Question

Show that the lines vectors r = (i + 2j + 3k) + λ(2i + 3j + 4k) and r = (4i + j) + μ(5i + 2j + k) intersect. Also, find their point of intersection.

Answer 1

Given equations :

3. Shortest distance between two lines :

The shortest distance between the skew lines \(\vec r=\bar a_1+\lambda \bar{b_1}\) and

\(\vec r=\bar a_2+\lambda \bar {b_2}\) is given by,

For given lines,

= - 15 – 18 + 33

= 0

Therefore, the shortest distance between the given lines is

\(\therefore\) d = 0 units

As d = 0

Hence, the given lines not intersect each other.

Now, to find point of intersection, let us convert given vector equations into Cartesian equations.

For that substituting \(\vec r=\text x\hat i+y\hat j+z\hat k\) in given equations,

⇒ 4λ – 6 = 15λ + 5

⇒ 11λ = -11

⇒ λ = -1

Therefore, x₁ = 2(-1)+1 , y₁ = 3(-1)+2 , z₁ = 4(-1)+3
⇒ x₁ = -1 , y₁ = -1 , z₁ = -1

Hence point of intersection of given lines is (-1, -1, -1).