Q.Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\vec{a} + \vec{b})$ and $(\vec{a} - 3\vec{b})$ externally in the ratio $1 : 2$. Also, show that P is the mid point of the line segment RQ.

Showing the 12 most recent of 62 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$R$ divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively in the ratio $2:1$ externally. $S$ divides $PQ$ internally in the ratio $2:1$. Then the position vector of the midpoint of the line joining $R$ and $S$ is ________ (A) $\frac{-5}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{5}{3}\hat{k}$ (B) $\frac{-5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k}$ (C) $\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{5}{3}\hat{k}$ (D) $\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k}$

   ›Reveal solutionSolution

   Applying the external and internal section formulas to find $R$ and $S$, then averaging them, gives $-\tfrac53\hat i+\tfrac23\hat j+\tfrac53\hat k$.

   Concept and Intuition

   The section formula for a point dividing $PQ$ in ratio $m:n$ is $\dfrac{mQ+nP}{m+n}$ internally and $\dfrac{mQ-nP}{m-n}$ externally. Applying both with $P=\hat i+2\hat j-\hat k$, $Q=-\hat i+\hat j+\hat k$ gives the two required points.

   Step-by-Step Solution

   1. External division $2:1$: $R=\dfrac{2Q-P}{2-1}=2Q-P$.
   2. $2Q=-2\hat i+2\hat j+2\hat k$; $R=(-2-1)\hat i+(2-2)\hat j+(2+1)\hat k=-3\hat i+0\hat j+3\hat k$.
   3. Internal division $2:1$: $S=\dfrac{2Q+P}{3}$.
   4. $2Q+P=(-2+1)\hat i+(2+2)\hat j+(2-1)\hat k=-\hat i+4\hat j+\hat k$, so $S=-\tfrac13\hat i+\tfrac43\hat j+\tfrac13\hat k$.
   5. Midpoint of $R$ and $S$: $M=\dfrac{R+S}{2}$. …

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2. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Let $\bar i-2\bar j+\bar k$, $\bar i+\bar j-2\bar k$, $2\bar i-\bar j-\bar k$ and $\bar i+\bar j+\bar k$ be the position vectors of four points A, B, C and D respectively. If a point P divides AB in the ratio $2:1$ internally and a point Q divides CD in the ratio $1:2$ externally, then the ratio in which the point with position vector $5\bar i-6\bar j-5\bar k$ divides PQ is (A) $2:1$ (B) $-2:1$ (C) $2:3$ (D) $-2:3$

   ›Reveal solutionSolution

   Compute $P$ (internal section of $AB$) and $Q$ (external section of $CD$) explicitly, then find in what ratio the given point divides $PQ$. Answer: $-2:1$.

   Concept and Intuition

   Section-formula problems are pure coordinate bookkeeping: internal division uses $\dfrac{nA+mB}{m+n}$ for ratio $m:n$; external division flips a sign, $\dfrac{mB-nA}{m-n}$ (equivalently substitute $n\to-n$ in the internal formula). Once $P,Q$ are known points, finding the ratio a third point divides $PQ$ in is a linear solve.

   Step-by-Step Solution

   1. $A=(1,-2,1)$, $B=(1,1,-2)$, $C=(2,-1,-1)$, $D=(1,1,1)$.
   2. $P$ divides $AB$ in ratio $2:1$ internally: $P=\dfrac{1\cdot A+2\cdot B}{2+1}=\dfrac{(1,-2,1)+(2,2,-4)}3=\dfrac{(3,0,-3)}3=(1,0,-1)$.
   3. $Q$ divides $CD$ in ratio $1:2$ externally: using the external form $Q=\dfrac{mD-nC}{m-n}$ with $m=1,n=2$: $Q=\dfrac{D-2C}{-1}=2C-D=(4,-2,-2)-(1,1,1)=(3,-3,-3)$.
   4. Let the point $R=(5,-6,-5)$ divide $PQ$ in ratio $m:n$ (i.e. $R=\dfrac{nP+mQ}{m+n}$). Using the $y$-coordinate (since $P_y=0$): $-6=\dfrac{-3m}{m+n}\Rightarrow -6(m+n)=-3m\Rightarrow -6n=3m\Rightarrow m=-2n$. …

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3. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.If P divides the line segment joining the points A $(1, 2, -1)$ and B $(-1, 0, 1)$ externally in the ratio $1 : 2$ and Q $= (1, 3, -1)$ then PQ $=$ (A) $\sqrt{10}$ (B) $3$ (C) $1$ (D) $\sqrt{13}$

   ›Reveal solutionSolution

   This tests the external-division section formula in 3D coordinate geometry; the answer is $PQ = 3$.

