Question 364482
If two of three unit vectors are the same then they cannot  form a triangle.

|a| = sqrt(4+16+1) = sqrt(21)       <-length of a 
|b| = sqrt(9+4+4 ) = sqrt(17)       <-length of b
|c| = sqrt(25+4+9) = sqrt(38)       <-length of c


â = a/|a|     <-unit vector for a
   = [ 2, -4, -1  ]/ sqrt( 21) 
b̂  = b/|b|     <-unit vector for b
     = [ 3, 2, -2  ]/sqrt(17)
ĉ  = c/|c|     <-unit vector for c
    = [ 5, -2, -3  ]/ sqrt(38)


If two vector are the same then the  difference is zero
â -  b̂  = [ 2, -4, -1  ]/ sqrt( 21) - [ 3, 2, -2  ]/sqrt(17) ≠ [0 0 0]  <--they are not the same 
â -  ĉ = [ 2, -4, -1  ]/ sqrt( 21) - [ 5, -2, -3  ]/ sqrt(38) ≠ [0 0 0] <--they are not the same 
b̂ - ĉ  = [ 3, 2, -2  ]/sqrt(17) - [ 5, -2, -3  ]/ sqrt(38) ≠ [0 0 0] <--they are not the same

To control the assumption, let d = [ 2*2, -4*2, -1*2  ]    i.e. 2*a
d̂ = [ 2*2, -4*2, -1*2  ]  / sqrt( 16+64+4) = [ 2*2, -4*2, -1*2  ]  / sqrt(84)
â -  d̂  = [ 2, -4, -1  ]/ sqrt( 21) - [ 2*2, -4*2, -1*2  ]  / sqrt(84) = [0 0 0] → â and  d̂  are the same. 
→ We cannot form a triangle by these three :  a,b and d or  a,c and d


Showing that it is right-angled
|a| = sqrt(4+16+1) = sqrt(21)    length of a 
|b| = sqrt(9+4+4 ) = sqrt(17)    length of b
|c| = sqrt(25+4+9) = sqrt(38)    length of c

Since |c|^2 = |a|^2 + |b|^2      ( using Pythagoras) 
→ they forms a right-angled.