Answer :

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 (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is,

So,

The equation of line passing through points A(0, -1, -1) and B(4, 5, 1) is

Where, x_{1} = 0, y_{1} = -1 and z_{1} = -1

And, x_{2} = 4, y_{2} = 5 and z_{2} = 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 L_{1} is (4λ, 6λ – 1, 2λ – 1).

The equation of line passing through points C(3, 9, 4) and D(-4, 4, 4) is

Where, x_{1} = 3, y_{1} = 9 and z_{1} = 4

And, x_{2} = -4, y_{2} = 4 and z_{2} = 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 L_{2} 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.**

Rate this question :

Find the equationMathematics - Board Papers

Find the vector aMathematics - Board Papers

Write the vector Mathematics - Board Papers

Find the value ofMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers