Question

How do you find the parametric equation of a line in R3 that passes through P(1,1,1) and is orthogonal to both the lines x=3-t, y= -1+4t, z = -2t and x=2+2t, y=1+t, z=-2+3t where t is any real number?

Answer 1

The answer is `x=1+14s ; y=1-s ;z=1-9s` ; `s in RR`

Explanation:

To find a line orthogonal to 2 other lines, we must perfom a cross product

The first line `L1` is

`x=3-t`

`y=-1+4t`

`z=-2t`

The vector parallel to line `L1` is `vecL_1=〈-1,4,-2〉`

The second line `L2` is

`x=2+2t`

`y=1+t`

`z=-2+3t`

The vector parallel to line `L2` is `vecL_2=〈2,1,3〉`

The vector orthogonal to `vecL_1` and `vecL_2=` is given by the cross product

`| (hati,hatj,hatk), (-1,4,-2), (2,1,3) |`

`=hati| (4,-2), (1,3) | -hatj| (-1,-2), (2,3) | +hatk| (-1,4), (2,1) |`

`=hati(12+2)-hatj(-3+4)+hatk(-1-8)`

`=〈14,-1,-9〉`

Verification by doing a dot product

`〈14,-1,-9〉.〈-1,4,-2〉=-14-4+18=0`

`〈14,-1,-9〉.〈2,1,3〉=28-1-27=0`

The equation of the line orthogonal to `L_1` and `L_2` and passing through `P(1,1,1)` is

`x=1+14s`

`y=1-s`

`z=1-9s`