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.


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.


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.