If 2 overrightarrow{a} + 3 overrightarrow{b} | vedclass

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(A) Given the equation: $2 \overrightarrow{a} + 3 \overrightarrow{b} - 5 \overrightarrow{c} = \overrightarrow{0}$Rearranging the terms,we get: $5 \ove

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$3: 2$ internally

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(A) Given the equation: $2 \overrightarrow{a} + 3 \overrightarrow{b} - 5 \overrightarrow{c} = \overrightarrow{0}$Rearranging the terms,we get: $5 \overrightarrow{c} = 2 \overrightarrow{a} + 3 \overrightarrow{b}$Dividing by $5$,we have: $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{5}$Since $2 + 3 = 5$,we can write this as: $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{2 + 3}$This is the section formula for internal division,which states that if a point $C$ divides the line segment $AB$ in the ratio $m:n$,its position vector is given by $\overrightarrow{c} = \frac{n \overrightarrow{a} + m \overrightarrow{b}}{m + n}$.Comparing $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{2 + 3}$ with the formula $\overrightarrow{c} = \frac{n \overrightarrow{a} + m \overrightarrow{b}}{m + n}$,we identify $n = 2$ and $m = 3$.Therefore,the point $C$ divides the segment $AB$ in the ratio $m:n = 3:2$ internally.

If $2 \overrightarrow{a} + 3 \overrightarrow{b} - 5 \overrightarrow{c} = \overrightarrow{0}$,then the ratio in which $\overrightarrow{c}$ divides $\overrightarrow{AB}$ is

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