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The following is the schematic diagram expressing the vectors and their relations.

We will assume that $ A $ is perpendicular to $ B $ .

Now, we will assume $ B $ along the west direction. We also assume $ B $ is perpendicular to $ C $ , such that $ A $ is along south direction and $ C $ is along north direction.

If $ A $ is perpendicular to $ B $ then their scalar product must be zero. This can be expressed as,

$ A \cdot B = \left| A \right|\,\left| B \right|\,\cos \theta $

Since, $ A $ and $ B $ are perpendicular, $ \theta $ is equal to $ {90^ \circ } $ . Substituting the value of $ \theta $ in the above expression, we get,

$ A \cdot B = 0 $

If $ B $ is perpendicular to $ C $ then their scalar product must be zero. This can be expressed as,

$ B \cdot C = \left| B \right|\,\left| C \right|\,\cos \theta $

Since, $ B $ and $ C $ are perpendicular, $ \theta $ is equal to $ {90^ \circ } $ . Substituting the value of $ \theta $ in the above expression, we get,

$ B \cdot C = 0 $

Hence we can say that $ A \cdot B = B \cdot C = 0 $

But $ A $ and $ C $ are in the direction south and north respectively, Hence $ A \ne C $ .

Hence, it Is proved.

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