Question

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

Answer 1

It is known that the equation of the line passing through the points, (x1,y1,z1) and (x2,y2,z2), is x-$\frac{x_1}{x_2}$-x1=y-$\frac{y_1}{y_2}$-y1=z-$\frac{z_1}{z_2}$-z1

The line passing through the points, (5,1,6), and (3,4,1) is given by,

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

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

⇒x=5-2k,y=3k+1,z=6-5k

Any point on the line is of the form (5-2k,3k+1,6-5k).

The equation of YZ-plane is x=0

Since the line passing through YZ-plane,
5-2k=0
⇒k=$\frac{5}{2}$
⇒3k+1
=3×$\frac{5}{2}$+1
=$\frac{17}{2}$ 6-5k
=6-5×$\frac{5}{2}$
=$\frac {-13}{2}$

Therefore, the required point is (0, 17/2, -13/2).