   Concept and Intuition

   Internal division of $AB$ in ratio $m:n$ gives $P = \dfrac{mB+nA}{m+n}$. External division uses the same idea but with a subtraction instead of addition (as if $n$ were negative): $P = \dfrac{mB-nA}{m-n}$. Geometrically, the external point lies on the line $AB$ extended, outside the segment.

   Step-by-Step Solution

   1. Here $A(1,2,-1)$, $B(-1,0,1)$, ratio $m:n = 1:2$.
   2. $P = \dfrac{1\cdot B - 2\cdot A}{1-2} = \dfrac{B-2A}{-1} = 2A - B$.
   3. $2A = (2,4,-2)$. So $P = (2-(-1),\ 4-0,\ -2-1) = (3,4,-3)$. …

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4. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $2\vec a+3\vec b-5\vec c=\vec 0$, then the ratio in which $\vec c$ divides $\vec{AB}$ is (A) 3 : 2 internally (B) 3 : 2 externally (C) 2 : 3 internally (D) 2 : 3 externally

   ›Reveal solutionSolution

   Rearranging the given vector equation into the section-formula shape shows $C$ divides $AB$ internally in the ratio $3:2$.

   Concept and Intuition

   The section formula says the point dividing $AB$ internally in ratio $m:n$ (from $A$ to $B$) has position vector $\dfrac{n\vec a+m\vec b}{m+n}$. So whenever a vector equation can be rearranged into that exact shape, the ratio can be read off directly from the coefficients.

   Step-by-Step Solution

   1. $2\vec a+3\vec b-5\vec c=\vec 0 \Rightarrow 5\vec c = 2\vec a+3\vec b \Rightarrow \vec c=\dfrac{2\vec a+3\vec b}{5}$.
   2. Compare with the section formula for a point dividing $AB$ internally in ratio $m:n$: $\dfrac{n\vec a+m\vec b}{m+n}$.
   3. Here $n=2$ (coefficient of $\vec a$) and $m=3$ (coefficient of $\vec b$), with $m+n=5$ matching the denominator. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Let $\overline{OA} = \bar{i} + 2\bar{j} - 4\bar{k}$ and $\overline{OB} = 3\bar{i} - 4\bar{j} - 2\bar{k}$ be the position vectors of two points $A$ and $B$. If a point $C$ divides the line segment $AB$ in the ratio 1:3 externally, then the position vector of a point which divides $OC$ in the ratio 4:1 internally is (A) $5(\bar{i} - \bar{j})$ (B) $\bar{i} - 4\bar{j} + 2\bar{k}$ (C) $4\bar{i} - 2\bar{j} + \bar{k}$ (D) $4(\bar{j} - \bar{k})$

   ›Reveal solutionSolution

   Apply the external section formula to locate $C$ on line $AB$, then apply the internal section formula on segment $OC$. Answer: $4(\bar j-\bar k)$.

   Concept and Intuition

   For points with position vectors $A,B$, the point dividing $AB$ internally in ratio $m:n$ is $\dfrac{mB+nA}{m+n}$, while the point dividing it externally in ratio $m:n$ is $\dfrac{mB-nA}{m-n}$ — the external version effectively places the dividing point beyond one of the endpoints. Once $C$ is found this way, dividing $OC$ internally is just the ordinary internal-section formula applied to the segment from the origin to $C$.

   Step-by-Step Solution

   1. $A=\overline{OA}=(1,2,-4)$, $B=\overline{OB}=(3,-4,-2)$.
   2. $C$ divides $AB$ externally in ratio $1:3$ ($m=1,n=3$): $C=\dfrac{mB-nA}{m-n}=\dfrac{1\cdot B-3\cdot A}{1-3}=\dfrac{B-3A}{-2}=\dfrac{3A-B}{2}$.
   3. Compute $3A=(3,6,-12)$, then $3A-B=(3-3,\,6-(-4),\,-12-(-2))=(0,10,-10)$.
   4. So $C=\dfrac{(0,10,-10)}{2}=(0,5,-5)$. …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $A=(1,2,3)$, $B=(3,4,7)$ and $C=(-3,-2,-5)$ are three points then the ratio in which the point C divides AB externally is (A) $2:3$ (B) $3:2$ (C) $4:3$ (D) $3:4$

   ›Reveal solutionSolution

   Tests external division of a segment in 3D using the section formula; the ratio is $2:3$.

   Concept and Intuition

   If $C$ divides $AB$ externally in ratio $m:n$, then $C=\dfrac{mB-nA}{m-n}$. Since $C$, $A$, $B$ are given, we can find $m:n$ from any one coordinate and confirm with the rest — a genuine external division must satisfy ALL three coordinates simultaneously.

   Step-by-Step Solution

   1. Let $C=\dfrac{mB-nA}{m-n}$. Using $x$-coordinates: $\dfrac{3m-n}{m-n}=-3 \Rightarrow 3m-n=-3m+3n \Rightarrow 6m=4n \Rightarrow \dfrac{m}{n}=\dfrac{2}{3}$.
   2. Check with $y$: $m=2,n=3\Rightarrow \dfrac{4(2)-2(3)}{2-3}=\dfrac{8-6}{-1}=-2$ ✓ (matches $C_y=-2$). …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.In $\triangle PQR$, $(4\overline{i} + 3\overline{j} + 6\overline{k}), (2\overline{i} + 2\overline{j} + 3\overline{k})$ and $(3\overline{i} + \overline{j} + 3\overline{k})$ are the position vectors of the vertices P, Q and R respectively. Then the position vector of the point of intersection of the angle bisector of P with QR is (A) $6\overline{i} + 5\overline{j} + 9\overline{k}$ (B) $2\overline{i} - \overline{j} + 3\overline{k}$ (C) $(5\overline{i} + 3\overline{j} - 2\overline{k})$ (D) $\dfrac{5}{2}\overline{i} + \dfrac{3}{2}\overline{j} + 3\overline{k}$

   ›Reveal solutionSolution

   This tests the angle-bisector-divides-opposite-side-in-ratio-of-adjacent-sides theorem in 3D vector form. Answer: $\dfrac52 i+\dfrac32 j+3k$.

   Concept and Intuition

   The internal bisector of angle $P$ in $\triangle PQR$ meets side $QR$ at a point dividing it in the ratio $PQ:PR$. Computing these two side lengths first tells us immediately whether the dividing point is the midpoint (when $PQ=PR$) or some other section point.

   Step-by-Step Solution

   1. $P=(4,3,6)$, $Q=(2,2,3)$, $R=(3,1,3)$.
   2. $\vec{PQ}=Q-P=(-2,-1,-3)$, so $PQ=\sqrt{4+1+9}=\sqrt{14}$.
   3. $\vec{PR}=R-P=(-1,-2,-3)$, so $PR=\sqrt{1+4+9}=\sqrt{14}$.
   4. Since $PQ=PR$, the bisector from $P$ divides $QR$ in ratio $1:1$ — i.e., it meets $QR$ at its midpoint.
   5. Midpoint $=\left(\dfrac{2+3}{2},\dfrac{2+1}{2},\dfrac{3+3}{2}\right)=\left(\dfrac52,\dfrac32,3\right)$. …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $\bar{i} + 2\bar{j} + \bar{k}$, $\alpha\bar{i} + 3\bar{j} + 2\bar{k}$, $-\bar{i} + 4\bar{j} + \beta\bar{k}$ are the position vectors of three points A, B, C, then the position vector of a point which divides BC in the ratio $\alpha + 1 : \beta$ is (A) $\left(\dfrac{-1}{4}, \dfrac{13}{4}, \dfrac{9}{4}\right)$ (B) $\left(\dfrac{-1}{3}, \dfrac{13}{3}, \dfrac{9}{3}\right)$ (C) $\left(\dfrac{5}{2}, \dfrac{7}{2}, \dfrac{6}{2}\right)$ (D) $\left(\dfrac{7}{3}, \dfrac{2}{3}, \dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   With A, B, C collinear, matching direction vectors pins down $\alpha$ and $\beta$, after which the section-formula point on BC is computed directly. The answer is (A).

