 ## V9 Intersecting planes Two 2-dimensional planes will slice through each other (unless they are parallel). Where they slice will be defined by a straight line. There will also be an angle between the two planes.

Learn how to determine the angle between two intersecting planes and the equation of the line of intersection.

### Introduction

When two planes in three dimensions intersect, the set of points in the intersection forms a line. However two planes do not form lines if they are parallel and hence never intersect.

This module considers the angle between two planes when they are not parallel.

### Definition of a Plane

A plane in three dimensions is a set of points satisfying the equation \begin{align*} ax+by+cz & =d \end{align*} where $$a,\,b,\,c$$ and $$d$$ are constants.

### Normal to a Plane

The normal $$\vec{N}$$, of a plane \begin{align*} ax+by+cz & =d \end{align*} is a vector at right angles to the plane and has the form \begin{align*} \vec{N} & =a\hat{i}+b\hat{j}+c\hat{k}. \end{align*} Of course, another normal is given by $$-\vec{N}=-a\hat{i}-b\hat{j}-c\hat{k}$$ as this just a vecor in the opposite direction of $$\vec{N}$$ and is still at right angles to the plane.

### The Angle Between Two Planes

The angle between two intersecting planes is the same as the angle between the normals of the planes.

If $$N_{1}$$ and $$N_{2}$$are the normals of two intersecting planes, we know that1 This follows from the definition of the dot or scalar product.

\begin{align*} \overrightarrow{N_{1}}.\overrightarrow{N_{2}} & =\left|\overrightarrow{N_{1}}\right|\left|\overrightarrow{N_{2}}\right|\cos\theta \end{align*} where $$\theta$$ is the angle between them.

Rearranging , we find :

\begin{align*} \cos\theta & =\frac{\overrightarrow{N_{1}}.\overrightarrow{N_{2}}}{\left|\overrightarrow{N_{1}}\right|\left|\overrightarrow{N_{2}}\right|}. \end{align*} Note that there are in general two possible angles. One is obtuse, the other is acute. Together they sum to $$180^{\circ}.$$

If we wish to find only the acute angle between the planes then we say that:

\begin{align*} \cos\theta & =\frac{\left|\overrightarrow{N_{1}}.\overrightarrow{N_{2}}\right|}{\left|\overrightarrow{N_{1}}\right|\left|\overrightarrow{N_{2}}\right|} \end{align*}

#### Example

Find the acute angle of intersection of the planes $$x+y+z=0$$ and $$x-3y+z=1$$.

The plane $$x+y+z=0$$ has the normal vector $$\overrightarrow{N_{1}}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$$.

The plane $$x-3y+z=1$$ has the normal vector $$\overrightarrow{N_{2}}=\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}.$$

Remember:

\begin{alignat*}{1} \cos\theta & =\frac{\left|\overrightarrow{N_{1}}.\overrightarrow{N_{2}}\right|}{\left|\overrightarrow{N_{1}}\right|\left|\overrightarrow{N_{2}}\right|} \end{alignat*}

so:

\begin{alignat*}{1} \left|\overrightarrow{N_{1}}.\overrightarrow{N_{2}}\right| & =\left|\left(1\times1\right)+\left(1\times-3\right)+\left(1\times1\right)\right|\\ & =\left|-1\right|\\ & =1 \end{alignat*}

and:

\begin{alignat*}{1} \left|\overrightarrow{N_{1}}\right| & =\sqrt{1^{2}+1^{2}+1^{2}}\\ & =\sqrt{3} \end{alignat*}

and:

\begin{alignat*}{1} \left|\overrightarrow{N_{2}}\right| & =\sqrt{1^{2}+3^{2}+1^{2}}\\ & =\sqrt{11} \end{alignat*}

Therefore:

\begin{align*} \cos\theta & =\frac{1}{\sqrt{3}\sqrt{11}}\simeq0.1741 \end{align*}

Rearranging formula and using inverse $$\cos$$ gives:

\begin{align*} \theta & =\cos^{-1}\left(0.1741\right)\simeq80^{o} \end{align*}

So the angle of intersection of the two planes is approximately $$80$$ degrees.

#### Exercise

Find the angle of intersection of the following planes

1. The plane $$x-y+z=1$$ and $$2x+y-z=3$$
Answer: $$90$$ degrees.

2. The plane $$2x+y-z=2$$ and $$3x+y-z=3$$
Answer: $$10$$ degrees.

### Line of Intersection of Two Planes

The planes, $$P_{1}$$and $$P_{2}$$ intersect along the line L, as in the diagram below: We want to find the equation of the line of intersection, $$L.$$

#### Example

If $$P_{1}$$: $$2x+4y-z=4$$ and $$P_{2}$$: $$x-2y+z=3$$ , find the parametric equations of the line of intersection of the two planes.
Solution:

Given $$2x+4y-z=4$$ and $$x-2y+z=3$$, we have two equations but three unknowns. This is a clue to introduce a parameter.2 We will set $$z=t$$ but you can set $$x=t$$ or $$y=t$$. This will generate a set of equations that may look different to what we show below, but they are correct. Let $$z=t$$ then the equations of the planes become

\begin{alignat*}{2} 2x+4y-t & =4 & \left(1\right)\\ x-2y+t & =3. & \left(2\right) \end{alignat*}

Multiplying $$\left(2\right)$$ by $$-2$$, the equations become:

\begin{alignat*}{1} 2x+4y-t & =4\\ -2x+4y-2t & =-6 \end{alignat*}

adding these two equations we get:

\begin{alignat*}{1} 8y-3t & =-2\\ 8y & =3t-2\\ y & =\frac{3}{8}t-\frac{1}{4}. \end{alignat*}

Substituting $$y$$ in equation $$\left(2\right)$$

\begin{alignat*}{1} x-2\left(\frac{3}{8}t-\frac{1}{4}\right)+t & =3\\ x-\frac{6}{8}t+\frac{2}{4}+t & =3\\ x+\frac{1}{4}t+\frac{1}{2} & =3\\ x & =3-\frac{1}{4}t-\frac{1}{2}\\ x & =\frac{5}{2}-\frac{1}{4}t. \end{alignat*}

Hence the parametric equations of the line of the intersection of the two planes are:

\begin{alignat*}{1} x=\frac{5}{2}-\frac{1}{4}t,\,y & =\frac{3}{8}t-\frac{1}{4}\text{ and z=t. } \end{alignat*}

#### Exercise

Find the parametric line of intersection and the angle of intersection of the planes $$x+y+2z=0$$ and $$2x-y+z=5$$.
Answer: Line is $$x=2-t$$ ; $$y=-1-t$$ ; $$z=t$$. Angle is $$60$$degrees.

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