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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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Lesson Transcript

Instructor:
*Nadia Beherens*

I have worked at several schools and universities as a Math teacher. I have a degree in Math teaching, and I am a master student.

In this lesson, we will study how lines and planes function in three-dimensional space, and learn how to calculate a line. We also address definitions, formulas and examples.

We can differentiate easily between a graph which shows a plane and one that shows a line. For example, if we observe a room, a wall in that room will be a graph model of a plane, while where two of the walls intersect gives us the idea of a line:

In a formal way, a line can be defined as a continuous and endless succession of points in one dimension, and a plane is an ideal object that only has two dimensions and contains infinite points and lines.

First, it's necessary to know how to determine a vector and how to do a **cross product**. A **vector** is a line segment oriented from its starting point, called its origin, to its end point, called the end. So, like you can see in the equation below, if we have two points *P*( *x*0 , *y*0 , *z*0 ) and *Q*( *x*1 , *y*1 , *z*1 ), we can calculate the vector v:

subtracting each coordinate of each point.

For example: *P* ( -2 , 3 , 1 ) and *Q* ( 5 , -2 , 4 )

On the other hand, a **cross product** defines a vector, perpendicular to the two vectors that originate it. To do that, we need two vectors, *v* and *w* and must do a procedure similar to a determinant:

If *v* = ( *a , b , c* ) and *w* = ( *d , e , f* ), we get a three by three determinant of:

That, in turn, gives us the formula here:

Now, we can determine the formulas for lines and planes in three dimensional space. There are several ways to calculate a line. However, we always need a point *P* of it and a vector *v* of it or parallel to it. Then, a **line** can be defined as:

*r* ( *t* ) = *r*o + *tv*

This is called a vector equation. This is where:

Find the vector equation of the line which contains the point *P* ( 1 , -2 , 1 ) and the vector *v* = ( 1 , 2 , 3 ).

So, *r*o = ( 1 , -2 , 1 ) and *v* = ( 1 , 2 , 3 )

If we replace in the equation, we get:

*r* ( *t* ) = ( 1 , -2 , 1 ) + *t* ( 1 , 2 , 3 )

Now, to calculate a plane, we also need a point *P* on the plane and a vector *n* = ( *A , B , C* ), normal to the plane:

Pi: *Ax* + *By* + *Cz* + *D* = 0, which is called a **general equation**.

Determinate the equation of the plane which contains to *P* ( 2 , 1 , 3 ) and is normal to the vector *n* = ( 1 , 1 , 4 )

Pi: *Ax* + *By* + *Cz* + D = 0

1*x* + 1*y* + 4*z* + *D* = 0

Replace the coordinates of *n*.

1(2) + 1(1) + 4(3) + *D* = 0

Replace the point coordinates to calculate *D*.

2 + 1 + 12 + *D* = 0

So *D* = -15

Then, Pi: 1*x* + 1*y* + 4*z* - 15 = 0

Find the equation of the plane containing the points *P*(1, 3, -1), *Q* (-2, 4, 0) and *R* (0, 0, 1).

To write the general equation we need a normal vector to the plane and a point. If there are two vectors and no parallels, a normal vector to them can be calculated using the cross product: *n* = *v x w*

So, in the example:

This leads us to getting a three-by-three determinant:

Which gives us the next formula:

Which we can simplify down further to this next formula:

Which eventually gives us the answer of:

n = ( 5 , 5 , 10 )

Then we have *P* (1 , 3 , -1 ) and *n* = ( 5 , 5 , 10 )

Pi: 5 *x* + 5 *y* + 10 *z* + *D* = 0

5 (1) + 5 (3) + 10 (-1) + *D* = 0

5 + 15 - 10 + *D* = 0

*D* = -10

So, therefore, Pi: 5 *x* + 5 *y* + 10 *z* - 10 = 0

Let's take a couple moments to review what we've learned in this lesson on lines and planes existing within a three-dimensional space. First of all, a **vector** is a line segment oriented from its starting point, called its origin, to its end point, called the end, which can be used in defining lines and planes in three-dimensional space. If we want to determine the equation of a line in 3-D we're going to need a point of the line and a vector. So *r* ( *t* ) = *r*o + *tv*, which is called vector equation.

Then, to determine the equation of a plane, we also need a point and a normal vector to the plane. So, Pi: *Ax* + *By* + *Cz* + *D* = 0, which is called **general equation**. We finally learned that using the general equation allows us to find a normal vector if there are two vectors and no parallels by using the **cross product**, which defines a vector perpendicular to the two vectors that originate it.

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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

- How to Calculate the Volumes of Basic Shapes 7:17
- How to Solve Visualizing Geometry Problems 10:41
- The Dot Product of Vectors: Definition & Application 6:21
- Cross Product & Right Hand Rule: Definition, Formula & Examples 5:02
- Triple Scalar Product: Definition, Formula & Example 5:54
- Lines & Planes in 3D-Space: Definition, Formula & Examples 6:00
- What Are Cylinders? - Definition, Area & Volume 5:09
- How to Find Surface Area of a Cylinder 4:26
- Cylindrical & Spherical Coordinates: Definition, Equations & Examples 7:38
- Go to Analytic Geometry in 3-Dimensions

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