16.2: Pythagorean Theorem
- Page ID
- 23683
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Here is an analog of the Pythagorean theorems (Theorem 6.2.1 and Theorem 13..1) in spherical geometry.
Let \(\triangle_sABC\) be a spherical triangle with a right angle at \(C\). Set \(a=BC_s\), \(b=CA_s\), and \(c=AB_s\). Then
\(\cos c=\cos a \cdot \cos b.\)
- Proof
-
Since the angle at \(C\) is right, we can choose the coordinates in \(\mathbb{R}^3\) so that \(v_C\z=(0,0,1)\), \(v_A\) lies in the \(xz\)-plane, so \(v_A=(x_A,0,z_A)\), and \(v_B\) lies in \(yz\)-plane, so \(v_B=(0,y_B,z_B)\).
Applying, 16.2.3, we get that
\(\begin{aligned} z_A&=\langle v_C,v_A\rangle =\cos b, \\ z_B&=\langle v_C,v_B\rangle =\cos a.\end{aligned}\)
Applying, 16.2.1 and 16.2.3, we get that
\(\begin{aligned} \cos c &=\langle v_A,v_B\rangle= \\ &=x_A\cdot 0+0\cdot y_B+z_A\cdot z_B= \\ &=\cos b\cdot\cos a.\end{aligned}\)
In the proof, we will use the notion of the scalar product which we are about to discuss.
Let \(v_A=(x_A,y_A,z_A)\) and \(v_B=(x_B,y_B,z_B)\) denote the position vectors of points \(A\) and \(B\). The scalar product of the two vectors \(v_A\) and \(v_B\) in \(\mathbb{R}^3\) is defined as
\[\langle v_A,v_B\rangle := x_A\cdot x_B+y_A\cdot y_B+z_A\cdot z_B.\]
Assume both vectors \(v_A\) and \(v_B\) are nonzero; suppose that \(\phi\) denotes the angle measure between them. Then the scalar product can be expressed the following way:
\[\langle v_A,v_B\rangle=|v_A|\cdot|v_B|\cdot\cos\phi, \]
where
\(\begin{aligned} |v_A|&=\sqrt{x_A^2+y_A^2+z_A^2}, & |v_B|&=\sqrt{x_B^2+y_B^2+z_B^2}.\end{aligned}\)
Now, assume that the points \(A\) and \(B\) lie on the unit sphere \(\Sigma\) in \(\mathbb{R}^3\) centered at the origin. In this case \(|v_A|=|v_B|=1\). By 16.2.2 we get that
\[\cos AB_s=\langle v_A,v_B\rangle.\]
AShow that if \(\triangle_s ABC\) is a spherical triangle with a right angle at \(C\), and \(AC_s=BC_s=\dfrac{\pi}{4}\), then \(AB_s=\dfrac{\pi}{3}\).
- Hint
-
Applying the Pythagorean theorem, we get that
\(\cos AB_s = \cos AC_s \cdot \cos BC_s = \dfrac{1}{2}.\)
Therefore, \(AB_s = \dfrac{\pi}{3}.\)
Alternatively, look at the tessellation of a hemisphere on the picture; it is made from 12 copies of \(\triangle_s ABC\) and yet 4 equilateral spherical triangles. From the symmetry of this tessellation, it follows that \([AB]_s\) occupies \(\dfrac{1}{6}\) of the equator; that is, \(AB_s = \dfrac{\pi}{3}\).