Equations of Planes

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

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., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

• The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a₁+b₁), (a₂+b₂), (a₃+b₃) >.

• 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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a₁-b₁), (a₂-b₂), (a₃-b₃) >.

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

• 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. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$

  It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$

• The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular (at a right angle) to both of the 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 following formula, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$, e.g., $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ 1 & 2 & 3\\ 4 & 5 & 6\end{smallmatrix}) =|\begin{smallmatrix}2 & 3\\ 5 & 6\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}1 & 3\\ 4 & 6\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}1 & 2\\ 4 & 5\end{smallmatrix}|\vec{k} = (2⋅6−3⋅5)\vec{i}-(1⋅6−3⋅4)+(1⋅5−2⋅4)\vec{j}\vec{k} = −3\vec{i}+6\vec{j}-3\vec{k} = ⟨-3, 6, -3⟩.$

Find the Equation of a Plane Given Three Points

To find the equation of a plane given three points P₁, P₂, and P₃ in the plane, we need to determine a condition that ensures a fourth point P lies in the same plane.

Consider the vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$. $\vec{P_1P}$ is a vector from point P₁ to point P.

For point P to be in the planed defined by P₁, P₂, and P₃, the parallelepiped spanned by these vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$ must be flat. This flatness condition is mathematically represented by the determinant of these vectors being zero: $det(\vec{P_1P}, \vec{P_1P_2},~ \vec{P_1P_3})$ = 0 (Figure vi).

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.

An alternative way to find the equation of the plane is by using a normal vector. For point P to be in the plane, the vector $\vec{P_1P}$ must be perpendicular to the normal vector $\vec{N}$, $\vec{P_1P}⊥\vec{N}$ where $\vec{N}$ is a normal vector to our plane. This can be expressed using the dot product: $\vec{P_1P}·\vec{N}=0$ [🚀]

Finding the Normal Vector

A normal vector $\vec{N}$ to the plane can be found using the cross product of the vectors $\vec{P_1P_2}$ and $\vec{P_1P_3}$: $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3}$

Thus, the condition that the point P lies in the plane can be rewritten as:

$\vec{P_1P}·\vec{N}=0 [🚀] ⇒ \vec{P_1P}·(\vec{P_1P_2}x\vec{P_1P_3})=0$ (In the accompanying figure vi, the vectors and the plane can be visualized to better understand the geometric relationships and the conditions derived). 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

• Determine the equation of the plane that contains the points P₁ = ⟨1, −2, 0⟩, P₂ = ⟨3, 1, 4⟩, and P₃ = ⟨0, -1, 2⟩ (Figure vi).

1. Find 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 one point from the coordinates of the other point.

   Let P₁ has coordinates (x₁, y₁, z₁) and P₂ has coordinates (x₂, y₂, z₂), then the coordinates of $\vec{P_1P_2}$ are ⟨x₂-x₁, y₂-y₁, z₂-z₁⟩. $\vec{P_1P_2} = ⟨2, 3, 4⟩, \vec{P_1P_3} = ⟨-1, 1, 2⟩$

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 $\vec{P_1P_3}$.

   $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3} = [\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}] = [\begin{smallmatrix}i & j & k\\ 2 & 3 & 4\\ -1 & 1 & 2\end{smallmatrix}] = (3⋅2−4⋅1)\vec{i}-((2⋅2−4⋅(−1)))\vec{j}+(2⋅1−3⋅(−1))\vec{k} = 2\vec{i}-8\vec{j}+5\vec{k}$

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 any vector $\vec{P_1P}$ from point P₁ to a point P must be zero, that is, $\vec{P_1P}⊥\vec{N} ↭ \vec{P_1P}·\vec{N}=⟨x-1, y+2, z-0⟩·(2\vec{i}-8\vec{j}+5\vec{k}) = (x-1)2-8(y+2)+5z = 0 ↭ 2x -2 -8y -16+5z = 0 ↭ 2x -8y +5z = 18$, and this is the equation of the plane.

Conclusion: The equation of the plane containing the points P₁ = ⟨1, −2, 0⟩, P₂ = ⟨3, 1, 4⟩, and P₃ = ⟨0, −1, 2⟩ is: 2x −8y +5z = 18.

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

Parallel planes have the same normal vectors. The normal vector of a plane ax + by + cz = d is $\vec{N}$ =⟨a,b,c⟩, hence any parallel to 2x +3y +z = 5, has a normal vector $\vec{N} = ⟨2, 3, 1⟩$.

A point P with coordinates (x, y, z) is in the plane ↭ $\vec{QP}·\vec{N} = 0 ↭ (x-1, y-1, z-1)·⟨2, 3, 1⟩ = 0 ↭ 2(x-1) +3(y-1)+z-1 = 0 ↭ 2x -2 + 3y -3 + z -1 = 0 ↭ 2x + 3y + z = 6.$

Therefore, the equation of the plane passing through the point Q (1, 1, 1) and parallel to the plane 2x +3y +z = 5 is 2x + 3y + z = 6.

