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 2^nd 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