Question

Determine the mutual positions of the line passing through (2,7,4) and (1,6,7) and the line x=1+t, y=z=t.

Answer 1

Equation of line passing through two points (a,b,c) and (p,q,r) is given by

$\frac{x-a}{a-p}=\frac{y-b}{q-b}=\frac{z-c}{r-c}$

Equation of line passing through (2,7,4) and (1,6,7) is given by

$\frac{x-1}{2-1}=\frac{y-6}{7-6}=\frac{z-7}{4-7}$

$\frac{x-1}{1}=\frac{y-6}{6}=\frac{z-7}{-3}$

Let , $\frac{x-1}{1}=\frac{y-6}{6}=\frac{z-7}{-3}=m$

Then, x= m+1, y= 6 m+6, z= -3 m +7

and another line is , x=1+t, y=z=t.

→ m+1 =1+t ,→ m=t   ................(1)

→ 6 m +6 =t → m = (t -6)/6  .....................(2)

→ -3 m +7=t → m= (7-t)/3............(3)

Equating (1) and (2), we get

→6 t+6=t

→5 t = -6

→ t = -6/5

m=t=-6/5

For different m and t we will get different lines and their point of intersection will be different.

Hence the two lines will intersect .