Let the three sides of a triangle ABC be given by the vectors 2 i- j+ k, i-3 j-5 k and 3 i-4 j-4 k. Let G be the centroid of the triangle ABC. Then 6 (| AG|2+| BG|2+| CG|2 ) is equal to .
Answer
164
Solution
Squares of the sides: |2 i- j+ k|2=6, | i-3 j-5 k|2=35, |3 i-4 j-4 k|2=41
(The first two add to the third, so they do form a triangle.)
Take G as origin with position vectors a, b, c of A, B, C: a+ b+ c= 0
| a+ b+ c|2=0 2( a b+ b c+ c a)=-(| a|2+| b|2+| c|2)
Sum of squares of sides =| a- b|2+| b- c|2+| c- a|2
=2(| a|2+| b|2+| c|2)-2( a b+ b c+ c a)=3(AG2+BG2+CG2)
AG2+BG2+CG2= 6+35+413= 823
6 (AG2+BG2+CG2 )=6 823=164
Answer: 164