Question

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

Answer 1

We have, $x_1=1,y_1=2,z_1=3,and{a}_1=2,b_1=3,c_1=4,$
Also, $x_2=4,y_2=1,z_2=0,and{a}_2=5,b_2=2,c_2=1$
If two lines intersect, then shortest distance between them should be zero.
$\therefore$ Shortest distance between two given lines
$\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}}$ = $\frac{\left|\begin{array}{ccc}4-1& 1-2& 0-3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|}{\sqrt{(3.1-2.4{)}^2+(4.5-1.2{)}^2+(2.2-5.3{)}^2}}$ = $\frac{\left|\begin{array}{ccc}3& -1& -3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|}{\sqrt{25+324+121}}$ $=\frac{3(3-8)+1(2-20)-3(4-15)}{\sqrt{470}}$ $=\frac{-15-18+33}{\sqrt{470}}=\frac{0}{\sqrt{470}}=0$

Therefore, the given two lines are intersecting.

For finding their point of intersection for first line. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

$\Rightarrow x=2\lambda +1,y=3\lambda +2andz=4\lambda +3$

Since, the lines are intersecting. So, let us put these values in the equation of another line.

Thus, $\frac{2\lambda +1-4}{5}=\frac{3\lambda +2-1}{2}=\frac{4\lambda +3}{1}\Rightarrow \frac{2\lambda -3}{5}=\frac{3\lambda +1}{2}=\frac{4\lambda +3}{-1}\Rightarrow \frac{2\lambda -3}{5}=\frac{4\lambda +3}{1}\Rightarrow 2\lambda -3=20\lambda +15\Rightarrow 18\lambda =-18=-1$

So, the required point of intersection is x=2(-1)+1=-1
y=3(-1)+2=-1
z=4(-1)+3=-1
Thus, the lines intersect at (-1,-1,-1).