We have,

Let $AB$ be the line segment joining the points

$A(1,2,3)$ and $B(-3,4,-5)$.

Let $XY-plane$ divide line $AB$.at $P(x,y,z)$ in the ratio $k:1$.

Co- ordinate of point $P(x,y,z)$ that divide line segment joining point $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $m:n$ is

$=({\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},{\displaystyle \frac{m{y}_2+n{y}_1}{m+n}},{\displaystyle \frac{m{z}_2+n{z}_1}{m+n}})$

Here,

$m:n=k:1$

$A(x_1,y_1,z_1)=(1,2,3)$

$B(x_2,y_2,z_2)=(-3,4,-5)$

Then the co-ordinate of

$P(x,y,z)=({\displaystyle \frac{k\times (-3)+1\left(1\right)}{k+1}},{\displaystyle \frac{k\times \left(4\right)+1\times \left(2\right)}{k+1}},{\displaystyle \frac{k\times (-5)+1\times \left(3\right)}{k+1}})$

$=({\displaystyle \frac{-3k+1}{k+1}},{\displaystyle \frac{4k+2}{k+1}},{\displaystyle \frac{-5k+3}{k+1}})$

Since, point $P(x,y,z)$ lie on the $XY-plane$

Then, Its Z-coordinate will be zero.

$P(x,y,0)=({\displaystyle \frac{-3k+1}{k+1}},{\displaystyle \frac{4k+2}{k+1}},{\displaystyle \frac{-5k+3}{k+1}})$

Then, Comparing and we get,

${\displaystyle \frac{-5k+3}{k+1}}=0$

$-5k+3=0$

$-5k=-3$

$k={\displaystyle \frac{3}{5}}$

Then, $k:1=3:5$

We can substitute the value of k in (1), we get,

$=({\displaystyle \frac{-0.8}{1.6}},{\displaystyle \frac{4.4}{1.6}},0)$

$=(-0.5,2.75,0)$ 

We get the positive vector of the point of division from (2). Thus, the vector is given by $-0.5\widehat{i}+2.75\widehat{j}\mathrm{.}$ Here, $\widehat{i}$ denotes unit vector to X-axis and $\widehat{j}$ denotes unit vector to Y-axis.