   Concept and Intuition

   For a division ratio expressed using unknown parameters $\alpha,\beta$ to yield one specific numeric point (as the answer choices demand), those parameters must be fixed by a geometric condition on A, B, C — here, that they are collinear (a standard setup for this style of vector problem). Once $\alpha,\beta$ are pinned down, the section formula $\dfrac{n\vec B+m\vec C}{m+n}$ for the point dividing $BC$ in ratio $m:n$ finishes the problem.

   Step-by-Step Solution

   1. $\vec{AB} = \vec B-\vec A = (\alpha-1)\bar i + (3-2)\bar j+(2-1)\bar k = (\alpha-1)\bar i+\bar j+\bar k$.
   2. $\vec{AC} = \vec C-\vec A = (-1-1)\bar i+(4-2)\bar j+(\beta-1)\bar k = -2\bar i+2\bar j+(\beta-1)\bar k$.
   3. Collinearity requires $\vec{AB}=t\,\vec{AC}$ for some scalar $t$. Matching the $\bar j$ components: $1 = 2t \Rightarrow t=\dfrac12$.
   4. Matching $\bar i$: $\alpha-1 = -2t = -1 \Rightarrow \alpha=0$.
   5. Matching $\bar k$: $1 = (\beta-1)t = \dfrac{\beta-1}{2} \Rightarrow \beta-1=2\Rightarrow \beta=3$.
   6. So the required ratio is $\alpha+1:\beta = 1:3$.
   7. With $\alpha=0$: $\vec B = (0,3,2)$; with $\beta=3$: $\vec C=(-1,4,3)$. …

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9. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are position vectors of 4 points such that $2\bar{a}+3\bar{b}+5\bar{c}-10\bar{d} = \bar{0}$, then the ratio in which the line joining $\bar{c}$ and $\bar{d}$ divides the line segment joining $\bar{a}$ and $\bar{b}$ is (A) $2:3$ (B) $-1:2$ (C) $2:1$ (D) $3:2$

   ›Reveal solutionSolution

   The given vector equation can be rearranged into a form that expresses one point as a weighted combination of the others, revealing the ratio in which the line joining $\bar{c}$ and $\bar{d}$ divides the segment joining $\bar{a}$ and $\bar{b}$. The ratio is $3:2$, so the correct option is (D).

   We start with the vector equation:

   $2\bar{a} + 3\bar{b} + 5\bar{c} - 10\bar{d} = \bar{0}$

   Concept and Intuition

   The Section Formula in vectors says: If a point $P$ divides the line segment joining $A$ and $B$ in the ratio $m:n$ (internally or externally), then its position vector is $\frac{m\bar{b} + n\bar{a}}{m+n}$ (if $P$ is between $A$ and $B$, both $m,n>0$; if external, one is negative).

   Here, we want the ratio in which the line joining $\bar{c}$ and $\bar{d}$ divides the segment joining $\bar{a}$ and $\bar{b}$. That means: there is some point $P$ on line $AB$ that also lies on line $CD$. We need to find the ratio $AP:PB$ (or $AP:PB$ with sign).

   The trick: Rearrange the given equation so that $\bar{a}$ and $\bar{b}$ appear on one side, and $\bar{c}$ and $\bar{d}$ on the other, then compare with the section formula.

   Step-by-step solution

   1. Rearrange the equation to isolate terms involving $\bar{a}$ and $\bar{b}$ on one side:

   $2\bar{a} + 3\bar{b} = 10\bar{d} - 5\bar{c}$

   2. Factor the right-hand side to express it as a combination of $\bar{c}$ and $\bar{d}$:

   $2\bar{a} + 3\bar{b} = 5(2\bar{d} - \bar{c})$

   But we want a form like $\frac{m\bar{b} + n\bar{a}}{m+n}$ for the left side, and something like $\frac{p\bar{d} + q\bar{c}}{p+q}$ for the right side, because the point where the lines intersect must satisfy both.

