Given: - Two lines equation: and

To find: - Intersection point

We have,

⇒ x = λ, y = 2λ + 2 and z = 3λ – 3

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

(λ, 2λ + 2, 3λ – 3)

The equation of the 2^nd line is

⇒ x = 2μ + 2, y = 3μ + 6 and z = 4μ + 3

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

(2μ + 2, 3μ + 6, 4μ + 3)

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

Therefore for some value of λ and μ, we have

⇒ λ = 2μ + 2 , 2λ + 2 = 3μ + 6, and 3λ – 3 = 4μ + 3

⇒ λ = 2μ + 2 ……(i)

⇒ 2λ – 3μ = 4 ……(ii)

and 3λ – 4μ = 6 ……(iii)

putting value of λ from eq i in eq ii, we get

⇒ 2(2μ + 2) – 3μ = 4

⇒ 4μ + 4 – 3μ = 4

⇒ μ = 0

Now putting value of μ in eq i, we get

⇒ λ = 2μ + 2

⇒ λ = 2(0) + 2

⇒ λ = 2

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

Check

⇒ 3λ – 4μ = 6

⇒ 3(2) = 6 ;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

λ, 2λ + 2, 3λ – 3

⇒ 2, 6, 3