If the lines `(x-1)/2=(y+1)/3=(z-1)/4 ` and `(x-3)/1=(y-k)/2=z/1` intersect each other then find value of k
Let `(x-1)/2=(y+1)/3=(z-1)/4 =u ` where is any constant.
So for any point on this line has co-ordinates in the form (2u+1,3u-1,4u+1)
So for any point on this line has co-ordinates in the form (v+3,2v+k,v).
Point of intersection of these two lines will have co-ordinates of the form
(2u +1, 3u −1,4u +1) and (v +3, 2v + k,v) .
Equating the x, y and z co-ordinates for both the forms we get three equations
Subtracting equation (1)from equation(3) we get,
2u = -3
Substitute value of u in equation (1) we get,
2(-3/2) - v=2
Substitute value of v and in equation (2) we get,
3(-3/2) - 2(-5)=k+1
the value of k is 9/2