0
$\begingroup$

Three vectors, $V_1,V_2,V_3$ are in the $\mathbb{R}^2$ plane where $V_1+V_2+V_3=\vec{0}$ and the magnitudes of these vectors are the same. Show that the angle between any two of these vectors is 120 degrees.

My try:

$V_1+V_2+V_3=\vec{0}$, so $V_1+V_2=-V_3$. This would mean that $(V_1+V_2)^2=V_3^2$ which is $|V_1|+|V_2| + 2V_1V_2 =|V_3|$. But since the magnitude is the same, we get $2|V_1|+2V_1V_2=|V_1|$ which is $2V_1V_2=-|V_1|$ so $V_1*V_2=-0.5|V_1|$.

The formula for the angle is $\cos(a)= \frac{V_1*V_2}{|V_1|*|V_2|} = -0.5*\frac{|V_1|}{|V_1|^2}=\frac{-0.5}{|V1|}$

I know that the solution would be a triangle and that $\cos(a)=-0.5$

But that would mean that my angle is dependent on $V_1$ which isn't the case. So what am I doing wrong?

$\endgroup$
1
  • $\begingroup$ With respect to the two answers provided, is either one useful to you? If not, please edit your query re (1) what is the background of the question - re what theorems or previously solved problems might be pertinent to you here? (2) please show your work, re what you have tried. $\endgroup$ Commented Oct 5, 2020 at 10:54

2 Answers 2

1
$\begingroup$

$V_3=-V_1-V_2$ implies that $\|V_3\|^2=\langle V_1+V_2,V_1+V_2\rangle =\|V_1\|^2+\|V_2\|^2+2\langle V_1,V_2\rangle =\|V\|^2$ where $\|V_1\|=\|V_2\|=\|v_3\|=\|V\|$

This implies that $2\|V\|^2 cos(V_1,V_2)=-\|V\|^2$.

$\endgroup$
0
$\begingroup$

You cannot "square" a vector: the statement "$(V_1 + V_2)^2 = V_3^2$" makes no sense.

I sense that you are trying to use the dot product. In that case, you will instead obtain:

$$|V_3|^2 = V_3 \cdot V_3 = (V_1 +V_2) \cdot (V_1 +V_2) = |V_1|^2 + 2 V_1 \cdot V_2 + |V_2|^2$$

Now we continue with your method. We have from $$|V_1|^2 = -2V_1\cdot V_2$$

$$\implies \cos (a) = \frac {V_1 \cdot V_2}{|V_1||V_2|} = \frac {V_1 \cdot V_2}{|V_1|^2} = -\frac12$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .