Answer:
Step-by-step explanation:
[tex]\frac{x-1}{2} =t\\\frac{y-2}{3} =t\\\frac{z-3}{4} =t\\so~eq.~of~line~L_{1}~is\\\frac{x-1}{2} =\frac{y-2}{3} =\frac{z-3}{4} \\its~d.r's~are~2,3,4\\again~\frac{x-2}{1} =s\\\frac{y-4}{2} =s\\\frac{z+1}{-4} =s\\so~eq. ~of~line~L_{2}~is\\\frac{x-2}{1} =\frac{y-4}{2} =\frac{z+1}{-4} \\its~d.r's ~are~1,2,-4\\let ~the ~d.r's~of~line~perpendicular~to~both~L_{1}~and~L_{2}~be~a,b,c,~then~\\2a+3b+4c=0\\1a+2b-4c=0\\solving\\\frac{a}{3*-4-4*2} =\frac{b}{4*1-2*-4} =\frac{c}{2*2-3*1} \\[/tex]
[tex]\frac{a}{-20} =\frac{b}{-4} =\frac{c}{1} \\d.r's~of ~reqd~line~is~-20,-4,1~or~20,4,-1[/tex]
now you find the point of intersection.
then calculate the angle.
