Question

Prove that the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}$ intersects and find the point of intersection.

Answer 1

Assume a point on line 1 ans calculate $\lambda$ for which this point is also on line 2.

$P(2\lambda +1,3\lambda +2,4\lambda +3)$

Line equation of line 2 can be wriiten as

$\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=\mu$

Put $x=2\lambda +1,y=3\lambda +2,z=4\lambda +3$

$\frac{2\lambda +1-4}{5}=\frac{3\lambda +2-1}{2}=\frac{4\lambda +3}{1}$

$\frac{2\lambda -3}{5}=\frac{3\lambda +1}{2}=\frac{4\lambda +3}{1}$

Find value of $\lambda$ for which are 3 values equate.

if you can find it then it will be intersection point of lines otherwise no intersection point.

$2(2\lambda -3)=5(3\lambda +1)$

$4\lambda -6=15\lambda +5\to 11\lambda =-11$

$\lambda =-1$

All 3 equates on $\lambda =-1$

$P(2\lambda +1,3\lambda +2,4\lambda +3)=P(-1,-1,-1)\to$ Intersection point