Question

Show that the lines given by $\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect. Also find the coordinate solutions.

Answer 1

Compare the given lines with $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

condition for two lines to intersect is

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Now,

$\left|\begin{array}{ccc}-10-(-1)& -1-(-3)& 1-4\\ -10& -1& 1\\ -1& -3& 4\end{array}\right|=0$

$\left|\begin{array}{ccc}-9& 2& -3\\ -10& -1& 1\\ -1& -3& 4\end{array}\right|=0$

$9+78-87=0$

$0=0$

Hence the given lines intersect to each other.

Coordinate solutions $=(x,y,z)$

comparing x,yand z termsin both the lines,then

$\frac{x+1}{-10}=\frac{x+10}{-1}$ $-->$ $x=-11$

$\frac{y+1}{-1}=\frac{y+1}{-3}$ $-->$ $y=-4$

$\frac{z-4}{1}=\frac{z-1}{4}$ $-->$ $z=5$

Hence the coordinate solutions$=(x,y,z)=(-11,-4,5)$