero. The volume of that parallelepiped is given by the absolute value of the scalar triple product: $\boxed{det(\vec{P_1P}, \vec{P_1P_2}, \vec{P_1P_3}) = \vec{P_1P} \cdot ( \vec{P_1P_2} \times \vec{P_1P_3}) = 0}$ (Figure vi).
5. Normal vector viewpoint. Definition. A normal vector to a plane is a vector that is perpendicular to the plane., A normal vector is perpendicular (orthogonal) to all vectors within the plane.
   A normal vector $\vec{N}$ to the plane can be found using the cross product of the vectors $\vec{P_1P_2}$ and $\vec{P_1P_3}$: $\vec{N}=\vec{P_1P_2} \times \vec{P_1P_3}$
6. For P to lie in the plane $\vec{P_1P}$ must be perpendicular to $\vec{N}$: $\vec{P_1P} \perp \vec{N} \iff \vec{P_1P} \cdot \vec{N} = 0$.
   Substituting $\vec{N}=\vec{P_1P_2} \times \vec{P_1P_3}$ gives: $\vec{P_1P} \cdot (\vec{P_1P_2} \times \vec{P_1P_3})$ which is exactly the scalar triple product condition above.
   This shows that the dot product of $\vec{P_1P}$ with the normal vector $\vec{N}$ must be zero for P to lie in the plane.

Solved exercises

• Plane from Three Points. Determine the equation of the plane containing points $P_1(1, -2, 0)$, $P_2(3, 1, 4)$, and $P_3(0, -1, 2)$ (Figure vi).

1. Find two vectors in the plane.
   To find vectors that lie in the plane, we need to create vectors using the given points. More specifically, those vectors are $\vec{P_1P_2}$ and $\vec{P_1P_3}$, and we just need to subtract the coordinates of $P_1$ from the coordinates of the others.
   $\vec{P_1P_2} = \langle 3-1, 1-(-2), 4-0 \rangle = \langle 2, 3, 4 \rangle$
   $\vec{P_1P_3} = \langle 0-1, -1-(-2), 2-0 \rangle = \langle -1, 1, 2 \rangle$

2. Calculate the Normal vector to the plane. The normal vector $\vec{N}$ can be found using the cross product of $\vec{P_1P_2}$ and $\vec{P_1P_3}$.
   $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3} = [\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}] = [\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\\ 2 & 3 & 4\\\ -1 & 1 & 2\end{smallmatrix}] = (3⋅2−4⋅1)\hat{\mathbf{i}} -(2⋅2−4⋅(−1))\hat{\mathbf{j}} +(2⋅1−3⋅(−1))\hat{\mathbf{k}} = 2\hat{\mathbf{i}} -8\hat{\mathbf{j}} +5\hat{\mathbf{k}} = = \langle 2, -8, 5 \rangle$

3. Form the Plane equation.
   A point P with coordinates ⟨x, y, z⟩ is in the plane if and only if the dot product of the normal vector $\vec{N}$ with $\vec{P_1P}$ is zero, that is, $\vec{P_1P} \perp \vec{N} \iff \vec{P_1P} \cdot \vec{N}=\langle x-1, y+2, z-0 \rangle \cdot \langle 2, -8, 5 \rangle = 0$
   $2(x-1) - 8(y+2) + 5(z) = 0 \iff 2x - 2 - 8y - 16 + 5z = 0 \iff 2x - 8y + 5z = 18$, and this is the equation of the plane.

4. Conclusion: The equation of the plane containing the points $P_1(1, -2, 0)$, $P_2(3, 1, 4)$, and $P_3(0, -1, 2)$ is: $2x - 8y + 5z = 18$.

• Parallel Planes. Find an equation for the plane passing through $Q(1, 1, 1)$ and parallel to the plane $2x + 3y + z = 5$.

1. Parallel planes share the same normal vector. The coefficients $A, B, C$ of a plane’s standard equation ($Ax + By + Cz = D$) directly form its normal vector.

2. The normal vector of $2x + 3y + z = 5$ is $\vec{N} = \langle 2, 3, 1 \rangle$, hence any parallel plane to $2x + 3y + z = 5$, has a normal vector $\vec{N} = ⟨2, 3, 1⟩$.

