Question

Prove that the line through A(0,-1,-1) and B(4,5,1) intersects the line through C(3,9,4) and D(-4,4,4).

Answer 1

We know that, the cartesian equation of a line that passesthrough two points $(x_1,y_1,z_1)$ and $(x_2,y_2{z}_2)$ is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Hence, the cartesian equation of line passes through A(0,-1,-1) and B(4,5,1)is
$\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}\Rightarrow \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}$ ...(i)
and cartesian equation of the line passes through C(3,9,4) and D(-4,4,4) is
$\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}\Rightarrow \frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}$ ...(ii)
If the lines intersect, then shortest distance between both of them should be zero.
$\therefore$ Shorttest distance between the lines $=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}}$ $=\frac{\left|\begin{array}{ccc}3-0& 9+1& 4+1\\ 4& 6& 2\\ -7& -5& 0\end{array}\right|}{\sqrt{(6.0+10{)}^2+(-14-0{)}^2+(-20+42{)}^2}}$ $=\frac{\left|\begin{array}{ccc}3& 10& 5\\ 4& 6& 2\\ -7& -5& 0\end{array}\right|}{\sqrt{100+196+484}}$ $=\frac{3(0+10)-10\left(14\right)+5(-20+42)}{\sqrt{780}}=\frac{30-140+110}{\sqrt{780}}=0$
So, the given lines intersect