Forces $3 \to O A$ , $5 \to O B$ act along OA and OB. If their resultant passes through C on AB, then :

C is mid-point of AB

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C divides AB in the ratio

2 : 1

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3 A C = 5 C B

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2 A C = 3 C B

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Solution

The correct option is C $3 A C = 5 C B$
Draw $O N$ perpendicular to $A B$
Let $^i$ be the unit vector along $O N$
The resultant force $\to R = 3 \to A + 5 \to B$ ...... $(1)$
The angles between $^i$ and the forces $\to R$ , $3 \to A$ and $5 \to B$ are $∠ C O N, ∠ A O N$ and $∠ B O N$ respectively.
$\to R .^i = 3 \to A .^i + 5 \to B .^i$
$\to R .1 . cos ∠ C O N = 3. - - \to O A .1 . cos ∠ A O N + 5 - - \to O B .1 . cos ∠ B O N$
$\to R . \frac{O N}{O C} = 3 - - \to O A \times \frac{O N}{O A} + 5 - - \to O B \times \frac{O N}{O B}$
$\Rightarrow \frac{\to R}{O C} = 3 + 5$
$\Rightarrow \to R = 8 - - \to O C$
We know that $- - \to O A = - - \to O C + - - \to C A$
$\Rightarrow 3 - - \to O A = 3 - - \to O C + 3 - - \to C A$ ....... $(1)$ by multiplying the complete equation by $3$
We know that $- - \to O B = - - \to O C + - - \to C B$
$\Rightarrow 5 - - \to O B = 5 - - \to O C + 5 - - \to C B$ ....... $(2)$ by multiplying the complete equation by $5$
Eqn $(1) + (2)$
$\Rightarrow 3 - - \to O A + 5 - - \to O B = 8 - - \to O C + 3 - - \to C A + 5 - - \to C B$
$\Rightarrow \to R = 8 - - \to O C + 3 - - \to C A + 5 - - \to C B$
$\Rightarrow 8 - - \to O C = 8 - - \to O C + 3 - - \to C A + 5 - - \to C B$
$\Rightarrow ∣ ∣ ∣ 3 - - \to C A ∣ ∣ ∣ = ∣ ∣ ∣ 5 - - \to C B ∣ ∣ ∣$
$∴ 3 A C = 5 C B$

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