3. Because our new plane must have the same normal vector, it takes the form $2x + 3y + z = C$. To find $C$, we just need to plug in our known point $Q(1, 1, 1)$: $2(1) + 3(1) + 1(1) = C \rightarrow C = 6$

4. Final Equation: $2x + 3y + z = 6$

• Equations of a Line in 3D Space. Find parametric and symmetric equations of the line passing through points $A(1, 4, -2)$ and $B(-3, 5, 0)$.

1. Find the Direction Vector. The direction vector $\vec{d}$ (a vector parallel to the line) can be found by subtracting the coordinates of point A from the coordinates of point B: $\vec{d} = \vec{AB} = \langle -3-1, 5-4, 0-(-2) \rangle = \langle -4, 1, 2 \rangle$

2. Parametric Equations. Parametric equations express the coordinates $\langle x, y, z \rangle$ of any point on the line as functions of a parameter t. The general form for the parametric equations is:
   $\begin{cases} x = x_0 + t·d_x \\ y = y_0 + t·d_y \\ z = z_0 + t·d_z \end{cases}$
   where $\langle x_0, y_0, z_0 \rangle$ is a point on the line (we can use point A) and $\vec{d} = ⟨d_x, d_y, d_z⟩$ is the direction vector.
   Substituting $A(1, 4, -2)$ and $\vec{d} = \langle -4, 1, 2 \rangle$, the parametric equations of the line are:
   $\begin{cases} x = 1 - 4t \\ y = 4 + t \\ z = -2 + 2t \end{cases}$

3. Symmetric equations eliminate the parameter t and relate x, y, and z directly. To find the symmetric equations, we solve each of the parametric equations for t: $t = \frac{x-1}{-4}, t = y - 4, t = \frac{x + 2}{2}$.
   Setting these expressions equal to each other gives the symmetric equations: $\frac{x-1}{-4} = y - 4 = \frac{x + 2}{2}$ (symmetric equation of the line).

• Shortest Distance from a Point to a Line. Let $\vec{a}=\hat{\mathbf{i}}+3\hat{\mathbf{j}}-\hat{\mathbf{k}}, \vec{b}=3\hat{\mathbf{i}}+6\hat{\mathbf{j}}$ and $\vec{c}=-2\hat{\mathbf{i}}+4\hat{\mathbf{j}}-3\hat{\mathbf{k}}$ be the position vectors of the points A, B and C respectively, and L be the line passing through A and B. Find the shortest distance from C to L (Figure B).

1. Define the line L:
   First, we need the direction vector of the line, which is simply the vector connecting $A$ and $B$.
   It is obtained by subtracting the coordinates of point A from the coordinates of point B: $\vec{AB} = \langle 3-1, 6-3, 0-(-1) \rangle = \langle 2, 3, 1 \rangle$
   Now we write the vector equation of the line. Using the point $A$ as our starting position, any point on the line can be reached by traveling some amount $t$ along the direction vector $\vec{AB}$: $\vec{r}(t) = \vec{a} + t\vec{AB} = \langle 1, 3, -1 \rangle + t\langle 2, 3, 1 \rangle$

2. To find the shortest distance from point C to the line, we need to determine where the perpendicular from C meets the line L.
   Let’s call this point D, which lies on the line L, $D = (2t+1, 3t+3, t-1)$.
   The vector connecting point $C$ to our mystery point $D$ is $\vec{CD}$. We find this by subtracting $C$’s coordinates from $D$’s coordinates: $\vec{CD} = \langle (2t+1)-(-2), (3t+3)-4, (t-1)-(-3) \rangle = \langle 2t+3, 3t-1, t+2 \rangle$

3. For $\vec{CD}$ to be the shortest path, it must be perpendicular to the line itself (which is moving in the direction of $\vec{AB}$). Two vectors are perpendicular if their dot product is zero: $\vec{CD} \cdot \vec{AB} = 0$
   $\langle 2t+3, 3t-1, t+2 \rangle \cdot \langle 2, 3, 1 \rangle = 0 \iff 2(2t+3) + 3(3t-1) + 1(t+2) = 0$.
   Simplify, $\iff 4t + 6 + 9t - 3 + t + 2 = 0 \iff 14t + 5 = 0 \implies t = -\frac{5}{14}$

