Question

Find the shortest distance between the lines.

`(x + 1)/7 = (y + 1)/(- 6) = (z + 1)/1` and `(x - 3)/1 = (y - 5)/(- 2) = (z - 7)/1`.

Answer 1

Equation of one line is  `(x + 1)/7 = (y + 1)/(- 6) = (z + 1)/1`

Comparing with `(x - x_1)/a_1 = (y - y_1)/(b_1) = (z - z_1)/c_1`; we have

x₁ = −1, y₁ = −1, z₁ = −1; a₁ = 7, b₁ = −6, c₁ = 1

∴ vector form of this line is `vecr = vec (a_1) + lambda  vec (b_1)`

Where `vec (a_1) = (x_1, y_1, z_1) = (-1, -1, -1) = - hati -hatj -hatk`

and `vec (b_1) = a_1 hati + b_1hatj + c_1hatk = 7 hati - 6 hatj + hatk`

Equation of second line is `(x - 3)/1 = (y - 5)/(- 2) = (z - 7)/1`

Comparing with `(x - x_2)/a_2 = (y - y_2)/(b_2) = (z - z_2)/c_2`; we have

x₂ = 3, y₂ = 2, z₂ = 7; a₂ = 1, b₂ = −2, c₂ = 1

∴ vector form of this second line is  `vecr = vec (a_2) + mu vec(b_2)`

Where `vec (a_2)  = (x_2, y_2, z_2) = (3, 5, 7) = 3 hati + 5 hatj + 7 hat k`

and `vec b_2 = a_2 hati + b_2 hatj + c_2 hat k = hati - 2 hatj + hatk`

We know that S. D. between two skew lines is given by

`d = (|(vec(a_2) - vec(a_1)) . (vec(b_1) xx vec (b_2))|)/(|vec(b_1) xx vec(b_2)|)`    ....(i)

Now `vec(a_2) - vec(a_1) = 3 hati + 5 hatj + 7 hatk - (- hati - hatj - hatk)`

= `4 hati + 6 hatj + 8 hatk`

`vec(b_1) xx vec(b_2) = |(hati, hatj, hatk),(7, -6, 1),(1, -2, 1)|`

= `(- 6 + 2)hati - (7 - 1)hatj + (-14 + 6) hatk`

= `- 4 hati - 6 hatj - 8 hatk`

∴ `|vec(b_1) xx vec(b_2)| = sqrt((-4)^2 + (-6)^2 + (-8)^2)`

= `sqrt(16 + 36 + 64)`

= `sqrt116`

again `(vec(a_2) - vec(a_1)) . (vec(b_1) xx vec(b_2)) = 4 (-4) + 6 (-6) + 8 (-8)`

= − 16 − 36 − 64

= −116

Putting these values in eqn. (i),

S.D. (d) = `|-116|/sqrt116`

= `116/sqrt116`

= `sqrt116`

= `sqrt(4 xx 29)`

= `2 sqrt29` 

Since distance is always non-negative, the distance between the given lines is `2sqrt29` units.

Answer 2

Compare the given equations:

Computing it with `(x - x_1)/a_1 = (y - y_1)/b_1 = (z - z_1)/c_1` and `(x - x_2)/a_2 = (y - y_2)/b_2 = (z - z_2)/c_2`,

x₁ = −1, y₁ = −1, z = −1; x₂ = 3, y₂ = 5, z₂ = 7;

a₁ = 7, b₁ = −6, c₁ = 1 and a₂ = 1, b₂ = 2, c₂ = 1

Hence, D = `(a_1b_2 - a_2b_1)^2 + (b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2`

= `(-14 + 6)^2 + (-6 + 2)^2 + (1 - 7)^2`

= 64 + 16 + 36

= 116

∴ Minimum Distance = `1/sqrtD |(x_2 - x_1, y_2 - y_1, z_2 -z_1), (a_1, b_1, c_1), (a_2, b_2, c_2)|`

= `1/sqrt116 |(3 + 1, 5 + 1, 7 + 1), (7, -6, 1), (1, -2, 1)|`

= `1/(4sqrt29) |(4, 6, 8), (7, -6, 1), (1, -2, 1)|`

= `1/(4sqrt29) [4 (-6 + 2) -6(7 - 1) + 8(-14 + 6)]`

= `1/(4sqrt29) [-16 - 36 -64]`

= `-1/(2sqrt29). 116`

= `(4 xx 29)/(2sqrt29)`        ...(omitting the minus sign)

= `2sqrt29` इकाई