# Halving a vector and halving an angle

An object is subjected to two equal forces along two different directions. If the magnitude of one of them is halved, the angle which the new resultant makes with the other component force is also halved. What is the angle between the forces? Well I'm not sure whether the vector notation in the second figure is right or not – if wrong please correct it (I'm learning vectors).

• "If one of them is halved": what is halved ? – Yves Daoust May 9 '17 at 7:34
• @Lovsovs: LOL, I am not asking the meaning of the verb ! I am helping the OP understand the problem statement. – Yves Daoust May 9 '17 at 7:37
• @YvesDaoust Isn't it clear that it is the length of the vector that is halved, both from the problem formulation and the diagram? – Bobson Dugnutt May 9 '17 at 7:40
• @Lovsovs no, and the resultant is very approximate. – Yves Daoust May 9 '17 at 7:44
• @Lovsovs one clearly needs to define their question properly. We can't assume OP means it. – Iti Shree May 9 '17 at 7:44

Let's set $$\left\{ \begin{array}{ll} \vec{p} &= p \cos{\alpha} \hat{i} + p \sin{\alpha}\hat{j} \\ \vec{p^*} &=p \cos{\alpha} \hat{i} - p \sin{\alpha}\hat{j} \\ \end{array} \right.$$ $\Rightarrow \vec{p} + \vec{p^*} = 2p \cos{\alpha}\hat{i}$. But $\frac{1}{2}\vec{p} + \vec{p^*} = \frac{3}{2}p \cos{\alpha} \hat{i} - \frac{1}{2} p \sin{\alpha} \hat{j} = \vec{p}_N$.
Now, it was defined that the angle between $\vec{p}_N$ and $\vec{p^*}$ is $\frac{\alpha}{2}$: $$\cos{\frac{\alpha}{2}} = \frac{\vec{p}_N \cdot \vec{p^*}}{\left|\vec{p}_N \right| \left|\vec{p^*} \right|} = \frac{\frac{3}{2}p^2 \cos^2{\alpha} + \frac{1}{2}p^2 \sin^2{\alpha}}{p^2 \sqrt{ \frac{9}{4}\cos^2{\alpha} + \frac{1}{4} \sin^2{\alpha} }} = \frac{ \frac{1}{2} + \cos^2{\alpha} }{\sqrt{ \frac{1}{4} + 2 \cos^2{\alpha}}} = \frac{ 1 + 2\cos^2{\alpha} }{\sqrt{ 1 + 8 \cos^2{\alpha}}}$$ We also note that $\cos{2x} = 2\cos^2{x} - 1 \Rightarrow \cos{\frac{\alpha}{2}} = \sqrt{ \frac{\cos{\alpha} + 1}{2} }$. Inserting this result into the previous equation, and taking the notation $\cos{\alpha} = \eta$, we obtain the equation $$\frac{\eta+1}{2} = \left( \frac{1+ 2\eta^2}{\sqrt{1+8\eta^2}} \right)^2 =\frac{(1+ 2\eta^2)^2}{1+8\eta^2} = \frac{4\eta^4 + 4\eta^2 +1}{8\eta + 1}$$
$$\Rightarrow 8\eta^3 + 3\eta^2 + \eta +1 = 8\eta^4 + 8 \eta^2 +2$$ $$\Rightarrow 8\eta^4 - 8\eta^3 - \eta + 1= 0$$ The solutions to this equation are $\eta = 1$ and $\eta= \frac{1}{2}$, giving $\alpha = 0$ (trivial) or $\alpha = 60^{\circ}$. The angle between the forces is therefore $2\alpha = 120^{\circ}$.
Fix the lengths of the original vectors as 1 and align one of them along the $+x$-axis, then call the angle between them $\theta$: When the non-axis-aligned vector is halved, the resultant vector is $(1+\frac12\cos\theta,\frac12\sin\theta)$ and we want its argument to be half the argument of the original resultant, which in turn is (by symmetry) $\frac\theta4$. In other words: $$\tan\frac\theta4=\frac{\frac12\sin\theta}{1+\frac12\cos\theta}$$ Let $x=\frac\theta4$ and expand: $$\tan x=\frac{\sin4x}{2+\cos4x}$$ $$\frac{\sin x}{\cos x}=\frac{4\sin x\cos x(2\cos^2x-1)}{1+2(2\cos^2x-1)^2}$$ If $\sin x=0$ we have the trivial $\theta=0$. Thus we can divide it out and let $\cos^2x=y$: $$\frac1{\sqrt y}=\frac{4\sqrt y(2y-1)}{1+2(2y-1)^2}$$ $$1+2(2y-1)^2=4y(2y-1)$$ $$1+8y^2-8y+2=8y^2-4y$$ $$y=\frac34\qquad\cos x=\pm\frac{\sqrt3}2$$ $$x=\pm\frac\pi6\lor x=\pm\frac{5\pi}6$$ $$\theta=\pm\frac{2\pi}3$$ Therefore the angle between the two vectors is 120°, as emphasised by the extra triangular grid I drew into the picture above.