4. Calculate the distance:
   Now plug $t = -5/14$ back into our $\vec{CD}$ vector: $\vec{CD} = \left\langle 2\left(\frac{-5}{14}\right)+3, 3\left(\frac{-5}{14}\right)-1, \left(\frac{-5}{14}\right)+2 \right\rangle$
   $\vec{CD} = \left\langle \frac{32}{14}, \frac{-29}{14}, \frac{23}{14} \right\rangle$
   The distance is simply the magnitude (length) of this vector: $\vert{}\vec{CD}\vert{} = \sqrt{\left(\frac{32}{14}\right)^2 + \left(\frac{-29}{14}\right)^2 + \left(\frac{23}{14}\right)^2} = \sqrt{\frac{2394}{196}} \approx 3.49$

• Graphing a Plane Using Intercepts. Graph the plane $3x + 4y + 6z = 12$ and identify its normal vector (Figure C).

1. The normal vector of a plane can be read directly from the coefficients of the standard equation. For the plane $3x + 4y + 6z = 12$, the normal vector $\vec{n}$ is: $\vec{n} = \langle 3, 4, 6 \rangle$.

2. To graph a plane by hand or visualize it mentally, the easiest method is to find where it slices through the $x$, $y$, and $z$ axes (the intercepts), and then connect those dots to form a triangle.
   x-intercept: Set y = 0 and z = 0 in the plane equation: $2x = 12 \implies x = 4$. So, the x-intercept is (4, 0, 0). Analogously, the y-intercept is (0, 3, 0) and the z-intercept is (0, 0, 2).
   To graph the plane, we plot these points on a 3D coordinate system. Then, we draw a triangular plane that passes through these points. This triangular plane represents the portion of the plane 3x + 4y + 6z =12 that intersects the positive x-, y-, and z-axes.

3. In addition to finding the intercepts, we can also graph the traces — the lines where the plane cuts through the 2D coordinate planes (the walls and floor of the 3D space):
   $\star$ xy-trace (floor): Set z = 0: 3x + 4y = 12. This is the equation of a line in the xy-plane.
   $\star$ xz-trace (wall): 3x + 6z = 12, the equation of a line in the xz-plane.
   $\star$ yz-trace (wall): 4y + 6z = 12, the equation of a line in the yz-plane.

• Plane Through a Point and a Line. Find the equation of the plane through the point $Q(-1, -3, 2)$ that contains the line defined parametrically by: $x = -1 - 2t$, $y = 4t$, and $z = 2 + t$ (Figure D).

1. To define a plane, we need a normal vector. To get a normal vector via the cross product, we need two vectors that sit flat inside the plane.

2. Find two vectors in the plane:
   $\star$, vector 1 (from the line): The direction vector of the line is already in the plane. We pull this directly from the coefficients of t in the parametric equations: $\vec{v} = \langle -2, 4, 1 \rangle$
   $\star$, vector 2 (from two points): We need another point on the line to connect to $Q$. Let’s plug $t=0$ into the line equations to find a starting point $P$: $P = (-1, 0, 2)$
   Now, create a vector from $P$ to our standalone point $Q$: $\vec{PQ} = \langle -1-(-1), -3-0, 2-2 \rangle = \langle 0, -3, 0 \rangle$. Obviously, this vector is in the plane, too.

3. Calculate the Normal vector. The normal vector to the plane can be found by taking the cross product of the direction vector $\vec{v}$ and the vector $\vec{PQ}$: $\vec{n} = \vec{v}x\vec{PQ} = (\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\\ -2 & 4 & 1\\\ 0 & -3 & 0\end{smallmatrix}) = (4⋅0−1⋅(−3))\hat{\mathbf{i}}-((−2)⋅0−1⋅0)\hat{\mathbf{j}}+((−2)⋅(-3)-0·1)\hat{\mathbf{k}} = \langle 3, 0, 6 \rangle$.

4. Form the equation of the plane:
   Use the normal vector $\langle 3, 0, 6 \rangle$ and our point $Q(-1, -3, 2)$: $3(x - (-1)) + 0(y - (-3)) + 6(z - 2) = 0$
   Simplify $3(x + 1) + 6(z - 2) = 0 \implies 3x + 3 + 6z - 12 = 0 \implies 3x + 6z = 9$.
   Divide the entire equation by 3 to simplify. Final Equation: $x + 2z = 3$ (Figure E)

Bibliography

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