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If $B \ne C$, prove that the perpendicular distance from $A$ to the line through $B$ and $C$ is $$\dfrac {|| (A-B)\times(C-B)||}{||B-C||} $$ where $\times$ means the vector cross product.

Attempt:

Let the perpendicular from point $A$ meet the vector joining $B$ and $C$ at the point $P$.

Let $X$ be the unit vector along $AP$. Then : $||AP|| = ||(B-A) \cdot X|| = ||(C-A) \cdot X ||$

I don't have much idea on how to proceed from here.

Could anyone please give me a direction on how to move ahead.

Thank you very much for your help.

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Hint: Use that $$||{\bf X} \times {\bf Y}|| = ||{\bf X}||\,||{\bf Y}|| \sin \theta,$$ where $\theta$ is the angle between ${\bf X}$ and ${\bf Y}$.

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  • $\begingroup$ Thank you for the hint. I got it. $\endgroup$ – MathMan Jan 30 '15 at 12:21
  • $\begingroup$ You're welcome, I'm glad it was useful. $\endgroup$ – Travis Jan 30 '15 at 12:22

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