Let L1 be the line passing through the points Q1(5, −3, 2) and Q2(−1, −12, −7) and let L2 be the line passing through the point P1(1, −19, −6) with direction vector d=[−2, 2, −2]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.

Therefore the lines L₁ and L₂ intersect each other.

The co-ordinate of the intersection point of lines L₁ and L₂ is (-3,-15,-10).

Step-by-step explanation:

The equation of line passes through the points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is

Given points are Q₁(5,-3,2) and Q₂(-1,-12,-7).

x₁= 5,y₁= -3,z₁=2 ,x₂= -1,y₂ = -12,z₂= -7

The equation of line is

.....(1)

Any point on the line is A(2λ+5,3λ-3,3λ+2)

[, similarly and  ]

The equation of line which passes through a point (x₁,y₁,z₁)  with direction ratio (x₂,y₂,z₂) is

The equation of line L₂ which passes through the point P₁(1,-19,-6) with direction vector d=[-2,2,-2] is

Any point on the line L₂ is B(-μ+1,μ-19,-μ-6)

Let the lines L₁ and L₂ intersect.

Consider the lines L₁ and L₂ intersect at A and B points.

Since there is only one intersect point. We compare the point A and B.

∴2λ+5=-μ+1

⇒2λ+μ+4=0 .......(2)

3λ-3=μ-19

⇒3λ-μ+16=0......(3)

and  

3λ+2=-μ-6

⇒3λ+μ+8=0....(4)

Adding (2) and (3) we get

2λ+μ+4+3λ-μ+16=0

⇒5λ+20=0

⇒5λ= - 20

Putting the value of λ in equation (2)

2(-4)+μ+4=0

⇒ -8+μ+4=0

⇒μ=4

Putting the values of μ and λ in equation (4)

3(-4)+4+8=0

⇒-12+12=0

The value of μ and λ satisfy the equation (4).

Therefore the line L₁ and L₂ intersect each other.

Therefore the co-ordinate of A is

=(2(-4)+5,3(-4)-3,3(-4)+2)  [ Putting the value of  λ]

=(-3,-15,-10)

The co-ordinate of the intersection point is (-3,-15,-10).