Question

Verify that lines given by `vecr = (1 - λ)hati + (λ - 2)hatj + (3 - 2λ)hatk and vecr = (μ + 1)hati + (2μ - 1)hatj - (2μ + 1)hatk` are skew lines. Hence, find shortest distance between the lines.

Answer 1

Given: `vecr = (1 - λ)hati + (λ - 2)hatj + (3 - 2λ)hatk`

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

and `vecr = (μ + 1)hati + (2μ - 1)hatj - (2μ + 1)hatk`

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

Check for parallel lines.

Check `vecb_1 = λvecb_2`

`(-hati + hatj - 2hatk) ≠ λ(hati + 2hatj - 2hatk)`

Since no such value of λ exists, the lines are not parallel; therefore, we check whether the lines intersect or not.

1 − λ = μ + 1    ...(i)

λ − 2 = 2μ − 1    ...(ii)

3 − 2λ = −2μ − 1    ...(iii)

From equation (i),

1 − μ = μ + 1

μ = − μ

From equation (ii), substitute the value of λ, we get

− μ − 2 = 2μ − 1

− 1 = 3μ

μ = `-1/3`

λ = `1/3`

Substitute λ and u in equation (iii),

`3 - 2(1/3) = -2(-1/3) - 1`

`7/3 ≠ (-1)/3`

Since the third equation is not satisfied, the lines do not intersect. Hence, the lines are skew.

Shortest distance = `|((veca_2 - veca_1).(vecb_1 xx vecb_2))/(|vecb_1 xx vecb_2|)|`

Now, `veca_2 - veca_1 = (hati - hatj - hatk) - (hati - 2hatj + 3hatk)`

= `hatj - 4 hatk`

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

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

= `2hati - 4hatj - 3hatk`

`|vecb_1 xx vecb_2| = |2hati - 4hatj - 3hatk|`

= `sqrt(4 + 16 + 9)`

= `sqrt(29)`

S.D. = `|(hatj - 4hatk).(2hati - 4hatj - 3hatk)|/(sqrt29)`

= `(|0 - 4 + 12|)/(sqrt29)`

= `8/(sqrt29)` unit