Question

Read the following passage and answer the questions given below.

Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines `vecr = λ(hati + 2hatj - hatk)` and `vecr = (3hati + 3hatj) + μ(2hati + hatj + hatk)` respectively.

Based on the above information, answer the following questions:

1. Find the shortest distance between the given lines.
2. Find the point at which the motorcycles may collide.

Answer 1

a. Given, lines are: `vecr = λ(hati + 2hatj - hatk)` and `vecr = (3hati + 3hatj) + μ(2hati + hatj + hatk)` 

We know that, shortest distance between the lines

`vecr_1 = veca + λb_1` and `vecr = veca_2 + λvecb_1` is d = `(|(veca_2 - veca_1).(vecb_1 xx vecb_2)|)/(|vecb_1 xx vecb_2|)`

Here, `veca_1 = 0, veca_2 = (3hati + 3hatj)`

`vecb_1 = hati + 2hatj - hatk`

and `vecb_2 = 2hati + hatj + hatk`

∴ `veca_2 - veca_1 = (3hati + 3hatj) - 0 = 3hati + 3hatj`

`vecb_1 xx vecb_2 = |(hati, hatj, hatk),(1, 2, -1),(2, 1, 1)|`

= `hati(2 + 1) - hatj(1 + 2) + hatk(1 - 4)`

= `3hati - 3hatj - 3hatk`

and `|vecb_1 xx vecb_2| = sqrt(3^2 + (-3)^2 + (-3)^2)`

= `sqrt(9 + 9 + 9)`

= `3sqrt(3)`

Also, `(veca_2 - veca_1).(vecb_1 xx vecb_2) = (3hati + 3hatj).(3hati - 3hatj - 3hatk)`

= 9 – 9

= 0

d = `0/(3sqrt(3))` = 0

Thus, distance between lines is 0.

b. We have, `vecr = λ(hati + 2hatj - hatk)`  ...(i)

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

or `vecr = (3 + 2μ)hati + (3 + μ)hatj + μhatk`  ...(ii)

Now, from equation (i) and equation (ii), we get

`λ(hati + 2hatj - hatk) = (3 + 2μ)hati + (3 + μ)hatj + μhatk`

On comparing both sides, we get

3 + 2µ = λ, 3 + µ = 2λ and µ = –λ

On solving for values of λ and µ, we get

λ = 1 and µ = –1

From equation (i), we get `vecr = hati + 2hatj - hatk`

`xhati + yhatj + zhatk = hati + 2hatj - hatk`

So, required point is (1, 2, –1).