What is the equation for a line in 3D space?
The line goes from (5, 5, 5) to (-8, -6, -3).
- 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.
- fdovertonLv 41 decade ago
To describe a line in space we use vectors and parametric equations:
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:
For t=0 we get the point P; and t=1 gives point QSource(s): Shenk Calculus and Analytic Geometry
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- 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!