AICBSE-CHEM practice

Geometry Level 5

In $\mathbb{R} ^3$ , 4 vectors are equally inclined to each other at an non-zero angle $\theta$ . What is $\cos(\theta)$ ?

The answer is -0.33333333.

1 solution

Marta Reece
Apr 7, 2017

For four vectors to be inclined at the same angle relative to each other, they have to be pointing from the center of a regular tetrahedron toward its vertices. For simplicity, we can use a unit tetrahedron.

Point $E$ is in the center of the base of the tetrahedron, so $AE$ is two thirds of a median of $\triangle ABC$ , therefore $AE=\frac{\sqrt{3}}{2}\frac{2}{3}=\frac{1}{\sqrt{3}}.$

From right $\triangle AED$ we get $\angle ADE=arcsin(\frac{1}{\sqrt{3}})=35.26^\circ$

If $O$ is the center of the tetrahedron, $\triangle AOD$ is isosceles and $\angle AOD=180^\circ-2\times 35.26^\circ=109.27^\circ$

Finally $cos(109.27^\circ)=-\frac{1}{3}\approx-0.333$ .

Can you please tell me how you rationalised the first point in your solution?

Rohith M.Athreya - 4 years, 1 month ago

The reason it has to be a tetrahedron: If you had only three vectors, you could use any angle between them and arrange them in a tent-like structure. Adding the fourth, however, is not possible with any angle. The fourth has to point away from the the other three, up from the top of the tent, and the angles in the tent have to be equal to the angles between the tent vectors and the one going up. The symmetry of the structure requires the vectors to be pointing from the center to the vertices of a tetrahedron.

Marta Reece - 4 years, 1 month ago

Thank you very much!

I was fixated on an algebraic solution earlier and was at loggerheads with intimidating equations before I finally resolved them.This approach is much more inspired,no doubt.

Thinking along the lines you have established,I began to wonder if there exists an upper bound for the number of vectors that can mutually be equally inclined to each other.I have however,not been able to establish much.Can you,when you find time,let me know if you have an approach I could go forward with?

Thank you!

Rohith M.Athreya - 4 years, 1 month ago

@Rohith M.Athreya – The upper bound for all the vectors mutually at the same angles is 4, as shown above. You could have a higher number, such as 6, if you allowed one of the vectors to be at a different angle and the rest have the same one.

Marta Reece - 4 years, 1 month ago

AICBSE-CHEM practice

The answer is -0.33333333.

1 solution

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