# What is the equation for a line in 3D space?

The line goes from (5, 5, 5) to (-8, -6, -3).

### 5 Answers

- รզlεսռց ☆Lv 61 decade agoBest Answer
I'll label your points A(5, 5, 5) and B(-8, -6, -3). Now, we need a direction vector, which will be AB (you can use BA if you want but I'll use AB):

AB = [-8-5 , -6-5 , -3-5]

AB = [-13, -11, -8]

Now that we have this, we'll use point A to find the equation of the line. I'm going to be finding the equation in parametric form (you can convert it to cartesian or vector later on if you wish):

The equation is given be:

x = x1 + at

y = y1 + bt

z = z1 + ct

Where the line goes through (x1, y1, z1) and has direction vector [a, b, c]

So, using your point (5, 5, 5) and direction vector [-13, -11, -8]:

x = 5 - 13t

y = 5 - 11t

z = 5 - 8t

Where t is the parameter and can take any real values.

- 4 years ago
This Site Might Help You.

RE:

What is the equation for a line in 3D space?

The line goes from (5, 5, 5) to (-8, -6, -3).

Source(s): equation line 3d space: https://tinyurl.im/cH4w9 - fdovertonLv 41 decade ago
To describe a line in space we use vectors and parametric equations:

x=initial(x)+at

y=inital(y)+bt

z=initial(z)+ct

where <a,b,c> represents the vector parallel to the line through the point (x,y,z)

in your case let's say P=(5,5,5) and Q=(-8,-6,-3)

Then vector PQ= (-8-5,-6-5,-3-5) = <-13,-11,-8>

Using vector and point P we have the equations:

x=5-13t

y=5-11t

z=5-8t

For t=0 we get the point P; and t=1 gives point Q

Source(s): Shenk Calculus and Analytic Geometry - How do you think about the answers? You can sign in to vote the answer.
- Anonymous6 years ago
All the answers bellow are incorrect. For the parametric equation of a line, you need a direction vector, which is the normal to the plane in which the line is located. Those answers are wrong!

You're thinking of the equation of a plane. For a line the vector parallel to the line and a point on the line is needed for the parametric equations.