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 each other. Find the co-ordinates of intersecting point also.

Answer 1

We know that, lines intersect if
$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a& b& c\\ a^{\prime }& b^{\prime }& c^{\prime }\end{array}\right|=0$
$\left|\begin{array}{ccc}2+1& 4+3& 6+5\\ 3& 5& 7\\ 1& 3& 5\end{array}\right|=0$
$3(25-12)-7(15-7)+11(9-5)=0$
$12-56+44=0$
$0=0$
Hence lines are intersecting
Now, Any point on line
$\frac{x+1}{3}=\frac{y+3}{5}\frac{z+5}{7}=r$
is $(3r-1,5r-3,7r-5)$
$\therefore$ It is intersecting point therefore, it must satisfy the equation of second line
$\frac{3r-1-2}{1}=\frac{5r-3-4}{3}=\frac{7r-5-6}{5}$
$r=\frac{1}{2}$
Therefore, the point of intersection is $(3\times \frac{1}{2},5\times \frac{1}{2},7\times \frac{1}{2}-5)$
$(\frac{1}{2},\frac{-1}{2},\frac{-3}{2})$