Answer :

Given: - Two lines equation: and


We have,



x = 3λ + 1, y = – λ + 1 and z = – 1


So, the coordinates of a general point on this line are


(3λ + 1, – λ + 1, – 1)


The equation of the 2nd line is



x = 2μ + 4, y = 0 and z = 3μ – 1


So, the coordinates of a general point on this line are


(2μ + 4, 0, 3μ – 1)


If the lines intersect, then they must have a common point.


Therefore for some value of λ and μ, we have


3λ + 1 = 2μ + 4 , – λ + 1 = 0, and – 1 = 3μ – 1


3λ – 2μ = 3 ……(i)


λ = 1 ……(ii)


and μ = 0 ……(iii)


from eq ii and eq iii, we get


λ = 1


and μ = 0


As we can see by putting the value of λ and μ in eq i, that it satisfy the equation.


Check


3λ – 2μ = 3


3(1) = 3


3 = 3


LHS = RHS ;Hence intersection point exist or line do intersects


We can find intersecting point by putting values of μ or λ in any one general point equation


Thus,


Intersection point


2μ + 4, 0, 3μ – 1


4, 0, – 1


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