Another way. A plane with normal vector $\vec{N} = ⟨2, 3, 1⟩$ has the form 2x + 3y + z = c, where c is a constant. Q(1, 1, 1) is in the plane. Substitute Q into the plane equation to find c: 2·1 + 3·1 + 1 = c ⇒ c = 6, so the plane is 2x + 3y + z = 6.

• Find parametric and symmetric equations of the line passing through points A = ⟨1, 4, −2⟩ and B = ⟨−3, 5, 0⟩.

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: $\vec{d}$ = ⟨−3−1, 5−4, 0+2⟩ = ⟨−4, 1, 2⟩.

Parametric Equations. Parametric equations express the coordinates (x, y, z) 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 (x₀, y₀,z₀) 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 $\vec{d} = ⟨-4, 1, 2⟩$, the parametric equations of the line are:

$ \begin{cases} x = 1 - 4t \\ y = 4 + t \\ z = -2 + 2t \end{cases}$

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: $\frac{x-1}{-4} = y - 4 = \frac{x + 2}{2}$ (symmetric equation of the 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).

Find the Direction Vector $\vec{AB}$. It is obtained by subtracting the coordinates of point A from the coordinates of point B: $\vec{AB} = (3-1)\hat{\mathbf{i}+(6-3)\hat{\mathbf{j}}}-\hat{\mathbf{k}} = 2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}}$

Find the vector equation of the line L. The vector equation of the line passing through A with direction vector $\vec{AB}$ is: $\vec{r} = \vec{a}+t\vec{AB} = \hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}} +t(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}-\hat{\mathbf{k}}) = (2t+1)\hat{\mathbf{i}} + (3t + 3)\hat{\mathbf{j}} + (t-1)\hat{\mathbf{k}}$

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.

$\vec{CD} = ⟨2t+1+2, 3t+3-4, t-1+3⟩ = ⟨2t+3, 3t-1, t +2⟩$

Find t by ensuring that $\vec{CD}$ is perpendicular to $\vec{AB}$, their dot product must be zero: $⟨2t+3, 3t-1, t +2⟩·⟨2, 3, 1⟩ = 0 ↭ 2(2t+3)+3(3t-1)+1·(t+2)=0 ↭ 4t+6+9t-3+t+2=0 ↭ 14t + 5 = 0 ↭ t = \frac{-5}{14}$.

$\vec{CD} = ⟨2·\frac{-5}{14}+3, 3(\frac{-5}{14})-1, \frac{-5}{14}+2⟩ = ⟨\frac{16}{7}, \frac{-29}{14}, \frac{23}{14}⟩$ The required distance is $|\vec{CD}| = \sqrt{(\frac{32}{14})^2 + (\frac{-29}{14})^2+ (\frac{23}{14})^2} = \sqrt{\frac{2394}{196}}≈ 3.5$

• Graph the plane 3x + 4y + 6z = 12 and identify its normal vector (Figure C).

The normal vector of a plane can be directly obtained from the coefficients of x, y, and z in the plane’s equation. For the plane 3x + 4y+ 6z = 12, the normal vector $\vec{n}$ is: $\vec{n}=⟨3, 4, 6⟩$.

To graph the plane, we need to find the intercepts where the plane intersects the x-, y-, and z-axes.

x-intercept: Set y = 0 and z = 0 in the plane equation: 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.

In addition to finding the intercepts, we can also graph the traces of the plane in the xy-, xz-, and yz-planes.

xy-trace: Set z = 0: 3x + 4y = 12. This is the equation of a line in the xy-plane. xz-trace: 3x + 6z = 12, the equation of a line in the xz-plane. yz-trace: 4y + 6z = 12, the equation of a line in the yz-plane.

• Find the equation of the plane through the point Q = (-1, -3, 2) and containing the line x = -1-2t, y = 4t, z = 2 + t.

From the equations, we can extract a point and a direction vector for the line: (Figure D)

1. The direction vector $\vec{v}=⟨-2, 4, 1⟩$ of the line comes from the coefficients of t in the parametric equations.
2. When t = 0, the parametric equations give us a specific point on the line: P = (-1, 0, 2).

Therefore, $\vec{PQ}=⟨−1−(−1),−3−0,2−2⟩ = ⟨0, -3, 0⟩$ is in the plane, too.

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

The equation of a plane with normal vector $\vec{n} = ⟨3, 0, 6⟩$ and a point Q(-1, -3, 2) in the plane is: 3x +6z = c, 3·(-1)+6·2 = c ⇒ c = 9 ⇒3x + 6z = 9 ⇒[Divide the entire equation by 3 to simplify:] x + 2z = 3 is the equation of the plane (Figure E).

Alternatively, 3(x + 1) +0(y + 3) + 6(z − 2) = 0 ⇒3x+6z=9 ⇒ x + 2z = 3.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
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