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$ABC$ is a triangle. $D$ divides $BC$ in the ratio $l:m$ and $G$ divides $AD$ in the ratio $\left( {l + m} \right):n$. Find the position vector of $D$ and $G$.

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Last updated date: 21st Jun 2024
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Answer
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Hint: Suppose the position vector of $B$ be $\overrightarrow b $ and position vector of $C$ be $\overrightarrow c $ and then use the formula of ratio $\overrightarrow d = \dfrac{{l\overrightarrow c + m\overrightarrow b }}{{l + m}}$ where $D$ divides $BC$ in the ratio $l:m$. Similarly find out the point $G$ and hence find the position vector of both the points.

Complete step-by-step answer:
According to the question, $ABC$ is a triangle and $D$ divides $BC$ in the ratio $l:m$ and $G$ divides $AD$ in the ratio $\left( {l + m} \right):n$ and we need to find the position vector of $D$ and $G$.
So first of all, we assume the position vector of $B$ be $\overrightarrow b $ and the position vector of $C$ be $\overrightarrow c $. According to the question, $D$ divides $BC$ in the ratio $l:m$ and hence we know the formula that if position vector of the two given points are given as $\overrightarrow a $ and $\overrightarrow b $ and the point $P$ divides it in the ratio $n:m$ and hence the formula for the position vector of point $\overrightarrow p = \dfrac{{\overrightarrow a m + \overrightarrow b n}}{{m + n}}$ .
So in this question, we have assumed that the position vector of $B$ and $C$ as $\overrightarrow b $ and $\overrightarrow c $ respectively and $BC$ is divided by $D$ in the ratio $l:m$. Hence we can get the position vector of $D$ as:
Position vector of $D = \dfrac{{l\overrightarrow c + m\overrightarrow b }}{{l + m}}$ .
Now let us assume that the position vector of point $A$ be and we proved that the position vector of $D$ be $\dfrac{{l\overrightarrow c + m\overrightarrow b }}{{l + m}}$ and $G$ divides $AD$ in the ratio $\left( {l + m} \right):n$ and we know the position vectors of $A$ and $D$. Hence the position vector of G will be:
Position vector of $G = \dfrac{{n\overrightarrow a + \left( {l + m} \right)\dfrac{{\left( {l\overrightarrow c + m\overrightarrow b } \right)}}{{\left( {l + m} \right)}}}}{{n + l + m}}$
                                    $ = \dfrac{{n\overrightarrow a + m\overrightarrow b + n\overrightarrow c }}{{n + m + l}}$
Hence position vector of $D = \dfrac{{l\overrightarrow c + m\overrightarrow b }}{{l + m}}$
And the position vector of $G$$ = \dfrac{{n\overrightarrow a + m\overrightarrow b + n\overrightarrow c }}{{n + m + l}}$.

Note: You should know that if $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is divided by the point $P\left( {x,y} \right)$ in the ratio $m:n$, then $P\left( {x,y} \right)$ is given by:
 $P\left( {x,y} \right)$$ = P\left( {\dfrac{{n{x_1} + m{x_2}}}{{n + m}},\dfrac{{n{y_1} + m{y_2}}}{{n + m}}} \right)$
Similarly if the position vector is given we can use a similar formula, just replacing the point $P\left( {x,y} \right)$ by the position vector namely $\overrightarrow a $.