Question

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

Answer 1

It is known that equation of ZX-plane is y = 0.
The equation of the line passing through the point $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)is\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
The equation of the line joining (5, 1, 6) and (3, 4, 1) is
$\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda \left(say\right)$
Any point on this line is $(5-2\lambda ,1+3\lambda ,6-5\lambda )$ . . . (i)
Since, the line passes through ZX-plane is (y = 0 ) form Eq. (i)
$1+3\lambda =0\Rightarrow \lambda =-\frac{1}{3}$
$\therefore$ Required point of intersection of planes (i) and (ii) is
$(5+\frac{2}{3},0,6+\frac{5}{3})\Rightarrow (\frac{17}{3},0,\frac{23}{3})$