Given:
A line through the points (7,1,-5) and (3,4,-2).
To find:
The parametric equations of the line.
Solution:
Direction vector for the points (7,1,-5) and (3,4,-2) is
![\vec {v}=\left](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D%3D%5Cleft%3Cx_2-x_1%2Cy_2-y_1%2Cz_2-z_1%5Cright%3E)
![\vec {v}=\left](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D%3D%5Cleft%3C3-7%2C4-1%2C-2-%28-5%29%5Cright%3E)
![\vec {v}=\left](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D%3D%5Cleft%3C-4%2C3%2C3%5Cright%3E)
Now, the perimetric equations for initial point
with direction vector
, are
![x=x_0+at](https://tex.z-dn.net/?f=x%3Dx_0%2Bat)
![y=y_0+bt](https://tex.z-dn.net/?f=y%3Dy_0%2Bbt)
![z=z_0+ct](https://tex.z-dn.net/?f=z%3Dz_0%2Bct)
The initial point is (7,1,-5) and direction vector is
. So the perimetric equations are
![x=(7)+(-4)t](https://tex.z-dn.net/?f=x%3D%287%29%2B%28-4%29t)
![x=7-4t](https://tex.z-dn.net/?f=x%3D7-4t)
Similarly,
![y=1+3t](https://tex.z-dn.net/?f=y%3D1%2B3t)
![z=-5+3t](https://tex.z-dn.net/?f=z%3D-5%2B3t)
Therefore, the required perimetric equations are
and
.