Vtyjkyuk

Let $|vecy|=|vecx+vecy|=|vecx+vec2y|$. How can I find $angle(vecx,vecy)$ without using the dot product?

HINT:

The segments $|vecy|,|vecx+vecy|,|vecx+2vecy|$ are the sides of an equilateral triangle. Observe the figure and decide whether this is the only solution.

By dot product we have that

• $langle y,yrangle=langle x+y,x+yrangle=langle y,yrangle+langle x,xrangle+2langle x,yrangle implies langle x,xrangle+2langle x,yrangle=0$




• $langle y,yrangle=langle x+2y,x+2yrangle=4langle y,yrangle+langle x,xrangle+4langle x,yrangle implies 3langle y,yrangle=langle x,xrangle$

then recall

$$cos theta =fraclangle x,yranglesqrtlangle x,xranglelangle y,yrangle$$

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

$$cos theta =-frac32 fracx$$

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

$$fracx2=|y|cos (pi-theta)$$

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

$$cos^2 theta=frac 3 4implies cos theta=-fracsqrt 32$$

that is $theta=frac56 pi$.

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 $2vec y$ 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 $150°$.

WLOG, consider $vecy(1,0)$ and $vecx(x_1,x_2)$. Then:
$$|vecx+vecy|=sqrt(x_1+1)^2+x_2^2=1=|y| Rightarrow x_1^2+x_2^2+2x_1=0;\
|vecx+2vecy|=sqrt(x_1+2)^2+x_2^2=1=|y| Rightarrow x_1^2+x_2^2+4x_1=-3.$$
Subtracting the two we get:
$$x_1=-frac32 Rightarrow x_2=pmfracsqrt32.$$
Hence:
$$cos theta =fracvecxcdot vecyvecx=frac-frac32+0sqrtfrac94+frac34cdot 1=-fracsqrt32 Rightarrow \
theta = frac5pi6 (textbecause the angle between vectors lies in $[0,2pi]$).$$
Alternatively:
$$|x||y|sin theta=textabsbeginvmatrixx_1&x_2 \ y_1& y_2endvmatrix;\
sqrt3cdot 1cdot sin theta=textabsbeginvmatrix-frac32&fracsqrt32 \ 1& 0endvmatrix Rightarrow \
theta = frac5pi6 text(because $vecx$ is in the 2nd quarter and $vecy$ on the $OX^+$);\
sqrt3cdot 1cdot sin theta=textabsbeginvmatrix-frac32&-fracsqrt32 \ 1& 0endvmatrix Rightarrow \
theta = frac5pi6 text(because $vecx$ is in the 3rd quarter and $vecy$ on the $OX^+$).$$