Planes: Equation, Distance, Angle, Parallelism & Intersection

TAKEAWAYS:

• A plane in 3D space can be defined by a point and a normal vector, or by three non-collinear points.
• The general equation of a plane is $Ax+By+Cz+D=0$, where $(A,B,C)$ is the normal vector.
• The distance from a point to a plane is given by $\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{A^2+B^2+C^2}}$.
• Two planes are parallel if their normal vectors are scalar multiples of each other.
• The angle between two planes is the angle between their normal vectors.
• The intersection of two planes is a line, which can be found by solving their equations simultaneously.

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What is a Plane?

A plane is a flat, two-dimensional surface that extends infinitely in all directions. In 3D space, a plane can be defined in several ways:

1. Using a point and a normal vector:

   • A normal vector $\vec{n}=(A,B,C)$ is perpendicular to the plane.
   • If $(x_0,y_0,z_0)$ is a point on the plane, the equation of the plane is: $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$
   • This can be rewritten in the general form: $$Ax+By+Cz+D=0$$ where $D=-(A{x}_0+B{y}_0+C{z}_0)$.

2. Using three non-collinear points:

   • If three points $P(x_1,y_1,z_1)$, $Q(x_2,y_2,z_2)$, and $R(x_3,y_3,z_3)$ lie on the plane, we can find two vectors in the plane: $$\overrightarrow{PQ}=(x_2-x_1,y_2-y_1,z_2-z_1)$$ $$\overrightarrow{PR}=(x_3-x_1,y_3-y_1,z_3-z_1)$$
   • The normal vector $\vec{n}$ is the cross product of $\overrightarrow{PQ}$ and $\overrightarrow{PR}$: $$\vec{n}=\overrightarrow{PQ}\times \overrightarrow{PR}$$
   • The equation of the plane can then be written using point $P$ and $\vec{n}$.

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Example 1: Equation of a Plane Using a Point and Normal Vector

Find the equation of the plane passing through the point $(2,-1,3)$ with normal vector $\vec{n}=(4,-2,5)$.

Solution: Using the point-normal form: $$4(x-2)-2(y+1)+5(z-3)=0$$ Simplify: $$4x-8-2y-2+5z-15=0$$ $$4x-2y+5z-25=0$$ So, the equation of the plane is: $$4x-2y+5z=25$$

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Example 2: Equation of a Plane Using Three Points

Find the equation of the plane passing through the points $A(1,2,3)$, $B(4,1,2)$, and $C(2,3,4)$.

Solution:

1. Find two vectors in the plane: $$\overrightarrow{AB}=(4-1,1-2,2-3)=(3,-1,-1)$$ $$\overrightarrow{AC}=(2-1,3-2,4-3)=(1,1,1)$$
2. Find the normal vector $\vec{n}$: $$\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 3& -1& -1\\ 1& 1& 1\end{array}\right|=\mathbf{i}((-1)(1)-(-1)(1))-\mathbf{j}((3)(1)-(-1)(1))+\mathbf{k}((3)(1)-(-1)(1))$$ $$=\mathbf{i}(-1+1)-\mathbf{j}(3+1)+\mathbf{k}(3+1)=(0,-4,4)$$ So, $\vec{n}=(0,-4,4)$. We can simplify this to $(0,-1,1)$ by dividing by 4.
3. Use point $A(1,2,3)$ and the simplified normal vector to write the equation: $$0(x-1)-1(y-2)+1(z-3)=0$$ Simplify: $$-y+2+z-3=0⟹-y+z-1=0$$ Or: $$-y+z=1$$ (We can also write it as $y-z=-1$.)

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Distance from a Point to a Plane

The distance $d$ from a point $P(x_0,y_0,z_0)$ to a plane $Ax+By+Cz+D=0$ is given by: $$d=\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{A^2+B^2+C^2}}$$

Example 3: Distance from a Point to a Plane

Find the distance from the point $(1,2,3)$ to the plane $2x-y+2z+3=0$.

Solution: Using the distance formula: $$d=\frac{|2(1)-1(2)+2(3)+3|}{\sqrt{2^2+(-1{)}^2+2^2}}=\frac{|2-2+6+3|}{\sqrt{4+1+4}}=\frac{|9|}{\sqrt{9}}=\frac{9}{3}=3$$ So, the distance is 3 units.

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Angle Between Two Planes

The angle $\theta$ between two planes is the angle between their normal vectors. If the normal vectors are $\overrightarrow{n_1}=(A_1,B_1,C_1)$ and $\overrightarrow{n_2}=(A_2,B_2,C_2)$, then: $$\cos \theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$$ The angle between the planes is the acute angle, so if $\theta >90^{\circ }$, we take $180^{\circ }-\theta$.

Example 4: Angle Between Two Planes

Find the angle between the planes $x+2y-z=4$ and $3x-y+2z=5$.

