Question

Show that the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ intersect. Also find their point of intersection.

Answer 1

Let $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=p,\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=q$

General points on the lines are

$(3p-1,5p-3,7p-5)\&(p+2,3q+4,5q+6)$

If the lines intersect, then

$3p-1=q+2,5p-3=3q+4,7p-5=5q+6$

or, 3p - q = 3--- (1)

5p - 3q = 7 ---(2)

7p - 5q = 11 --- (3)

Solving equation (1) and (2), we get

$p=\frac{1}{2},q=\frac{-3}{2}$

Putting the values of p and q in equtaion (3)

$7.\frac{1}{2}-5.\frac{-3}{2}=11$.

Therefore, lines intersect.

Point of intersection of lines is: $(\frac{1}{2},-\frac{1}{2},\frac{-3}{2}).$