Determine vector, parametric, and if possible, symmetric equations of the line through A(-3, 5, -5) and B(9, 2, -1).

direction vector AB = <12, -3, 4>

so a possible vector equation is
<x,y,z> = <-3,5,-5> + t<12,-3,4>

from that, it is a piece of cake to form the other two versions.
Let me know what you get.

{ vector } = < -3 + 12t, 5 - 3t , -5 + 4t > { parametric } ====> [ x + 3 ] / 12 = [ y - 5 ] / - 3 = [ z + 5 ] / 4 {= t ; symmetric }

for parametric, the more common form would be:

x = -3 + 12t
y = 5 - 3t
z = -5 + 4t

your symmetric version is correc

could i also write the symmetric equation like this?

(x + 3) / 12 = (y - 5) / -3 = (z + 5) / 4

Isn't that what you first had in your original post ?

Another version would use the other point (9, 2, -1)
It would be :
(x-9)/12 = (y-2)/-3 = (z+1)/4

To determine the vector equation, we need to find a vector parallel to the line. This can be done by subtracting the position vectors of two points on the line. Let's calculate the vector equation:

Vector AB = Vector B - Vector A
= <9, 2, -1> - <-3, 5, -5>
= <9 + 3, 2 - 5, -1 + 5>
= <12, -3, 4>

So, a vector parallel to the line is <12, -3, 4>.

To find the parametric equations, we can express each coordinate of any arbitrary point on the line in terms of a parameter, typically denoted by t. Let's call the parametric equations:

x = -3 + 12t
y = 5 - 3t
z = -5 + 4t

These equations represent the coordinates (x, y, z) of any point on the line in terms of the parameter t.

Now, let's find the symmetric equations. Symmetric equations involve setting all three coordinates equal to some constant value. Let's set them equal to k and form the equation:

(x + 3) / 12 = (y - 5) / (-3) = (z + 5) / 4 = k

These equations are referred to as the symmetric equations and represent the line passing through points A and B.