Prove that the line through A (0, –1, –1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (– 4, 4, 4).
Given: Points A(0, -1, -1), B(4, 5, 1), C(3, 9, 4) and D(-4, 4, 4)
To Prove: Line through A and B intersects the line through C and D.
Proof: We know that, the equation of a line passing through two points (x1, y1, z1) and (x2, y2, z2) is,
So,
The equation of line passing through points A(0, -1, -1) and B(4, 5, 1) is
Where, x1 = 0, y1 = -1 and z1 = -1
And, x2 = 4, y2 = 5 and z2 = 1
Let
We need to find the value of x, y and z. So,
Take 
.
⇒ x = 4λ
Take 
.
⇒ y + 1 = 6λ
⇒ y = 6λ – 1
Take 
⇒ z + 1 = 2λ
⇒ z = 2λ – 1
This means, any point on the line L1 is (4λ, 6λ – 1, 2λ – 1).
The equation of line passing through points C(3, 9, 4) and D(-4, 4, 4) is
Where, x1 = 3, y1 = 9 and z1 = 4
And, x2 = -4, y2 = 4 and z2 = 4
Let
We need to find the value of x, y and z. So,
Take 
.
⇒ x – 3 = -7μ
⇒ x = -7μ + 3
Take 
.
⇒ y – 9 = -5μ
⇒ y = -5μ + 9
Take 
.
⇒ z – 4 = 0
⇒ z = 4
This means, any point on the line L2 is (-7μ + 3, -5μ + 9, 4).
If lines intersect then there exist a value of λ, μ for which
(4λ, 6λ – 1, 2λ – 1) ≡ (-7μ + 3, -5μ + 9, 4)
⇒ 4λ = -7μ + 3 …(i)
6λ – 1 = -5μ + 9 …(ii)
2λ – 1 = 4 …(iii)
From equation (iii), we get
2λ – 1 = 4
⇒ 2λ = 4 + 1
⇒ 2λ = 5
Putting the value of λ in equation (i), we get
⇒ 2 × 5 = -7μ + 3
⇒ 10 = -7μ + 3
⇒ 7μ = 3 – 10
⇒ 7μ = -7
⇒ μ = -1
Putting the values of λ and μ in equation (ii), we get
⇒ 3 × 5 – 1 = 5 + 9
⇒ 15 – 1 = 14
⇒ 14 = 14
Since, these values of λ and μ satisfy equation (ii), this implies that the lines intersect.
Hence, the lines through A and B intersects line through C and D.