How to find $\angle(\vec{x},\vec{y})$ if $|\vec{y}|=|\vec{x}+\vec{y}|=|\vec{x}+\vec{2y}|$?

Question

Let $|\vec{y}|=|\vec{x}+\vec{y}|=|\vec{x}+\vec{2y}|$. How can I find $\angle(\vec{x},\vec{y})$ without using the dot product?

Answer 1

By dot product we have that

• $\langle y,y\rangle=\langle x+y,x+y\rangle=\langle y,y\rangle+\langle x,x\rangle+2\langle x,y\rangle \implies \langle x,x\rangle+2\langle x,y\rangle=0$

• $\langle y,y\rangle=\langle x+2y,x+2y\rangle=4\langle y,y\rangle+\langle x,x\rangle+4\langle x,y\rangle \implies 3\langle y,y\rangle=\langle x,x\rangle$

then recall

$$\cos \theta =\frac{\langle x,y\rangle}{\sqrt{\langle x,x\rangle\langle y,y\rangle}}$$

After OP editing, as an alternative by law of cosines we obtain

• $|x+y|^2=|x|^2+|y|^2+2|x||y|\cos \theta$
• $|x+2y|^2=|x|^2+|2y|^2+2|x||2y|\cos \theta$

from which by the given conditions we obtain

$$\cos \theta =-\frac32 \frac{|y\ |}{|x|}$$

and from the triangle with sides $|x|$,$|y|$,$|x+y|$ isosceles by the condition $|y|=|x+y|$ we have

$$\frac{|x|}{2}=|y|\cos (\pi-\theta)$$

and finally, since for the above condition we need $\cos \theta <0$, we obtain

$$\cos^2 \theta=\frac 3 4\implies \cos \theta=-\frac{\sqrt 3}2$$

that is $\theta=\frac56 \pi$.

Answer 2

HINT:

The segments $|\vec{y}|,|\vec{x}+\vec{y}|,|\vec{x}+2\vec{y}|$ are the sides of an equilateral triangle. Observe the figure and decide whether this is the only solution.

Answer 3

WLOG, consider $\vec{y}(1,0)$ and $\vec{x}(x_1,x_2)$. Then: $$|\vec{x}+\vec{y}|=\sqrt{(x_1+1)^2+x_2^2}=1=|y| \Rightarrow x_1^2+x_2^2+2x_1=0;\\ |\vec{x}+2\vec{y}|=\sqrt{(x_1+2)^2+x_2^2}=1=|y| \Rightarrow x_1^2+x_2^2+4x_1=-3.$$ Subtracting the two we get: $$x_1=-\frac32 \Rightarrow x_2=\pm\frac{\sqrt{3}}{2}.$$ Hence: $$\cos \theta =\frac{\vec{x}\cdot \vec{y}}{|\vec{x}|\cdot |\vec{y}|}=\frac{-\frac32+0}{\sqrt{\frac94+\frac34}\cdot 1}=-\frac{\sqrt{3}}{2} \Rightarrow \\ \theta = \frac{5\pi}{6} \ \ (\text{because the angle between vectors lies in $[0,2\pi]$}).$$ Alternatively: $$|x||y|\sin \theta=\text{abs}\begin{vmatrix}x_1&x_2 \\ y_1& y_2\end{vmatrix};\\ \sqrt{3}\cdot 1\cdot \sin \theta=\text{abs}\begin{vmatrix}-\frac32&\frac{\sqrt{3}}{2} \\ 1& 0\end{vmatrix} \Rightarrow \\ \theta = \frac{5\pi}{6} \ \ \text{(because $\vec{x}$ is in the 2nd quarter and $\vec{y}$ on the $OX^+$)};\\ \sqrt{3}\cdot 1\cdot \sin \theta=\text{abs}\begin{vmatrix}-\frac32&-\frac{\sqrt{3}}{2} \\ 1& 0\end{vmatrix} \Rightarrow \\ \theta = \frac{5\pi}{6} \ \ \text{(because $\vec{x}$ is in the 3rd quarter and $\vec{y}$ on the $OX^+$)}.$$

Answer 4

A bit of geometry:

For simplicity orient $\vec y$ along the positive $x$-axis in a Cartesian coordinate system .

Add any vector $\vec x$ , i.e consider $\vec y + \vec x$.

The resulting $\vec y +\vec x$ is a vector that starts from the origin $O$ and points to the tip of $\vec y +\vec x$, and has length $|\vec y|$, i.e it lies on the Thales circle about O with radius $r:= |\vec y|$.

Let $A(0,r)$, and $B$, on the Thales circle, the tip of $\vec x + \vec y$.

$\triangle OAB$ is isosceles : $|OA|=|OB|=r.$

Let $C(-r,0)$.

Start from $C$.

The tip of $2\vec y$ is at $A$, then add $\vec x$.

Resultant is a a vector $CB$ of length $r$.

$\triangle COB$ is equilateral , side length $r$, every interior angle is $60°$,

$\triangle OAB$ is isosceles with $|OA|=|OB|=r.$

$\angle BOA =120°$,

the acute $\angle OAB =30°$,

the obtuse angle between $\vec y, \vec x$ is $150°$.