Solution:

1. Identify the normal vectors: $$\overrightarrow{n_1}=(1,2,-1),\overrightarrow{n_2}=(3,-1,2)$$
2. Compute the dot product and magnitudes: $$\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(1)(3)+(2)(-1)+(-1)(2)=3-2-2=-1$$ $$|\overrightarrow{n_1}|=\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{1+4+1}=\sqrt{6}$$ $$|\overrightarrow{n_2}|=\sqrt{3^2+(-1{)}^2+2^2}=\sqrt{9+1+4}=\sqrt{14}$$
3. Compute $\cos \theta$: $$\cos \theta =\frac{-1}{\sqrt{6}\cdot \sqrt{14}}=\frac{-1}{\sqrt{84}}=\frac{-1}{2\sqrt{21}}$$ Since $\cos \theta$ is negative, the angle between the planes is the supplement: $$\theta ={\cos }^{-1}\left(\frac{1}{2\sqrt{21}}\right)\approx 97.1^{\circ }$$ The acute angle between the planes is: $$180^{\circ }-97.1^{\circ }=82.9^{\circ }$$ (Alternatively, we can take the absolute value of the dot product for the acute angle.)

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Parallelism and Intersection of Planes

Parallel Planes

Two planes are parallel if their normal vectors are scalar multiples of each other. That is: $$\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}$$ If the planes are parallel and have the same constant term $D$, they are identical (the same plane).

Intersection of Two Planes

The intersection of two planes is a line (unless the planes are parallel and distinct, in which case they do not intersect). To find the line of intersection:

1. Solve the two plane equations simultaneously to find a relationship between two variables.
2. Express one variable in terms of the other (e.g., $y$ in terms of $x$) and let the third variable be a parameter (e.g., $z=t$).
3. The parametric equations of the line can be written as: $$\{\begin{array}{l}x=x_0+at\\ y=y_0+bt\\ z=z_0+ct\end{array}$$ where $(x_0,y_0,z_0)$ is a point on the line and $(a,b,c)$ is the direction vector (found from the cross product of the normal vectors of the two planes).

Example 5: Intersection of Two Planes

Find the line of intersection of the planes $x+y+z=1$ and $2x-y+z=2$.

Solution:

1. Solve the system of equations: $$\{\begin{array}{l}x+y+z=1(1)\\ 2x-y+z=2(2)\end{array}$$ Subtract (1) from (2): $$(2x-y+z)-(x+y+z)=2-1⟹x-2y=1⟹x=2y+1$$ Substitute $x=2y+1$ into (1): $$(2y+1)+y+z=1⟹3y+z=0⟹z=-3y$$
2. Let $y=t$ (a parameter). Then: $$x=2t+1,y=t,z=-3t$$ So, the parametric equations of the line are: $$\{\begin{array}{l}x=1+2t\\ y=t\\ z=-3t\end{array}$$ The direction vector is $(2,1,-3)$.

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Comparison Table: Key Concepts

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Applications of Planes

1. Computer Graphics: Planes are used to model 3D objects and surfaces in rendering.
2. Architecture and Engineering: Planes help in designing flat surfaces like walls, floors, and roofs.
3. Physics: Planes are used to describe wavefronts, mirrors, and other flat surfaces in optics.
4. Navigation: Planes (literally!) are used in aviation to describe flight paths and altitudes.

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Exam Tip

1. Memorize the General Form: Always write the equation of a plane in the form $Ax+By+Cz+D=0$.
2. Normal Vector is Key: For most problems, finding the normal vector is the first step.
3. Distance Formula: Practice calculating distances from points to planes—this is a common exam question.
4. Angle Between Planes: Remember that the angle between planes is the angle between their normal vectors.
5. Parallelism Check: Two planes are parallel if their normal vectors are scalar multiples.
6. Intersection: To find the intersection, solve the system of equations. If the planes are parallel, check if they are identical or distinct.
7. Parametric Equations: For the line of intersection, express variables in terms of a parameter $t$.
8. Significant Figures: In calculations, keep intermediate steps precise and round only the final answer if needed.

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NEB Board-Style Questions

Short Answer Questions

1. Write the equation of the plane passing through the point $(1,-1,2)$ with normal vector $\vec{n}=(3,-2,1)$.
2. Find the distance from the point $(4,-1,2)$ to the plane $2x-3y+z+5=0$.
3. Determine whether the planes $x+2y-z=3$ and $2x+4y-2z=6$ are parallel or identical.
4. Find the angle between the planes $3x-y+2z=4$ and $x+2y-z=5$.

Long Answer Questions

5. Find the equation of the plane passing through the points $A(1,2,3)$, $B(4,1,2)$, and $C(2,3,4)$. Also, find the distance from the origin to this plane.
6. Two planes are given by $x+y+z=1$ and $2x-y+z=2$. Find:
   • The angle between them.
   • The parametric equations of their line of intersection.
7. A plane has the equation $4x-3y+2z=12$. Find:
   • The distance from the point $(1,-1,2)$ to the plane.
   • The equation of a plane parallel to the given plane and passing through the origin.
8. Prove that the planes $x+2y-z=4$ and $2x+4y-2z=8$ are identical. What does this imply about their intersection?

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