Question

$EFGH$ is a rhombus such the angle $EFG$ is ${60}^o$. The magnitude of vectors $\overrightarrow{FH}$ and $\left\{m\overrightarrow{EG}\right\}$ are equal where $m$ is a scalar. What is the value of $m$?

• A. $3$
• B. $1.5$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

Answer: (D)

$\sqrt{3}$

Let the side of rhombus

$\overrightarrow{EF}=\hat{i}+\hat{j}$

$\overrightarrow{FG}=\hat{j}+\hat{k}$

Here the angle between $\overrightarrow{EF},\overrightarrow{FG}$ is 60

$\overrightarrow{EF}\cdot \overrightarrow{FG}=\left|\overrightarrow{EF}\right|\left|\overrightarrow{FG}\right|\cos \theta$

$(\hat{i}+\hat{j})\cdot (\hat{j}+\hat{k})=\sqrt{1^2+1^2}\sqrt{1^2+1^2}\cos \theta$

$1=\sqrt{2}\sqrt{2}\cos \theta$

$\frac{1}{2}=\cos \theta$

$\theta =60$

Hence our vector satisfying the given condition

EG is the diagonal so by triangle law in triangle EFG

$\overrightarrow{EG}=\overrightarrow{EF}+\overrightarrow{FG}$

$\overrightarrow{EG}=\hat{i}+\hat{j}+\hat{j}+\hat{k}$

$\overrightarrow{EG}=\hat{i}+2\hat{j}+\hat{k}$

FH is also diagonal so in triangle FGH

$\overrightarrow{FH}=\overrightarrow{FG}+\overrightarrow{GH}$

but $\overrightarrow{GH}=-\overrightarrow{EF}$

$\overrightarrow{FH}=\overrightarrow{FG}-\overrightarrow{EF}$

$\overrightarrow{FH}=\hat{j}+\hat{k}-\hat{i}-\hat{j}$

$\overrightarrow{FH}=-\hat{i}+\hat{k}$

$m\left|\overrightarrow{FH}\right|=\left|\overrightarrow{EG}\right|$

$m\sqrt{(-1{)}^2+1^2}=\sqrt{1^2+2^2+1^2}$

$m\sqrt{1+1}=\sqrt{1+4+1}$

$m\sqrt{2}=\sqrt{6}$

$m=\frac{\sqrt{6}}{\sqrt{2}}$

$m=\sqrt{3}$