Vector Problem

|3→x+2→y|Formula: |→v|=√→v⋅→v =√(3→x+2→y)(3→x+2→y)=√9→x2+4→y2+2⋅3⋅2⋅(→x⋅→y)Formula: |→x2|=|→x|2=x2|→y2|=|→y|2=y2 =√9x2+4y2+12⋅(→x⋅→y)unit vectors: |→x|=x=1|→y|=y=1=√9+4+12⋅(→x⋅→y)Formula: →x⋅→y=|→x|⋅|→y|⋅cos(120∘)=1⋅1⋅(−0.5)=−0.5 =√9+4+12⋅(−0.5)=√13−6=√7
