91影视

$\text{Consider the distinct points}P=(a,b,c);Q=(伪,尾,纬)\text{in}{\mathrm{鈩�}}^3\text{.}$

First we will find the direction vector for the line $\stackrel{鈫�}{QP}$

The points are $Q=(伪,尾,纬)$and $P=(a,b,c)$

$\stackrel{鈫�}{QP}=(a-伪,b-尾,c-纬)$

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

$r\left(t\right)=P_0+td$Where, $P_0$is the point and $d$is the direction vector.

Here$Q=(伪,尾,纬)$and $\stackrel{鈫�}{QP}=d=(a-伪,b-尾,c-纬)$then the equation is,

$r\left(t\right)=(伪,尾,纬)+t(a-伪,b-尾,c-纬)$

$r\left(t\right)=(伪+(a-伪)t,尾(b-尾)t,纬+(c-纬\left)t\right)$

The equation is written as follows,

$r\left(t\right)=(伪+(a-伪)t,尾(b-尾)t,纬+(c-纬\left)t\right)$

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

The line segment starts at $Q$and ends at $P$

Thus $t$is from $0$to $1$that is $0鈮�t鈮�1$.

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

$r\left(t\right)=(伪+(a-伪)t,尾(b-尾)t,纬+(c-纬\left)t\right)\text{where}0鈮�t鈮�1$

$\text{Therefore, the required equation is}r\left(t\right)=(伪+(a-伪)t,尾(b-尾)t,纬+(c-纬\left)t\right);0鈮�t鈮�1\text{.}$