The position vectors of the points A and B are 3 vec i +5 vec j and vec i -2 vec j respectively. If the point M divides the line segment AB internally in the ratio of 5/3 then find the position vector of M.

 The position vectors of the points A and B are 3 vec i +5 vec j and vec i -2 vec j respectively. If the point M divides the line segment AB internally in the ratio of 5/3 then find the position vector of M.

Solutions. 

To find the position vector of point M, which divides the line segment AB internally in the ratio of $\frac{5}{3}$, we can use the section formula. The section formula states that if a line segment AB is divided by a point M internally in the ratio $𝑚:𝑛$, then the position vector of M is given by:

$\text{Position vector of M}=\frac{𝑛\cdot \text{Position vector of A}+𝑚\cdot \text{Position vector of B}}{𝑚+𝑛}$

 M = n.a vector + m.b vec

/M+N

Given: Position vector of A ($\overrightarrow{𝑎}$) = $3\overrightarrow{𝑖}+5\overrightarrow{𝑗}$ Position vector of B ($\overrightarrow{𝑏}$) = $\overrightarrow{𝑖}-2\overrightarrow{𝑗}$ Ratio m:n = 5:3

Let's substitute these values into the formula:

$\text{Position vector of M}=\frac{3\cdot (3\overrightarrow{𝑖}+5\overrightarrow{𝑗})+5\cdot (\overrightarrow{𝑖}-2\overrightarrow{𝑗})}{5+3}$

$\text{Position vector of M}=\frac{9\overrightarrow{𝑖}+15\overrightarrow{𝑗}+5\overrightarrow{𝑖}-10\overrightarrow{𝑗}}{8}$

$\text{Position vector of M}=\frac{14\overrightarrow{𝑖}+5\overrightarrow{𝑗}}{8}$

Therefore, the position vector of point M is $\frac{14\overrightarrow{𝑖}+5\overrightarrow{𝑗}}{8}$.