# How can I prove that the angles of the tetrahedral structure is $109.5^\circ$ with calculus. I could do it with geometry. [duplicate]

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.

• Does the dot product count as "calculus"? Feb 4, 2014 at 13:30
• Yes, sure...actually I was looking for new methods.
– Hawk
Feb 4, 2014 at 13:32
• Yes, Certainly...thank you for pointing out!
– Hawk
Feb 4, 2014 at 13:40
• angle subtended by the vertices of a regular tetrahedral structure to the centre
– Hawk
Feb 4, 2014 at 14:05
• I changed $109.5^o$ to $109.5^\circ.$ That is standard usage. $\qquad$ Jul 26, 2017 at 21:13

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 $-1/3.$

• 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 $1 + 3\cos(\theta) = 0$. Jul 26, 2017 at 21:21

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.

 Gordon Louis Gombert, The valence angle of the carbon atom, Journal of Chemical Education 18 #7 (July 1941), 336-337.

 Walter Harrington Dore, Calculation of the valence angle, Journal of Chemical Education 19 #1 (January 1942), 29-30.

 Philip Francis Weatherill, The valence angle of the carbon atom, Journal of Chemical Education 19 #1 (January 1942), 35.

 Garrett William Thiessen, The carbon valence angle, Letters section, Journal of Chemical Education 19 #4 (April 1942), 198.

 Wesley Emil Brittin, Valence angle of the tetrahedral carbon atom, Journal of Chemical Education 22 #3 (March 1945), 145.

 Thomas McCullough, Simple calculation of the tetrahedral bond angle, Journal of Chemical Education 39 #9 (September 1962), 476.

 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 $$r_1 + r_2 + r_3 + r_4 = 0.]$$

 Bruce Lindsey Cockburn, [Letter to the Editor], Letters section, Journal of Chemical Education 40 #2 (February 1963), 94.

 Christopher James Kawa, Finding the bond angle in a tetrahedral-shaped molecule, Journal of Chemical Education 65 #10 (October 1988), 884-885.

 George Henry Duffey, Employing vector algebra to obtain the tetrahedral bond angle, Journal of Chemical Education 67 #1 (January 1990), 35-36.

 Dylan Gow and Amites Sarkar, [Solution to Problem 22.5], Mathematical Spectrum 23 #1 (September 1990), 27-28.

 Resat Mustafa Apak and Izzet Tor, Finding the bond angle, Letters section, Journal of Chemical Education 68 #11 (November 1991), 970.

 Brian Terence Sutcliffe and Stephen J. Smith, Computing the tetrahedral angle, Letters section, Journal of Chemical Education 69 #2 (February 1992), 171.

 George Henry Duffey, Computing the tetrahedral angle, Letters section, Journal of Chemical Education 69 #2 (February 1992), 171.

 Paul Glaister, Rotational symmetry of a methane molecule and the bond angle, Journal of Chemical Education 70 #5 (May 1993), 351.

 Paul Glaister, Calculating the tetrahedral bond angle using spherical polars and the dot product, Journal of Chemical Education 70 #7 (July 1993), 546-547.

 Alfred Alexander Woolf, Tetrahedral geometry made simple, Journal of Chemical Education 72 #1 (January 1995), 19-20.

 Paul Glaister, Two comments on bond angles, Journal of Chemical Education 74 #9 (September 1997), 1086.

 Ricardo de Carvalho Ferreira, Tetrahedral bond angle, Letters section, Journal of Chemical Education 75 #9 (September 1998), 1087.

 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.

 Marten J. ten Hoor, An evergreen: the tetrahedral bond angle, Journal of Chemical Education 79 #8 (August 2002), 956-957.

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: There is an amazing geometric interpretation for figuring out the angle in that link ->

http://maze5.net/?page_id=367

• Welcome to math.se. Please describe the content of the link rather than just posting the link. May 6, 2017 at 18:33