91Ó°ÊÓ

the line containing the points P$\left(x_0,y_0,z_0\right)$and Q$(x_1,y_1,z_1)$

$\text{The point are}P\left(x_0,y_0,z_0\right)\text{and}Q\left(x_1,y_1,z_1\right)\text{.}$

$\stackrel{→}{PQ}=\left(x_1-x_0,y_1-y_0,z_1-z_0\right)$

$\text{The formula to find the line}\mathcal{L}\text{equation is as follows,}$

$\text{Here}P\left(x_0,y_0,z_0\right)\text{and}\stackrel{→}{PQ}=d=\left(x_1-x_0,y_1-y_0,z_1-z_0\right)\text{then the equation is,}$

$r\left(t\right)=\left(x_0,y_0,z_0\right)+t\left(x_1-x_0,y_1-y_0,z_1-z_0\right)$

$r\left(t\right)=\left(x_0+\left(x_1-x_0\right)t,y_0+\left(y_1-y_0\right)t,z_0+\left(z_1-z_0\right)t\right)$

The equation is written as follows,

$r\left(t\right)=\left(\left(x_1-x_0\right)t+x_0,\left(y_1-y_0\right)t+y_0,\left(z_1-z_0\right)t+z_0\right)$

The equation of a line $\mathcal{L}$in the form of vector parametrization is,

$r\left(t\right)=\left(\left(x_1-x_0\right)t+x_0,\left(y_1-y_0\right)t+y_0,\left(z_1-z_0\right)t+z_0\right).$

$\text{the line containing the points}P\left(x_0,y_0,z_0\right)\text{and}Q\left(x_1,y_1,z_1\right)$

The equation of line$\mathcal{L}$in the form of vector parametrization is,

$r\left(t\right)=\left(\left(x_1-x_0\right)t+x_0,\left(y_1-y_0\right)t+y_0,\left(z_1-z_0\right)t+z_0\right)$

Here the range is restricted so that the line segment starts and ends at the given points.

The line segment starts at P and ends at Q.

Thus $t$is from 0 to 1 that is $0≤t≤1$.

$\text{Therefore, the answer is}r\left(t\right)=\left(\left(x_1-x_0\right)t+x_0,\left(y_1-y_0\right)t+y_0,\left(z_1-z_0\right)t+z_0\right);0≤t≤1\text{.}$