Question

Find the coordinates of the point where the line joining A(3, 4, 1) and B(5, 1, 6) crosses the xy-plane.

Answer 1

Let given point be $A(x_1,y_1,z_1)=(3,4,1)$ and $B(x_2,y_2,z_2)=(5,1,6)$.
The equation of a line in a space joining the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
$\Rightarrow \frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}$
$\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$
Since, xy plane z ordinate is zero,
Thus, this happens only if $\frac{z-1}{5}=-\frac{1}{5}$
$\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=-\frac{1}{5}$
$\Rightarrow \frac{x-3}{2}=-\frac{1}{5}$ and $\frac{y-4}{-3}=-\frac{1}{5}$
$\Rightarrow x-3=-\frac{2}{5}$ and $y-4=\frac{3}{5}$
$\Rightarrow x=\frac{-2}{5}+3$ and $y=\frac{3}{5}+4$
$\Rightarrow x=\frac{-2+15}{3}$ and $y=\frac{3+20}{4}$
$\therefore x=\frac{13}{3}$ and $y=\frac{23}{4}$
Hence, the required coordinate where the given line crossed the xy plane is $(\frac{13}{3},\frac{23}{4},0)$