Two-Dimensional Unit Vectors!

Let $\vec x$ , $\vec y$ , $\vec z$ be two-dimensional unit vectors such that $\vec x+\vec y+\vec z$ is also a unit vector. Prove that two of the vectors must be opposite to each other (i.e. two of $\vec x$ , $\vec y$ , $\vec z$ add up to zero).

Bonus: Prove that this does not hold for higher dimensions.

There are many ways to solve this problem. Try to experiment with different methods.

#Algebra

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let $\vec{AB} = \vec{x}$ , $\vec{BC} = \vec{y}$ , and $\vec{CD} = \vec{z}$ . Then $\vec{AD} = \vec{x} + \vec{y} + \vec{z}$ .

Since $\vec{x}$ , $\vec{y}$ , $\vec{z}$ , and $\vec{x} + \vec{y} + \vec{z}$ are unit vectors, $AB = BC = CD = AD = 1$ , so $ABCD$ is a regular rhombus or a degenerated rhombus where at least one pair of vertices coincide.

If $ABCD$ is a regular rhombus, then $AB || CD$ so that $\vec{x}$ and $\vec{z}$ are opposite to each other.

If $ABCD$ is a degenerated rhombus where at least one pair of vertices coincide, then there are two pairs of sides with the same endpoints which means two of $\vec{x}$ , $\vec{y}$ , and $\vec{z}$ are opposite to each other.

Either way, two of $\vec{x}$ , $\vec{y}$ , and $\vec{z}$ are opposite to each other.

This does not hold for higher dimensions because rhombus $ABCD$ can be "bent" in the third dimension so that no two vectors are opposite to each other. A counterexample is $\vec{x} = (\frac{4}{5}, \frac{3}{5}, 0)$ , $\vec{y} = (-\frac{4}{5}, \frac{3}{5}, 0)$ , and $\vec{z} = (0, -\frac{3}{5}, \frac{4}{5})$ , which makes $\vec{x} + \vec{y} + \vec{z} = (0, \frac{3}{5}, \frac{4}{5})$ , all unit vectors where no two are opposite to each other.

David Vreken - 1 year, 8 months ago

Great solution! My solution is pretty similar (although slightly more complicated than yours), and I will post it when I have time.

Sam Zhou - 1 year, 8 months ago

Thanks! I first tried to prove it algebraically but it got too complicated for my liking.

David Vreken - 1 year, 8 months ago

@David Vreken – Yeah. Many people would try using algebra to solve this question without realizing that there is a very simple solution using geometry. This question would be harder if I actually translated it into a set of algebraic equations.

Sam Zhou - 1 year, 8 months ago

It is clear that the locus of the endpoint of $\vec x$ is on the black circle, which we will call $C_{0}$ . $\vec x+\vec y$ can lie on any point $P$ within the circle of radius $2$ centered at the origin. In the diagram above, $P=\vec x+\vec y=(1,1)$ . The red circle, $C_{1}$ , represents the locus of $\vec x+\vec y+\vec z$ . The points of intersection of the two circles are the possible values of $\vec x+\vec y+\vec z$ such that it is a unit vector.

Note that the two points are the endpoints of $\vec x$ and $\vec y$ , since they are the only two points whose distances from the centers of $C_{0}$ and $C_{1}$ are both $1$ . Therefore, $\vec x+\vec y+\vec z=\vec x$ or $\vec y$ , meaning that $\vec z=-\vec x$ or $-\vec y$ .

The cases where the two circles don't intersect at two points are (1) $\vec x=\vec y$ , in which case the two circles are tangent and (2) $\vec x=-\vec y$ . In case (1), it can be deduced that the only value of $\vec z$ is $-\vec x$ . Case (2) is trivial.

This does not hold for higher dimensions since in $\mathbb{R}^{3}$ , the intersection of the loci (two spheres) is a circle, so there are infinite possible values of $\vec z$ , meaning that it does not have to be $-\vec x$ or $-\vec y$ . It can be generalized to higher dimensions as $\mathbb{R}^{3} \in \mathbb{R}^{3+k}$ where $k \in \mathbb{N}$ .

Sam Zhou - 1 year, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$