Plane Geometry

Let's say you're in class and want to take notes. You pull out a sheet of paper from your notebook to write on: this sheet of paper is similar to a geometric plane in that it is a **two-dimensional space** that provides a canvas to hold the information you draw or write on it.

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Jetzt kostenlos anmeldenLet's say you're in class and want to take notes. You pull out a sheet of paper from your notebook to write on: this sheet of paper is similar to a geometric plane in that it is a **two-dimensional space** that provides a canvas to hold the information you draw or write on it.

Planes in geometry provide a space for defining lines and points. Unlike a piece of paper, however, geometric planes extend infinitely. In real life, any flat two-dimensional surface can be considered mathematically as a plane, such as, for example, the surface of a desk. On the other hand, the block of wood that forms the top of the desk cannot be considered a two-dimensional plane, as it has three dimensions (length, width, and **depth**).

This article will explain the topic of planes in geometry and will go into detail about the **definition** of planes, some **examples** of planes, how planes **intersect**, and the **equation** of planes.

Let's begin our discussion with a formal definition of a plane.

In geometry, a **plane** is a flat two-dimensional surface that extends infinitely. Planes are defined as having zero thickness or depth.

For example, a **Cartesian coordinate system** represents a plane, since it is a flat surface that extends infinitely. The two dimensions are given by the the x- and the y-axis:

Since a plane is two-dimensional, this means that **points** and **lines** can be defined as existing within it, as they have less than two dimensions. In particular, points have 0 dimension, and lines have 1 dimension. Additionally, all two-dimensional shapes like quadrilaterals, triangles, and polygons are part of plane geometry and can exist in a plane.

The figure below shows a plane with points and a line. When points and lines exist within a plane, we say that the plane is the **ambient space **for the point and the line.

So, small geometrical objects like points and lines can "live" in bigger ones, like planes. These bigger objects hosting smaller ones are called **ambient spaces**. According to this same logic, can you guess what the ambient space that hosts a plane is?

It takes a three-dimensional space to provide ambient space for a two-dimensional plane. In fact, a three-dimensional Cartesian coordinate system can contain an infinite number of planes, lines, and points. Similarly, a plane can contain an infinite number of lines and points.

We know that the equation of a line in a two-dimensional Cartesian system is typically given by the equation \(y=mx+b\). On the other hand, the equation of a plane must be defined in three-dimensional space. Thus, it is a bit more complex. The equation to define a plane is given by:

\[ax+by+cz=d\]

Now that we have seen the equation, how can we build a plane in geometry? Some methods include:

- Three non-collinear points
- A normal vector and a point

We can define a plane by using 3 points that are **non-collinear **and** coplanar**. But what does it mean to be non-collinear and coplanar? Let's look at the definitions.

**Non-collinear points** occur when 3 or more points do not exist on a shared straight line.

**Coplanar points** are points that lie on the same plane.

If 3 given points are non-collinear and coplanar, we can use them to define the plane they share. The figure below shows a plane ABC which is defined and formed by the coplanar points \(A\), \(B\), and \(C\).

Next, let's take a second look at the figure which now includes a new point, \(D\).

Is \(D\) a coplanar point as well? From the figure, we can see that point \(D\) doesn't lie on plane \(ABC\) like the points \(A\), \(B\), and \(C\) do. Rather, it appears to be lying above the plane. So, point \(D\) is **non-coplanar**. Let's take a look at an example about defining a plane using three points.

Define the plane shown below using three points.

**Solution:** From the figure, we see that \(Q\), \(R\), and \(S\) are non-collinear and coplanar. Therefore, we can define a plane \(QRS\) using these three points. Although point \(T\) is also non-collinear with the other points, it is **not **coplanar because it is **not** at the same level or depth as points \(Q\), \(R\), and \(S\). Rather, it floats above the points \(Q\), \(R\), and \(S\). Therefore, point \(T\) cannot help us define the plane \(QRS\).

Does point \(D\), given by \((3,2,8)\), lie on plane \(ABC\), given by \(7x+6y-4z=1\)?

**Solution:**

To check whether a point lies on a plane, we can insert its coordinates into the plane equation to verify. If the point's coordinates are able to satisfy the plane equation mathematically, then we know the point lies on the plane.

\[7x+6y-4z=7(3)+6(2)-4(8)=21+12-32=1\]

Therefore, point \(D\) lies on plane \(ABC\).

A point in a three-dimensional Cartesian coordinate system is denoted by \((x,y,z)\).

Of all the infinite planes that can exist in a three-dimensional Cartesian coordinate system, three are particularly important:

- The \(xy\) plane that is given by the equation \(z=0\) (red in the figure below).
- The \(yz\) plane that is given by the equation \(x=0\) (green in the figure below).
- The \(xz\) plane that is given by the equation \(y=0\) (blue in the figure below).

Each plane is split into **four quadrants**, based on the values of the coordinates. For example in the \(xy\) plane, we have the following four quadrants:

- The first quadrant has a positive \(x\) and \(y\) coordinate.
- The second quadrant has a negative \(x\) and positive \(y\) coordinate.
- The third quadrant has a negative \(x\) and negative \(y\) coordinate.
- The fourth quadrant has a positive \(x\) and negative \(y\) coordinate.

Determine which of the following points lies in the \(xy\) plane: \((3,-7,4)\), \((4,8,0)\), \((2,3,-4)\).

We know that points that lie in the \(xy\) plane will have a z-value of \(0\), as they are only defined by the \(x\)- and \(y\)- axes. This means that the point \((4,8,0)\) lies in the \(xy\) plane.

Recall that a vector is a quantity that is defined by two elements: a magnitude (size or length) and a direction (orientation in space). Vectors are typically represented in geometry as arrows.

