Question

Find the angle between the following pair of lines:

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

Answer 1

The standard symmetric form of a line is `(x - x_1)/a = (y - y_1)/b = (z - z_1)/c` where (a, b, c) is the direction vector.

Line 1: `(x - 2)/3 = (y + 5)/2 = (1 - z)/-6`

We need to rewrite the z-term:

`((1 - z)/-6) = ((-(z - 1))/-6) = (z - 1)/6`

So, the direction vector is `vec b_1` = (3, 2, 6)

Line 2: `(x - 7)/1 = y/2 = (6 - z)/-2`

Rewrite the z-term:

`(6 - z)/-2 = ((-(z - 6))/-2) = (z - 6)/2`

So, the direction vector is `vec b_2` = (1, 2, 2)

The angle θ between two lines with direction vectors `vec b_1 and vec b_2` is given by the formula:

cos θ = `|vec b_1 . vec b_2|/(|vec b_1| |vecb_2|)`

Find the dot product `(vec b_1 . vec b_2)`:

`vec b_1 . vec b_2` = (3)(1) + (2)(2) + (6)(2)

= 3 + 4 + 12

= 19

Find the magnitudes:

`|vec b_1| = sqrt(3^2 + 2^2 + 6^2)`

= `sqrt(9 + 4 + 36)`

= `sqrt49`

= 7

`|vec b_2| = sqrt(1^2 + 2^2 + 2^2)`

= `sqrt(1 + 4 + 4)`

= `sqrt9`

= 3

Substitute into the formula:

cos θ = `19/(7 xx 3)`

= `19/21`

θ = `cos^-1 (19/21)`