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Two forces are equal to $2 \overrightarrow{O A}$ and $3 \overrightarrow{B O}$, their resultant
being $\lambda \overrightarrow{O G}$, where $G$ is the point on $A B$ such that
$\frac{B G}{A G}=-\frac{2}{3} .$ What is the value of $\lambda ?$
Options:
being $\lambda \overrightarrow{O G}$, where $G$ is the point on $A B$ such that
$\frac{B G}{A G}=-\frac{2}{3} .$ What is the value of $\lambda ?$
Solution:
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Verified Answer
The correct answer is:
$-1$
$\overrightarrow{O G}=\frac{2 \overrightarrow{O A}-3 \overrightarrow{O B}}{2-3}$
$\overrightarrow{O G}=\frac{2 \overrightarrow{O A}-3 \overrightarrow{O B}}{-1}$
$-\overrightarrow{O G}=2 \overrightarrow{O A}-3 \overrightarrow{O B}$ ... (1)
$\lambda \overrightarrow{O G}=2 \overrightarrow{O A}+3 \overrightarrow{O B} \quad \cdots$ (2)
Adding (1) and (2)
$(\lambda-1) \overrightarrow{O G}=4 \overrightarrow{O A}$
$\Rightarrow \overrightarrow{O A}=\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}$ ... (3)
Subtracting (2) from (1)
$(-1-\lambda) \overrightarrow{O G}=-6 \overrightarrow{O B}$
$\overrightarrow{O B}=\frac{(1+\lambda)}{6} \overrightarrow{O G}$ ... (4)
From equ (2), (3) and (4)
$\lambda \overrightarrow{O G}=2\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}+3\left(\frac{\lambda+1}{6}\right) \overrightarrow{O G}$
$\Rightarrow \lambda=\frac{\lambda-1}{2}=\frac{\lambda+1}{3}$
$\therefore \lambda=\frac{\lambda-1}{2}$ or $\lambda=\frac{\lambda+1}{3}$
$\Rightarrow 2 \lambda-\lambda=-1$ or $3 \lambda-\lambda=1$
$\Rightarrow \lambda=-1$ or $\lambda=\frac{1}{2}$
$\overrightarrow{O G}=\frac{2 \overrightarrow{O A}-3 \overrightarrow{O B}}{-1}$
$-\overrightarrow{O G}=2 \overrightarrow{O A}-3 \overrightarrow{O B}$ ... (1)
$\lambda \overrightarrow{O G}=2 \overrightarrow{O A}+3 \overrightarrow{O B} \quad \cdots$ (2)
Adding (1) and (2)
$(\lambda-1) \overrightarrow{O G}=4 \overrightarrow{O A}$
$\Rightarrow \overrightarrow{O A}=\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}$ ... (3)
Subtracting (2) from (1)
$(-1-\lambda) \overrightarrow{O G}=-6 \overrightarrow{O B}$
$\overrightarrow{O B}=\frac{(1+\lambda)}{6} \overrightarrow{O G}$ ... (4)
From equ (2), (3) and (4)
$\lambda \overrightarrow{O G}=2\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}+3\left(\frac{\lambda+1}{6}\right) \overrightarrow{O G}$
$\Rightarrow \lambda=\frac{\lambda-1}{2}=\frac{\lambda+1}{3}$
$\therefore \lambda=\frac{\lambda-1}{2}$ or $\lambda=\frac{\lambda+1}{3}$
$\Rightarrow 2 \lambda-\lambda=-1$ or $3 \lambda-\lambda=1$
$\Rightarrow \lambda=-1$ or $\lambda=\frac{1}{2}$
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