In a three-dimensional Cartesian space, vectors are denoted by a linear combination of **components **\((i,j,k)\). For example a vector with component 1 in the \(x\) direction, 2 in the \(y\) direction, and 3 in the \(k\) direction is denoted by:

\[v=i+2j+3k\]

A vector perpendicular to a plane is said to be **normal** to the plane. Such a vector has a very special property: the values of \(a\), \(b\), and \(c\) in the plane equation (\(ax+by+cz = d\)) are given by the components of the vector normal to the plane!

This means that we can find the equation of a plane if we know both:

- The coordinates of one point on the plane, and
- The vector normal to the plane.

Let's take a look at some examples.

A plane \(P\) has a normal vector \(7i+6j-4k\). The point \((3,2,8)\) lies on plane \(P\). Find the equation of the plane \(P \) in the form \(ax+by+cz=d\).

**Solution:**

The normal vector gives us our values for \(a\), \(b\), and \(c\):

- The \(i\) component of the vector is \(a\), so \(a=7\),
- the \(j\) component is \(b\), so \(b=6\),
- and the \(k\) component is \(c\), so \(c=-4\).

This gives us: \(7x+6y-4z=d\).

Next, we now need to find the value of \(d\). How can we do this? Well, we know the coordinates of a point that lies on the plane, so if we substitute these values into the equation, it will give us \(d\). Remember, the coordinates of the point is in the form \((x,y,z)\).

\[7(3)+6(2)-4(8)=d\]

\[21+12-32=d\]

\[d=1\]

Now we have our value for \(d\), so we can put this back into the equation to give us our answer:\[7x+6y-4z=1\]

Find an equation for the plane that passes through the point \((1,1,1)\) and is parallel to the plane \(3x+y+4z=6\).

**Solution:**

The plane is parallel to the plane \(3x+y+4z=6\). This means that they share the same normal, and a plane written in the form \(ax+by+cz=d\) has normal vector, \(ai+bk+ck\). Thus, the plane has normal \(3i+j+4k\). This gives us part of the equation for the plane: \(3x+y+4z=d\). We must now find a value for \(d\). As the plane passes through the point \((1,1,1)\), we know that the point lies on the plane. Therefore, we can substitute these values into our plane equation to give us a value for \(d\):

\[3(1)+1+4(1)=8\]

Our value for d gives us our complete plane equation:

\[3x+y+4z=8\]

If we have two planes in a three-dimensional space they are either parallel planes, meaning they never intersect (meet), or they are intersecting planes. When two lines intersect they intersect at a singular point, as lines are one-dimensional. When planes intersect, they intersect at a line that extends infinitely; this is because planes are two-dimensional. Imagine you had two pieces of paper that could pass through each other, these two sheets of paper each represent planes. When you pass them through each other, they will intersect once and form a line.

As you can see in the above image, intersecting planes form a line.

When we define a plane and a line, there are three possible cases:

- The plane and the line are parallel, meaning that they will never intersect.
- The plane and the line intersect at a single point in three-dimensional space.
- The line lies on the plane.

In the case that a line intersects perpendicular to (at a right angle) a plane, there are more properties we can utilize:

- Two lines that are perpendicular to the same plane are parallel to each other.
- Two planes that are perpendicular to the same line are parallel to each other.

Let's consider a couple more examples involving planes in geometry.

Define the plane:

This plane can be defined as \(CAB\), since a plane is made up of three non-collinear and coplanar points: \(C\), \(A\) and, \(B\) are non-collinear and coplanar.

A plane \(P\) has a normal vector \(2i+8j-3k\). The point \((3,9,1)\) lies on plane \(P\). Find the equation of the plane \(P\) in the form \(ax+by+cz=d\).

**Solution:**

The normal vector gives us our values for \(a\), \(b\) and \(c\):

- The \(i\) component of the vector is \(a\), so \(a=2\),
- the \(j\) component is \(b\), so \(b=8\),
- and the \(k\) component is \(c\), so \(c=-3\).

This gives us: \(2x+8y-3z=d\).

Now we can use the given point to find the value of \(d\). Since we have been given the coordinates, we can substitute them into the equation to solve for \(d\).

\[2(3)+8(9)-2(1)=d\]

\[21+72-2=d\]

\[d=91\]

Therefore:

\[2x+8y-2z=91\]

- A
**plane**is a flat two-dimensional surface that extends infinitely. - The
**equation of a plane**is given by: \(ax+by+cz=d\) - 3 non-collinear points can be used to define a plane in three-dimensional space.
- In coordinate geometry, we typically define points and lines in the \(xy\), \(xz\) and \(yz\) planes. If a point lies in one of these planes, then they have a coordinate of \(0\) in the remaining axis.
- When planes intersect, they intersect at a line that extends infinitely.
- A plane and a line are either parallel, intersect at a point, or the line lies in the plane.
- Two lines that are perpendicular to the same plane are parallel.
- Two planes that are perpendicular to the same line are parallel.

A plane is a flat two-dimensional surface that extends infinitely.

The intersection of two planes is called a line.

What is a plane?

A plane is a flat two-dimensional surface that extends infinitely .

How many dimensions does a plane have?

2

How can you define a plane using points?

3 non collinear points define a plane in three-dimensional space.

What are the quadrants of the xy coordinate plane?

- Quadrant 1 has a positive x and positive y
- Quadrant 2 has a negative x and positive y
- Quadrant 3 has a negative x and negative y
- Quadrant 4 has a positive x and negative y

What is the z coordinate of a point in the xy plane?

0

What is the x coordinate of a point in the yz plane?

0

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