55.2k views
0 votes
A line passes through the points P(1,-6,7) and Q(-9,10,-5) find the standard parametric equations for the line, written using the base point P(1,-6,7) and the components of the vector PQ rightarrow.

x = _________, y = _________, z = __________.

User Nena
by
5.5k points

1 Answer

2 votes

Answer:


x = 1-10t\\y = -6+16t\\z = 7-12t

Explanation:

We are given the coordinates of points P(1,-6,7) and Q(-9,10,-5).

The values in the form of (
x,y,z) are:


x_1=1\\x_2=-9\\y_1=-6\\,y_2=10\\z_1=7\\z_2=-5


$\vec{PQ}$ can be written as the difference of values of x, y and z axis of the two points i.e. change in axis.


\vec{PQ}=<x_2-x_1,y_2-y_1,z_2-z_1>


\vec{PQ} = <(-9-1), 10-(-6),(-5-7)>\\\Rightarrow \vec{PQ} = <-10, 16,-12>

The equation of line in vector form can be written as:


\vec{r} (t) = <1,-6,7> + t<-10,16,-12>

The standard parametric equation can be written as:


x = 1-10t\\y = -6+16t\\z = 7-12t

User Smartse
by
5.7k points