Lines & Planes in 3D-Space: Definition, Formula & Examples Video

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• 0:03 Lines Versus Planes
• 1:55 Examples
• 5:01 Lesson Summary
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Lesson Transcript

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.

Lines Versus Planes

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( x0 , y0 , z0 ) and Q( x1 , y1 , z1 ), 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:

Examples

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 ) = ro + tv

This is called a vector equation. This is where:

Example 1

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

So, ro = ( 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.

Example 2

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

1x + 1y + 4z + 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: 1x + 1y + 4z - 15 = 0

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