If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.

If the magnitude of the resultant of two equal vectors is equal to that of either vector, find the angle between them.

My Attempt: Let $\vec {a}$ and $\vec {b}$ be two vectors such that $\|\vec {a}\|=\|\vec {b}\|$

Magnitude of Resultant: $$=\sqrt {a^2+b^2+2ab\cos \theta}$$ $$=\sqrt {2a^2+2ab\cos \theta}$$

How do I proceed further?

• Hint: which rhombus has a diagonal equal to the sides? – dxiv Apr 18 '17 at 16:57
• It is better to say the original two vectors $a,b$ have equal magnitudes than to say these vectors are equal (which we reserve for $a=b$). – hardmath Apr 18 '17 at 17:23

From $\|a+b\|^2=\|a\|^2$ and $\|a\|=\|b\|$ we have $$\|a\|^2=\|a+b\|^2=\|a\|^2+\|b\|^2+2\langle a,b\rangle=2\|a\|^2+2\langle a,b\rangle,$$ so $\langle a,b\rangle=-\|a\|^2/2$. Hence the cosine of the angle between $a$ and $b$ using $\|a\|=\|b\|$ again is $$\frac{\langle a,b\rangle}{\|a\|\|b\|}=\frac{-1}{2}.$$

• How did you get $(a,b)=-|b|^2/2$? – pi-π Apr 19 '17 at 1:31
• See my edit, please. – Michael Hoppe Apr 19 '17 at 6:54

The resultant vector $\mathbf{OC}$ is the diagonal of the rhombus whose adjacent sides are $\mathbf{OA}, \mathbf{OB}$. Since its length is equal to either, we have an equilateral triangle $\mathbf{OBC}$. Thus the angle between the vectors is $120^\circ$

Let,$R$ be the resultant of the sum of the two vectors $\vec A$ and $\vec B$ such that $|\vec A|=|\vec B|$.So,$$R=\sqrt{|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta}$$

where $\theta$ is the angle between $\vec A$ and $\vec B$.So,

$$R=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$.

Now magnitude of resultant $R$ is equal to either of the vectors $\vec A$ and $\vec B$.So,

$$|A|=\sqrt{2|\vec A|^2+2|\vec A|^2\cos\theta}$$ $$\implies|\vec A|^2=2|\vec A|^2+2|\vec A|^2\cos\theta$$ $$\implies\frac{-|\vec A|^2}{2|\vec A|^2}=\cos \theta$$ $$\implies\theta=\cos^{-1}-\frac{1}{2}$$.

Hope this helps!!