Question

If $\overrightarrow{a}=\hat{i},\overrightarrow{b}=-\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$ and $\hat{c}=-\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}$ are three coplanar unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$. If three other vectors $\overrightarrow{p},\overrightarrow{q}$ and $\overrightarrow{r}$ are parallel to $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$, respectively, and have integral but different magnitudes, then among the following options, $|\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}|$ can take a value equal to

• A. $1$
• B. $0$
• C. $\sqrt{3}$
• D. $2$

Answer 1

Answer: (C), (D)

$2$
$\sqrt{3}$

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ lie in the x-y plane.

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

Therefore,

$|\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}|=|\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\nu \overrightarrow{c}|$

$=|\lambda \hat{i}+\mu (-\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})+\nu (-\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j})|$

$=|(\lambda -\frac{\mu }{2}-\frac{\nu }{2})\hat{i}+\frac{\sqrt{3}}{2}(\mu -\nu \left)\hat{j}\right|$

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

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

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

$\ge \frac{1}{\sqrt{2}}\sqrt{1+1+4}=\sqrt{3}$

Hence, $|\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r}|$ can take a value equal to $\sqrt{3}$ and 2.