Since a,b,c are unit vectors:
∣a∣2=∣b∣2=∣c∣2=1
The given equation a+2b−3c=0 can be rearranged in three ways and squared to find each dot product.
a+2b=3c
Taking the dot product of each side with itself:
∣a∣2+4(a⋅b)+4∣b∣2=9∣c∣2
1+4(a⋅b)+4=9
a⋅b=1⋯(i)
a=3c−2b
Taking the dot product of each side with itself:
∣a∣2=9∣c∣2−12(b⋅c)+4∣b∣2
1=9−12(b⋅c)+4
12(b⋅c)=12
b⋅c=1⋯(ii)
2b=3c−a
Taking the dot product of each side with itself:
4∣b∣2=9∣c∣2−6(c⋅a)+∣a∣2
4=9−6(c⋅a)+1
6(c⋅a)=6
c⋅a=1⋯(iii)
Substituting (i), (ii), and (iii):
2a⋅b−6b⋅c−3c⋅a
=2(1)−6(1)−3(1)
=2−6−3
=−7