Question

If line 1 has parametric equations `x=s, y=2-s, z=-2+s, s inRR`. Line 2 has parametric eqns `x=1+3t, y=-2-2t, z=6+2t, t in RR`. Can you find shortest distance between the points? And, the coord of points where the common perpendicular meets the lines?

Answer 1

See below.

Explanation:

Given

`L_1->p_1 = p_(01) + lambda_1 vec v_1` and
`L_2->p_2 = p_(02) + lambda_2 vec v_2`

with `p_1 = (x_1,y_1)` and `p_2 = (x_2,y_2)` the distance between two generic points `p_1` and `p_2` is given by

`d(lambda_1,lambda_2)=norm(p_1-p_2) = norm(p_(01)-p_(02)+lambda_1 vec v_1-lambda_2 vec v_2)` or

`d^2(lambda_1,lambda_2) = norm(p_(01)-p_(02))^2+lambda_1^2norm(vec v_1)^2+lambda_2^2norm(vec v_2)^2+2lambda_1 << p_(01)-p_(02), vec v_1 >> - 2lambda_2 << p_(01)-p_(02), vec v_2 >>-2lambda_1 lambda_2 << vec v_1, vec v_2 >>`

Now the condition for minimum in `d^2(lambda_1,lambda_2)` which is a continuous `C^2` function, is

`{((partial d^2(lambda_1,lambda_2))/(partial lambda_1) = 2lambda_1 norm(vec v_1)^2+2 << p_(01)-p_(02), vec v_1 >> -2lambda_2 << vec v_1, vec v_2 >> =0),((partial d^2(lambda_1,lambda_2))/(partial lambda_2) = 2lambda_2 norm(vec v_2)^2-2 << p_(01)-p_(02), vec v_2>> -2lambda_1 << vec v_1, vec v_2 >> =0):}`

solving now the system

`{(lambda_1 norm(vec v_1)^2+ << p_(01)-p_(02), vec v_1 >> -lambda_2 << vec v_1, vec v_2 >> =0),(lambda_2 norm(vec v_2)^2- << p_(01)-p_(02), vec v_2>> -lambda_1 << vec v_1, vec v_2 >> =0):}`

or

`((norm(vec v_1)^2, - << vec v_1, vec v_2 >>),(- << vec v_1, vec v_2 >>, norm(vec v_2)^2))((lambda_1),(lambda_2)) = ((- << p_(01)-p_(02), vec v_1 >>),(<< p_(01)-p_(02), vec v_2 >>) )`

Now with

`p_01 = (0, 2, -2)`
`p_02 = (1, -2, 6)`
`vec v_1 = (1, -1, 1)`
`vec v_2 = (3, -2, 2)`

or

`((3, -7),(-7, 17))((lambda_1),(lambda_2))=((13), (-27))`

giving

`lambda_1 = 16, lambda_2 = 5` so the distance is

`d(16,5) = 2 sqrt 2` and the points at which this condition is verified are

`p_1 = (16, -14, 14)`
`p_2 = (16, -12, 16)`