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, (x_1,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).