If $A(\bar a)$ and $B(\bar b)$ are any two points in space and $R(\bar r)$ be a point on the line segment $AB$ dividing internally in the ratio $m:n$ then prove that $\bar r=\dfrac{m\bar b+n\bar a}{m+n}$.

If A(bar a) and B(bar b) are any two points in space and R(bar r) be a point on the line segment AB dividing i

Question 215 of 215

A(\bar a)

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 4mImportance★★★★★

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Use $\overrightarrow{AR}:\overrightarrow{RB}=m:n$ with position vectors, and solve for $\bar r$ .

Let $O$ be the origin, and let $\bar a,\bar b,\bar r$ be the position vectors of $A,B,R$ respectively, where $R$ lies on segment $AB$ such that $AR:RB=m:n$ .

Since $A,R,B$ are collinear with $R$ between $A$ and $B$ :

$\overrightarrow{AR}=\frac{m}{m+n}\overrightarrow{AB}$

(because $R$ divides $AB$ in ratio $m:n$ , so $R$ is at fraction $\dfrac{m}{m+n}$ of the way from $A$ to $B$ ).

Now, $\overrightarrow{AR}=\bar r-\bar a$ and $\overrightarrow{AB}=\bar b-\bar a$ , so:

$\bar r-\bar a=\frac{m}{m+n}(\bar b-\bar a)$

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL2 marks
$\bar a, \bar b, \bar c$
›Reveal solutionSolution

Rewrite $5\bar a-3\bar b-2\bar c=0$ as $\bar c=\dfrac{5\bar a-3\bar b}{5-3}$ , the external-division form.

$5\bar a-3\bar b-2\bar c=\bar0 \Rightarrow \bar c=\dfrac{5\bar a-3\bar b}{2}$

…

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CBSE 2023Set ANNUAL2 marks
$\bar a, \bar b, \bar c$
›Reveal solutionSolution

Rewrite $5\bar a+3\bar b-8\bar c=0$ as $\bar c=\dfrac{5\bar a+3\bar b}{5+3}$ and read off the section-formula weights.

$5\bar a+3\bar b-8\bar c=\bar0 \Rightarrow \bar c=\dfrac{5\bar a+3\bar b}{8}$

…

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CBSE 2016Set ANNUAL2 marks
$\bar{p} = \hat{i} - 2\hat{j} + \hat{k}$
›Reveal solutionSolution

Use the internal section formula $R=\dfrac{m\bar q+n\bar p}{m+n}$ for ratio $m:n=2:1$ .

Given $\bar p=\hat i-2\hat j+\hat k$ (P.V. of P) and $\bar q=\hat i+4\hat j-2\hat k$ (P.V. of Q). R divides PQ internally in ratio $2:1$ .

By the section formula, the position vector of R is:

$\bar r=\frac{2\bar q+1\cdot\bar p}{2+1}$

$=\frac{2(\hat i+4\hat j-2\hat k)+(\hat i-2\hat j+\hat k)}{3}$

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Q.If $A(\bar a)$ and $B(\bar b)$ are any two points in space and $R(\bar r)$ be a point on the line segment $AB$ dividing internally in the ratio $m:n$ then prove that $\bar r=\dfrac{m\bar b+n\bar a}{m+n}$ .

Concept understanding — Section Formula for Vectors

Q.If A(aˉ)A(\bar a)A(aˉ) and B(bˉ)B(\bar b)B(bˉ) are any two points in space and R(rˉ)R(\bar r)R(rˉ) be a point on the line segment ABABAB dividing internally in the ratio m:nm:nm:n then prove that rˉ=mbˉ+naˉm+n\bar r=\dfrac{m\bar b+n\bar a}{m+n}rˉ=m+nmbˉ+naˉ​.

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Q.If $A(\bar a)$ and $B(\bar b)$ are any two points in space and $R(\bar r)$ be a point on the line segment $AB$ dividing internally in the ratio $m:n$ then prove that $\bar r=\dfrac{m\bar b+n\bar a}{m+n}$ .

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