Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that point $P$ outside plane $\alpha$. Plane $\beta$ is a plane that contained point $P$ and perpendicular to one of the line in plane $\alpha$. If line $l$ is the intersection line between plane $\alpha$ and plane $\beta$, then $d(P,\alpha) = d(P,l)$

share|improve this question
1  
A terminological comment: one does not prove problems. One may prove a theorem, or solve a problem. –  Gerry Myerson Feb 20 '13 at 0:08
    
edited.., thanks... :) –  chihiroasleaf Feb 20 '13 at 0:12
add comment

1 Answer

Consider the point $R$ on $\alpha$ minimizing $d(P,R)$. By definition, this is $d(P,\alpha)$, and a basic fact about distances between points and planes tells us that the segment $PR$ will be perpendicular to $\alpha$. Moreover, $R$ is unique, and so is the segment $PR$

The plane $\beta$ is uniquely defined by the point $P$ and the line $l$ where it intersects $\alpha$. There is a point $L$ on $l$ minimizing the distance $d(P,L)$. $L$ is unique, and it is the only point such that the segment $PL$ is perpendicular to $L$.

Because $\alpha$ and $\beta$ are perpendicular, any segment in $\beta$ perpendicular to $l$ (their intersection) must be perpendicular to alpha. Therefore $PL$ is perpendicular to $\alpha$.

This is enough to deduce that $PL$ and $PR$ are on the same line. Because $L$ and $R$ are both in $\alpha$, they must be the same. Therefore the lengths of $PL$ and $PR$ are equal, and by definition $d(P,\alpha) = d(P,l)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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