Since #vec{a}# and #vec{b}# are unit vectors,
#|vec{a}|=|vec{b}|=1#,
and since the angle between them is #60^circ#,
#vec{a} cdot vec{b}=|vec{a}||vec{b}|cos(60^circ)=1cdot1cdot1/2=1/2#.
Now, let us compute:
#(6 vec{a}+ vec{b})cdot(vec{a}-2vec{b})#
by multiplying out,
#=6vec{a}cdot vec{a}-12vec{a}cdot vec{b}+vec{b}cdot vec{a}-2vec{b}cdot vec{b}#
by #vec{a}cdot vec{a}=|vec{a}|^2#, #vec{b}cdot vec{b}=|vec{b}|^2#, and #vec{a}cdot vec{b}=vec{b}cdot vec{a}#,
#=6|vec{a}|^2-11vec{a}cdot vec{b}-2|vec{b}|^2#
by plugging in the values we found above,
#=6(1)^2-11(1/2)-2(1)^2=-3/2#
I hope that this was helpful.
