Can you please help me with this:
1. Find the fourth vertex of the parallelogram:
A(2,1,0) B(3,2,-1) C(0,-1,1)
2. Find the shortest distance between point A (3,5,6) and line (x-7)/2=y-5=(z-12)/6
Thanks in advance...
AD=BC=(0,-1,1)-(3,2,-1)=(-3,-3,2) so D=A+(-3,-3,2)=(-1,-2,2)
2.eequation of line=(7,5,12)+t(2,1,6). Suppose they intersect at (a,b,c) then (a-3,b-5,c-6) is at right angles to (2,1,6) but (a,b,c) is on line so (a,b,c)= (7+2t,5+t,12+6t)
and (4+2t,t,6+6t).(2,1,6)=0 ie 44+41t=o so t=-44/41 sthe sub into (7+2t,5+t,12+6t) to find a and use pythagoras theorem in 3D to find distance between A and (a,b,c)