Question

Find the coordinates of the point where the line through (5,1,6) and (3,4,1) crosses the YZ -plane.

Answer 1

We know that the line passing through two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

So, the equation of line passing through (5,1,6) and (3,4,1) is

$\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$

$\Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k\left(say\right)$

Any point on this line is
$x=-2k+5,y=3k+1,z=-5k+6$

The line crosses YZ -plane when x = 0

$\Rightarrow -2k+5=0$

$\Rightarrow k=\frac{5}{2}$

So, the required point is

$x=-2\left(\frac{5}{2}\right)+5$,

$y=3\left(\frac{5}{2}\right)+1$,

$z=-5\left(\frac{5}{2}\right)+6$

$\Rightarrow x=0,y=\frac{17}{2},z=-\frac{13}{2}$

Hence, the required coordinates of the point are $(0,\frac{17}{2},\frac{-13}{2})$