Line L is 3i-5j+5k + t(2i+j-k)
Point A has position vector i+2j+2k
Point B is on the line L at 5i-4j+4k
The line through A and B is perpendicular to L.
Points C and D are on line L and 4*sqrt(5) units from A.
Find the position vectors for C,D and the area of triangles ABC, ABD
I got the solutions to be 9i-2j+2k and i-6j+6k and I got the area to be 4*sqrt(21)
My problem is that I am not really happy with my method and I intuitively feel there is some nice simple way to do this. I also think that the question is looking for the use of the cross product in working out the area. I did not use it.