Question

Show that the lines `vecr = (hati + hatj - hatk) + lambda(2hati - 2hatj + hatk)` and `vecr = (4hati - 3hatj + 2hatk) + µ(hati - 2hatj + 2hatk)` intersect each other.

Answer 1

Let the two lines be

`vecr = (hati + hatj - hatk) + lambda(2hati - 2hatj + hatk)` 

`vecr = (4hati - 3hatj + 2hatk) + µ(hati - 2hatj + 2hatk)`

Step 1: Write in component form

Line 1:

`vecr = (1, 1, -1) + λ(2, -2, 1)`

⇒ `vecr = (1 + 2λ, 1 - 2λ, -1 + λ)`

Line 2:

`vecr = (4, -3, 2) + µ(1, -2, 2)`

⇒ `vecr = (4 + µ, -3 - 2µ, 2 + 2µ)`

Step 2: Equate components for intersection the position vectors must be equal

1 + 2λ = 4 + µ   ...(1)

1 – 2λ = –3 – 2µ   ...(2)

–1 + λ = 2 + 2µ   ...(3)

From (1):

µ = 2λ – 3

Substitute into (2):

1 – 2λ = –3 – 2(2λ – 3)

= –3 – 4λ + 6

= 3 – 4λ

1 – 2λ = 3 – 4λ

⇒ 1 + 2λ = 3

⇒ 2λ = 2

⇒ λ = 1

Then

µ = 2(1) – 3

= –1

Check in (3):

–1 + 1 = 0

2 + 2(–1) = 0

So the same λ, µ satisfy all three equations.

Step 3: Find the point of intersection

Put λ = 1 in Line 1:

(1 + 2, 1 – 2, –1 + 1) = (3, –1, 0)

Since a common point exists, the lines intersect and they intersect at `vecr = 3hati - hatj + 0hatk` i.e., (3, –1, 0).