Question in vector algebra regarding minimum value of modulus.

Question

If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are three coplanar unit vectors such that $\vec{a} +\vec{b} +\vec{c} =0$. If three vectors $\vec{p}$ , $\vec{q}$ , $\vec{r}$ are parallel to $\vec{a}$ , $\vec{b} $ and $\vec{c}$ respectively and have integral but different magnitudes, then $\left| \vec{p} +\vec{q} +\vec{r} \right|$ can take a value equal to

1. $\sqrt{3}$
2. $2$
3. $1$
4. $0$

Approach:

Let $\vec{a}$ , $\vec{b}$ , $\vec{c}$ lie in the xy-plane.

Let $\vec{a} =-\frac{1}{2}\hat{i} +\frac{\sqrt{3} }{2} \hat{j},\; \vec{b} =- \frac{1}{2} \hat{i} -\frac{\sqrt{3} }{2} \hat{j}$ and $\vec{c} =\hat{i}$.

Then,

\begin{align} \bigl\lvert \vec{p} + \vec{q} + \vec{r}\bigr\rvert &= \bigl\lvert \lambda\vec{a} +\mu\vec{b} +\nu\vec{r}\bigr\rvert\\ &= \biggl\lvert \lambda\biggl({ -\frac{1}{2} \hat{i} +\frac{\sqrt{3} }{2} \hat{j} }\biggr) +\mu\biggl({ -\frac{1}{2} \hat{i} -\frac{\sqrt{3} }{2} \hat{j}}\biggr) +\nu\hat{i}\biggr\rvert\\ &= \biggl\lvert \hat{i}\biggl(-\frac{1}{2}\lambda-\frac{\mu}{2} +\nu\biggr)+\hat{j}\biggl(\frac{\sqrt{3} }{2} \lambda- \frac{\sqrt{3} }{2}\mu\biggr)\biggr\rvert\\ &= \sqrt{\biggl(-\frac{\lambda}{2}-\frac{\mu}{2}+\nu\biggr)^{2}+\biggl(\frac{\sqrt{3}}{2}\lambda-\frac{\sqrt{3}}{2}\mu\biggr)^{2}}\\ &= \sqrt{{\lambda}^{2}+ {\mu}^{2}+ {\nu}^{2}- {\lambda\mu} - {\lambda\nu} - {\nu\mu}}\\ &= \frac{1}{\sqrt{2}}.\sqrt{(\lambda-\mu)^{2}+ (\mu-\nu)^{2}+ (\nu-\lambda)^{2}} \end{align}

After this I'm not able to find it's minimum value to guess the answer from options.

Any hint or suggestion will be of great help.

Moreover, Is there any other method to solve this problem?

Can we solve this without assuming the values for $\vec{a} ,\vec{b}$ and $\vec{c}$?

Thanks.

Answer 1

I don't believe you need to find the minimum value of your expression. Instead, I would ask myself, if the expression

$$\frac{1}{\sqrt{2}}\sqrt{(\lambda-\mu)^{2}+ (\mu-\nu)^{2}+ (\nu-\lambda)^{2}}$$

can somehow attain the value $\sqrt{3}$, $2$, $1$, and $0$?

Starting backwards, if it is supposed to be equal to $0$, then we must have $\lambda=\mu=\nu$, but this is not allowed by the assumption that the magnitudes are different.

Can it be equal to $1$, then? In that case, you must have that

$$\sqrt{(\lambda-\mu)^{2}+ (\mu-\nu)^{2}+ (\nu-\lambda)^{2}}=\sqrt{2},$$ or $$(\lambda-\mu)^{2}+ (\mu-\nu)^{2}+ (\nu-\lambda)^{2}=2.$$

Now I need to consider the second assumption that the magnitudes are integral. Then each square is at least $1$, because the magnitudes are different, but then the sum of the squares is at least $3$: The equality is not possible.

We now know that both $0$ and $1$ are impossible values for the expression to obtain, and maybe you now have an idea on how to approach the problem for the other cases.

Answer 2

I found a way to calculate the minimum value of

$\sqrt{(\lambda-\mu)^{2}+ (\mu-\nu)^{2}+ (\nu-\lambda)^{2}}$

Sharing for other users who may need in future.

If we take $\lambda$, $\mu$, $\nu $ as three consecutive integers Example $1, 2,3$.

$\lambda-\mu=1$ $\Rightarrow$ $(\lambda-\mu)^{2}=1$

$\mu-\nu=1$ $\Rightarrow$$(\mu-\nu)^{2}=1$

$\nu-\lambda=2$ $\Rightarrow$ $(\nu-\lambda)^{2}=4$

$\frac{1}{\sqrt{2} } \sqrt{(\lambda -\mu )^{2}+(\mu -\nu )^{2}+(\nu -\lambda )^{2}}$$ \geq$$ \frac{1}{\sqrt{2} } \sqrt{1+1+4}$=$\sqrt {3}$

$\Rightarrow$ $\left| \vec{p} +\vec{q} +\vec{r} \right| $ can take values equal to $\sqrt{3}$ and $2$.