5.5: Perpendicular is shortest
- Page ID
- 23609
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Assume \(Q\) is the foot point of \(P\) on the line \(\ell\). Then the inequality
\(PX > PQ\)
holds for any point \(X\) on \(\ell\) distinct from \(Q\).
If \(P, Q\), and \(\ell\) are as above, then \(PQ\) is called the distance from \(P\) to \(\ell\).
- Proof
-
If \(P \in \ell\), then the result follows since \(PQ = 0\). Further we assume that \(P \not\in \ell\).
Let \(P'\) be the reflection of \(P\) across the line \(\ell\). Note that \(Q\) is the midpoint of \([PP']\) and \(\ell\) is the perpendicular bisector of \([PP']\). Therefore
\(PX = P'X\) and \(PQ = P'Q = \dfrac{1}{2} \cdot PP'\)
Note that \(\ell\) meets \([PP']\) only at the point \(Q\). Therefore, \(X \not\in [PP']\); by triangle inequality and Corollary 4.4.1,
\(PX + P'X > PP'\)
and hence the result: \(PX > PQ\).
Assume \(\angle ABC\) is right or obtuse. Show that
\(AC > AB.\)
- Hint
-
If \(\angle ABC\) is right, the statement follows from Lemma \(\PageIndex{1}\). Therefore, we can assume that \(\angle ABC\) is obtuse.
Draw a line \((BD)\) perpendicular to \((BA)\). Since \(\angle ABC\) is obtuse, the angles \(DBA\) and \(DBC\) have opposite signs.
By Corollary 3.4.1, \(A\) and \(C\) lies on opposite sides of \((BD)\). In particular, \([AC]\) intersects \((BD)\) at a point; denote it by \(X\).
Note that \(AX < AC\) and by Lemma \(\PageIndex{1}\), \(AB \le AX\).
Suppose that \(\triangle ABC\) has right angle at \(C\). Show that for any \(X \in [AC]\) the distance from \(X\) to \((AB)\) is smaller than \(AB\).
- Hint
-
Let \(Y\) be the foot point of \(X\) on \((AB)\). Apply Lemma \(\PageIndex{1}\) to show that \(XY < AX \le AC < AB\).