Q.Consider two points P and Q with position vectors $\overrightarrow{OP}=3\vec{a}-2\vec{b}$ and $\overrightarrow{OQ}=\vec{a}+\vec{b}$. Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1,

Concept: Section Formula — the position vector of a point dividing a segment in a given ratio is a weighted average of the endpoints.

Let $\vec{p} = 3\vec{a} - 2\vec{b}$ and $\vec{q} = \vec{a} + \vec{b}$.

(i) Internal division (ratio $2:1$)

Using $\vec{r} = \frac{m\vec{q} + n\vec{p}}{m+n}$ with $m:n = 2:1$:

(ii) External division (ratio $2:1$)

Using $\vec{r} = \frac{m\vec{q} - n\vec{p}}{m-n}$ with $m:n = 2:1$:

The core idea here is the section formula — a tool that tells us exactly where a point lies on a line joining two given points, based on the ratio in which it divides the segment. Think of it like a weighted average: if you want a point that is closer to P than to Q, you give more "weight" to P's position vector.

Why does this work? When dividing internally, R lies between P and Q. The vector from P to R is a fraction of the vector from P to Q, proportional to the ratio. When dividing externally, R lies beyond Q (or beyond P) on the extended line — one of the weights becomes negative to "push" the point outside the segment.

Let's apply this to our specific vectors.

1. Identify the given vectors and ratio.

   We have $\vec{p} = 3\vec{a} - 2\vec{b}$ and $\vec{q} = \vec{a} + \vec{b}$. The ratio is $2:1$, so $m=2$ and $n=1$.

2. Internal division (i).

   Using the internal formula:

Simplify the numerator:

$3\vec{a} - 2\vec{b} + 2\vec{a} + 2\vec{b} = (3+2)\vec{a} + (-2+2)\vec{b} = 5\vec{a}$

So:

Notice the $\vec{b}$ terms cancelled out — that's fine; it just means R lies along the direction of $\vec{a}$ from the origin.

3. External division (ii). Using the external formula:

Simplify the numerator:

$-3\vec{a} + 2\vec{b} + 2\vec{a} + 2\vec{b} = (-3+2)\vec{a} + (2+2)\vec{b} = -\vec{a} + 4\vec{b}$

Since $m-n = 1$, we get:

Method: Section formula (internal and external division) in vector form

Use this to locate the point $R$ dividing the segment $PQ$ (position vectors $\vec{p},\vec{q}$) in a ratio $m:n$.

Steps

Step 1: Choose internal or external and write the right formula.

Note the cross-weighting (the far endpoint $\vec{q}$ carries $m$) and that external division uses a minus sign and denominator $m-n$.

Step 2: Substitute the position vectors and the ratio.

Put in $\vec{p},\vec{q}$ (which may themselves be combinations like $3\vec{a}-2\vec{b}$) and the numbers $m,n$, then expand the numerator.

Step 3: Simplify by collecting like terms.

Group the coefficients of each base vector; some terms may cancel. The midpoint $\dfrac{\vec{p}+\vec{q}}{2}$ is just the internal case with $m=n$.

Common Mistakes

Mistake 1: Mixing up which endpoint carries $m$ and which carries $n$.

Why it's wrong: the section formula cross-weights — for ratio $PR:RQ=m:n$ the far point $\vec{q}$ gets $m$ and the near point $\vec{p}$ gets $n$; swapping them places $R$ on the wrong side. Correct approach: use $\vec{r}=\frac{m\vec{q}+n\vec{p}}{m+n}$ internally, keeping the cross-pairing.

Mistake 2: Using the internal formula (with $+$ and $m+n$) for external division.

Why it's wrong: external division needs a minus sign and denominator $m-n$: $\vec{r}=\frac{m\vec{q}-n\vec{p}}{m-n}$. Correct approach: switch to the external form, giving $-\vec{a}+4\vec{b}$ here, not the internal $\frac{5}{3}\vec{a}$.

Mistake 3: Being alarmed when a base vector cancels.

Why it's wrong: the $\vec{b}$-terms cancelling in the internal case (leaving $\frac{5}{3}\vec{a}$) is legitimate, not an error. Correct approach: collect like terms and accept a simplified result even if one vector disappears.