Can There Be Five Vectors in ( mathbb{R}^3 ) with Negative Dot Products Between Any Two?

In this article, we explore the possibility of having five vectors in ( mathbb{R}^3 ) such that the dot product between any two of these vectors is negative. We use both geometric and algebraic methods to demonstrate that it is impossible to achieve this configuration.

Understanding the Dot Product

The dot product is defined for two vectors ( mathbf{u} ) and ( mathbf{v} ) as:

where ( theta ) is the angle between the vectors. The dot product is negative if and only if the angle between the vectors is greater than ( 90^circ ). Geometrically, this means the vectors must be pointing in directions that are more than ( 90^circ ) apart.

Geometric Arrangement in ( mathbb{R}^3 )

Visualize vectors in ( mathbb{R}^3 ) as arrows originating from the origin. For any vector, there exists a hemisphere where the angle between it and another vector is more than ( 90^circ ). This hemisphere is the set of all vectors whose dot product with the given vector is negative.

Maximum Number of Vectors with Negative Dot Products

We attempt to place five vectors such that each pair has a negative dot product. Here is the step-by-step reasoning:

( text{For the first vector, } mathbf{v}_1, text{ it can be placed anywhere.} )

( mathbf{v}_2 text{ must be placed in the hemisphere opposite to } mathbf{v}_1. )

( mathbf{v}_3 text{ must be placed in the hemispheres opposite to both } mathbf{v}_1 text{ and } mathbf{v}_2. )

( text{However, in } mathbb{R}^3, text{ the intersection of these hemispheres leaves only one hemisphere where } mathbf{v}_3 text{ can go.} )

( text{Continuing this reasoning, adding a fourth vector will further restrict the available space.} )

( text{Fitting a fifth vector while maintaining the negative dot product condition becomes impossible.} )

Conclusion

In ( mathbb{R}^3 ), it is not possible to have more than four vectors such that the dot product between any two of them is negative. Therefore, it is impossible to find five vectors in ( mathbb{R}^3 ) that satisfy the condition that the dot product between any two vectors is negative.

Thus, the answer is No, there cannot be five such vectors in ( mathbb{R}^3 ).

Further Exploration on the Unit Sphere

To explore this problem from another angle, consider all vectors as unit vectors. The problem can be reworded as placing five points on a unit sphere such that the arc connecting any two points is more than ( 90^circ ).

Without loss of generality, let us place one point, ( A_1 ), on the sphere. ( A_1 ) divides the sphere into two hemispheres. The other four points must be in the hemisphere opposite to ( A_1 ). Let us place a second point, ( A_2 ), in this hemisphere, and consider the half-sphere with the center ( A_2 ). The other three points must be outside the union of the two half-spheres, those with centers ( A_1 ) and ( A_2 ), leaving less than a quarter of the sphere!

Now, let us prove that even a full quarter of the sphere cannot hold three points. The quarter of the sphere is cut by two perpendicular planes passing through the poles. The equator divides the quarter of the sphere into two eighths. We can place two points in these eighths but they must be in different eighths. There is no room for a third point because no two points can share an eighth of the sphere, making the arc between them less than ( 90^circ ).

Conclusion on Four Points

Finally, it is easy to prove that four points can be placed on the sphere making all ( c(4, 2) 6 ) arcs more than ( 90^circ ). To do this, we place three points on the equator of the Northern half-sphere such that all three arcs are ( 120^circ ). Move the points slightly North of the equator; the arcs will become slightly less than ( 120^circ ), but still more than ( 90^circ ). Finally, put the fourth point in the South pole. Thus, it is possible to place four points on the sphere with the required conditions.