Question

Let ( vec{c}=3(vec{A}-vec{B}) )
( vec{c} cdot vec{B}=3(vec{A}-vec{B}) cdot vec{B}=3(vec{A} cdot vec{B}-vec{B} cdot vec{B}) )
( because vec{A} perp vec{B}, vec{A} cdot vec{B}=A B(0) 90^{prime}=0 )
( vec{c} cdot vec{B}=-3 B^{2} )
( Rightarrow|vec{a}| vec{B} mid cos theta=-3 B^{2} )
( Rightarrow quad 3|vec{A}-vec{B}||vec{B}| cos theta=-3 B^{2} )
( Rightarrow 3 sqrt{A^{2}+B^{2}} cdot B G D O=-3 B^{2} quad|cdot| vec{A}|=| vec{B} mid )
( leqslant) sqrt{2} B^{2} cdot B cos theta=-B^{2} )
( Rightarrow quad cos theta=-frac{1}{sqrt{2}} )
( theta=-45^{circ} )

# If A=1B and A I B , then find the angle between 3(A-B) with B.

Solution