Prove that the li

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.

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