Question

Give an example for which \[\vec{A} \cdot \vec{B} = \vec{C} \cdot \vec{B} \text{ but } \vec{A} \neq \vec{C}\].

Answer 1

To prove: 
\[\vec{A} \cdot \vec{B} = \vec{C} \cdot \vec{B,} \text{ but } \vec{A} \neq \vec{C}\]
Suppose that  
\[\vec{A}\]  is perpendicular to
\[\vec{B};\vec{B}\] is along the west direction.
Also, \[\vec{B}\] is perpendicular to `vecC ; vecA`
\[\vec{C}\] are along the south and north directions, respectively.

`vecA` is perpendicular to \[\vec{B}\], so there dot or scalar product is zero.
i.e .,

\[\vec{A} \cdot \vec{B} = \left| \vec{A} \right|\left| \vec{B} \right|\cos\theta = \left| \vec{A} \right|\left| \vec{B} \right|\cos90^\circ= 0\] 
\[\vec{B}\] is perpendicular to \[\vec{C}\],  so there dot or scalar product is zero.
i.e., \[\vec{C} \cdot \vec{B} = \left| \vec{C} \right|\left| \vec{B} \right|cos\theta = \left| \vec{C} \right|\left| \vec{B} \right|cos90\ = 0\]
\[\therefore \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{C,} but \vec{A} \neq \vec{C}\]

Hence, proved.