How can I prove that the dihedral angles of the regular tetrahedral structure is $109.5^\circ$ with calculus or any other technique. I could do it with geometry by considering a cube.
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$\begingroup$ Does the dot product count as "calculus"? $\endgroup$– Andrew D. HwangFeb 4, 2014 at 13:30
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$\begingroup$ Yes, sure...actually I was looking for new methods. $\endgroup$– HawkFeb 4, 2014 at 13:32
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$\begingroup$ Yes, Certainly...thank you for pointing out! $\endgroup$– HawkFeb 4, 2014 at 13:40
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1$\begingroup$ angle subtended by the vertices of a regular tetrahedral structure to the centre $\endgroup$– HawkFeb 4, 2014 at 14:05
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1$\begingroup$ I changed $109.5^o$ to $109.5^\circ.$ That is standard usage. $\qquad$ $\endgroup$– Michael HardyJul 26, 2017 at 21:13
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4 Answers
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Four unit vectors pointing outward from the origin have equal angles between them.
By symmetry, their average, and hence their sum, is $0.$
Let a coordinate axis go out from the origin in the direction of one of those four vectors.
The component of that one vector on that axis is $1.$
Therefore sum of the components, on that axis, of the other three, must be $-1.$
Since there are three of them, the component of each on that axis is $-1/3.$
So the cosine of the angle is $-1/3.$
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6$\begingroup$ Or a more algebraic version of this answer: If the center of the tetrahedron is at the origin and the four vertices are at unit vectors $\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{t}$, then $\mathbf{x} + \mathbf{y} + \mathbf{z} + \mathbf{t} = \mathbf{0}$, and taking dot products we get $\mathbf{x} \cdot \mathbf{x} + \mathbf{x} \cdot \mathbf{y} + \mathbf{x} \cdot \mathbf{z} + \mathbf{x} \cdot \mathbf{t} = 0$, so $1 + 3\cos(\theta) = 0$. $\endgroup$ Jul 26, 2017 at 21:21
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What you're looking for is the (theoretical) angle associated with sp$^{3}$ hybrid orbitals. ("Theoretical" because in any molecule where this shows up, there are other forces at work that cause the angle differ by various amounts.) As Ian Mateus said in a comment, you just need the Law of Cosines for a triangle.
Assume the cube has edge length $1$. (The angles will be the same for each value you use for the edge length, so just pick a nice number like $1$.) Then the face diagonals have length $\sqrt{2}$ and the space diagonals have length $\sqrt{3}.$ You can find the space diagonal length by two appropriate applications of the Pythagorean theorem, or by using the distance formula in ${\mathbb R}^3$ to find the distance between $(0,0,0)$ and $(1,1,1).$
By drawing a picture (maybe someone can post a nice diagram), it follows that you want to find the angle $\theta$ formed by the equal-length sides of a triangle whose sides have lengths $\frac{1}{2}\sqrt{3}$ and $\frac{1}{2}\sqrt{3}$ and $\sqrt{2}.$ To do this, use the Law of Cosines with $\theta$ as the angle and $\sqrt{2}$ as the side opposite $\theta$:
$$c^2 \; = \; a^2 + b^2 - 2ab\cos \theta$$
$$\left(\sqrt{2}\right)^2 \;\; = \;\; \left(\frac{1}{2}\sqrt{3}\right)^2 \; + \; \left(\frac{1}{2}\sqrt{3}\right)^2 \; - \; 2\left(\frac{1}{2}\sqrt{3}\right)\left(\frac{1}{2}\sqrt{3}\right)\cos \theta$$
$$ 2 \; = \; \frac{3}{4} + \frac{3}{4} - \frac{3}{2}\cos \theta $$
$$ \frac{1}{2} \; = \; - \frac{3}{2} \cos \theta $$
$$ - \frac{1}{3} = \cos \theta$$
$$ \theta = \arccos \left( -\frac{1}{3} \right) $$
Now use a calculator.
(ADDED 4 YEARS 5 MONTHS LATER) A few months ago (from the date of this edit) I came across several papers in Journal of Chemical Education that give various ways of deriving the tetrahedral bond angle. I put the papers aside at the time, intending to include them here at some later time, and now is that later time.
[1] Gordon Louis Gombert, The valence angle of the carbon atom, Journal of Chemical Education 18 #7 (July 1941), 336-337.
