Given: – Two lines having vector notion and

The position vectors of arbitrary points on the given lines are

1^st line

2^nd line

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, 1 – λ = 0, – 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) – 2(0) = 3

⇒ 3 = 3

⇒ LHS = RHS ; Hence intersection point exists or line do intersect

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

Thus,

Intersection point

⇒

⇒

⇒

Hence, Intersection point is (4,0, – 1)