# Two forces vecF_1=hati+5hatj and vecF_2=3hati-2hatj act at points with two position vectors respectively  hati and -3hati +14hatj How will you find out the position vector of the point at which the forces meet?

Jun 16, 2016

$3 \hat{i} + 10 \hat{j}$

#### Explanation:

The support line for force ${\vec{F}}_{1}$ is given by

${l}_{1} \to p = {p}_{1} + {\lambda}_{1} {\vec{F}}_{1}$

where $p = \left\{x , y\right\}$, ${p}_{1} = \left\{1 , 0\right\}$ and ${\lambda}_{1} \in \mathbb{R}$.
Analogously for ${l}_{2}$ we have

${l}_{2} \to p = {p}_{2} + {\lambda}_{2} {\vec{F}}_{2}$

where ${p}_{2} = \left\{- 3 , 14\right\}$ and ${\lambda}_{2} \in \mathbb{R}$.
The intersection point or ${l}_{1} \cap {l}_{2}$ is obtained equating

${p}_{1} + {\lambda}_{1} {\vec{F}}_{1} = {p}_{2} + {\lambda}_{2} {\vec{F}}_{2}$

and solving for ${\lambda}_{1} , {\lambda}_{2}$ giving

$\left\{{\lambda}_{1} = 2 , {\lambda}_{2} = 2\right\}$

so ${l}_{1} \cap {l}_{2}$ is at $\left\{3 , 10\right\}$ or $3 \hat{i} + 10 \hat{j}$

Jun 17, 2016

$\textcolor{red}{3 \hat{i} + 10 \hat{j}}$

#### Explanation:

Given

• $\text{The 1st force } {\vec{F}}_{1} = \hat{i} + 5 \hat{j}$
• $\text{The 2nd force } {\vec{F}}_{2} = 3 \hat{i} - 2 \hat{j}$
• ${\vec{F}}_{1} \text{ acts at point A with position vector } \hat{i}$
• ${\vec{F}}_{2} \text{ acts at point B with position vector } - 3 \hat{i} + 14 \hat{j}$

We are to find out the position vector of the point where the two given forces meet.

Let that point where the two given forces meet, be P with

position vector $\textcolor{b l u e}{x \hat{i} + y \hat{j}}$

$\text{Now displacement vector } \vec{A P} = \left(x - 1\right) \hat{i} + y \hat{j}$

$\text{And displacement vector } \vec{B P} = \left(x + 3\right) \hat{i} + \left(y - 14\right) \hat{j}$

$\text{Since " vec (AP) and vecF_1 " are collinear we can write}$

$\frac{x - 1}{1} = \frac{y}{5} \implies 5 x - y = 5. \ldots . . \left(1\right)$

$\text{Again " vec (BP) and vecF_2 " are collinear , so we can write}$

$\frac{x + 3}{3} = \frac{y - 14}{-} 2 \implies 2 x + 3 y = 36. \ldots . . \left(2\right)$

Now multiplying equation (1) by 3 and adding with equation (2) we get

$15 x + 2 x = 3 \times 5 + 36 \implies x = \frac{51}{17} = 3$

Inserting the value of x in equation (1)

$5 \times 3 - y = 5 \implies y = 10$

$\text{Hence the position vector of the point where the two given forces meet is } \textcolor{red}{3 \hat{i} + 10 \hat{j}}$