[2] Walter Harrington Dore, Calculation of the valence angle, Journal of Chemical Education 19 #1 (January 1942), 29-30.
[3] Philip Francis Weatherill, The valence angle of the carbon atom, Journal of Chemical Education 19 #1 (January 1942), 35.
[4] Garrett William Thiessen, The carbon valence angle, Letters section, Journal of Chemical Education 19 #4 (April 1942), 198.
[5] Wesley Emil Brittin, Valence angle of the tetrahedral carbon atom, Journal of Chemical Education 22 #3 (March 1945), 145.
[6] Thomas McCullough, Simple calculation of the tetrahedral bond angle, Journal of Chemical Education 39 #9 (September 1962), 476.
[7] Günther Snatzke, [Letter to the Editor], Letters section, Journal of Chemical Education 40 #2 (February 1963), 94. [Note: There is a typo in the first displayed equation, and it should be $r_1 + r_2 + r_3 + r_4 = 0.]$
[8] Bruce Lindsey Cockburn, [Letter to the Editor], Letters section, Journal of Chemical Education 40 #2 (February 1963), 94.
[9] Christopher James Kawa, Finding the bond angle in a tetrahedral-shaped molecule, Journal of Chemical Education 65 #10 (October 1988), 884-885.
[10] George Henry Duffey, Employing vector algebra to obtain the tetrahedral bond angle, Journal of Chemical Education 67 #1 (January 1990), 35-36.
[11] Dylan Gow and Amites Sarkar, [Solution to Problem 22.5], Mathematical Spectrum 23 #1 (September 1990), 27-28.
[12] Resat Mustafa Apak and Izzet Tor, Finding the bond angle, Letters section, Journal of Chemical Education 68 #11 (November 1991), 970.
[13] Brian Terence Sutcliffe and Stephen J. Smith, Computing the tetrahedral angle, Letters section, Journal of Chemical Education 69 #2 (February 1992), 171.
[14] George Henry Duffey, Computing the tetrahedral angle, Letters section, Journal of Chemical Education 69 #2 (February 1992), 171.
[15] Paul Glaister, Rotational symmetry of a methane molecule and the bond angle, Journal of Chemical Education 70 #5 (May 1993), 351.
[16] Paul Glaister, Calculating the tetrahedral bond angle using spherical polars and the dot product, Journal of Chemical Education 70 #7 (July 1993), 546-547.
[17] Alfred Alexander Woolf, Tetrahedral geometry made simple, Journal of Chemical Education 72 #1 (January 1995), 19-20.
[18] Paul Glaister, Two comments on bond angles, Journal of Chemical Education 74 #9 (September 1997), 1086.
[19] Ricardo de Carvalho Ferreira, Tetrahedral bond angle, Letters section, Journal of Chemical Education 75 #9 (September 1998), 1087.
[20] Sara Noemí Mendiara and Luis José Perissinotti, Tetrahedral geometry and the dipole moment of molecules, Journal of Chemical Education 79 #1 (January 2002), 64-66.
[21] Marten J. ten Hoor, An evergreen: the tetrahedral bond angle, Journal of Chemical Education 79 #8 (August 2002), 956-957.
answered Feb 4, 2014 at 17:59
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3$\begingroup$ If you are stuck in an island, you can compute it using $$\cos^{-1}(-1/3)=\frac{\pi}{2}-\sum_{k=0}^\infty{2k\choose k}\frac{(-1/3)^{2k+1}}{4^{k}(2k+1)}$$ $\endgroup$ Feb 4, 2014 at 18:33
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With dot products: Taking the axis-oriented cube of side length $2$ centered at the origin, you're looking for the angle $\theta$ between (say) the vectors $\mathbf{u} = (1, -1, 1)$ and $\mathbf{v} = (-1, 1, 1)$: $$ \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\, \|\mathbf{v}\|} = -\frac{1}{3}. $$ In response to Dave Renfro's request for a diagram (his triangles are bounded by a pair of vectors and a dashed line), here's a crossed-eyes stereogram:
answered Feb 4, 2014 at 23:35
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There is an amazing geometric interpretation for figuring out the angle in that link ->
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$\begingroup$ Welcome to math.se. Please describe the content of the link rather than just posting the link. $\endgroup$– CouchyMay 6, 2017 at 18:33