   3. Divide both sides by the sum of coefficients on the left (which is $2+3=5$):

   $\frac{2\bar{a} + 3\bar{b}}{5} = \frac{10\bar{d} - 5\bar{c}}{5}$

   Simplify the right side:

   $\frac{2\bar{a} + 3\bar{b}}{5} = 2\bar{d} - \bar{c}$

   4. Interpret the left side using the section formula: $\frac{2\bar{a} + 3\bar{b}}{5}$ is the position vector of a point $P$ that divides $AB$ in the ratio $3:2$ (since the coefficient of $\bar{b}$ is $3$ and of $\bar{a}$ is $2$, and the denominator is the sum). Specifically, $P = \frac{3\bar{b} + 2\bar{a}}{3+2}$, so $AP:PB = 3:2$ (with $A$ at $\bar{a}$, $B$ at $\bar{b}$). …

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10. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.The position vectors of $A$ and $B$ are $(\hat{i} + \hat{j} + \hat{k})$ and $\left(\dfrac{1}{3}\hat{j} + \dfrac{1}{3}\hat{k}\right)$. If 'B' divides the line $AC$ in the ratio $2 : 1$, then position vector of 'C' is (A) $\left(\dfrac{1}{2}, 0, 0\right)$ (B) $\left(0, \dfrac{1}{3}, 0\right)$ (C) $\left(\dfrac{-1}{2}, \dfrac{-1}{2}, 0\right)$ (D) $\left(\dfrac{-1}{2}, 0, 0\right)$

    ›Reveal solutionSolution

    Using the section formula with B dividing AC in ratio 2:1, we solve for C and get $\left(-\frac12,0,0\right)$.

    Concept and Intuition

    "B divides AC in ratio $2:1$" means $AB:BC=2:1$, so B is closer to C. The section formula for a point dividing a segment in ratio $m:n$ (from the first point to the second) is $P=\dfrac{n\cdot(\text{first})+m\cdot(\text{second})}{m+n}$. Here we invert this to solve for the unknown endpoint C.

    Step-by-Step Solution

    1. $B$ divides $AC$ in ratio $2:1$ ($AB:BC=2:1$), so $B=\dfrac{1\cdot\vec A+2\cdot\vec C}{2+1}=\dfrac{\vec A+2\vec C}{3}$.
    2. Rearranged: $3\vec B = \vec A+2\vec C \Rightarrow \vec C=\dfrac{3\vec B-\vec A}{2}$. …

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.Let 'O' be the origin, $A$ and $B$ be two points with position vectors $-3\bar{i}-3\bar{j}+4\bar{k}$ and $4\bar{i}-4\bar{j}-3\bar{k}$ respectively. Let $P$ be a point such that the line drawn through $P$ parallel to $\overline{OB}$ meets $OA$ in $L$ and another line through $P$ parallel to $\overline{OA}$ meets $OB$ in $M$. If $L$ divides $OA$ in the ratio $2:3$ and $M$ divides $OB$ in the ratio $3:2$, then the distance from $O$ to $P$ is (A) $\dfrac{19}{5}$ (B) $\dfrac{\sqrt{389}}{5}$ (C) $\dfrac{\sqrt{341}}{5}$ (D) $\dfrac{21}{5}$

    ›Reveal solutionSolution

    The construction describes a parallelogram $OLPM$, so $\overrightarrow{OP}$ is simply the vector sum of $\overrightarrow{OL}$ and $\overrightarrow{OM}$. Answer: $19/5$.

    Concept and Intuition

    When a line through $P$ parallel to $OB$ meets $OA$ at $L$, and another line through $P$ parallel to $OA$ meets $OB$ at $M$, the quadrilateral $OLPM$ has $OL\parallel MP$ and $OM\parallel LP$ — exactly the definition of a parallelogram. So $\overrightarrow{OP}=\overrightarrow{OL}+\overrightarrow{OM}$ (parallelogram law), and we just need the position of $L$ on $OA$ and $M$ on $OB$.

    Step-by-Step Solution

    1. $L$ divides $OA$ in ratio $2:3$ (from $O$), so $\overrightarrow{OL}=\dfrac{2}{5}\overrightarrow{OA}=\dfrac25(-3,-3,4)=\left(-\dfrac65,-\dfrac65,\dfrac85\right)$.
    2. $M$ divides $OB$ in ratio $3:2$ (from $O$), so $\overrightarrow{OM}=\dfrac{3}{5}\overrightarrow{OB}=\dfrac35(4,-4,-3)=\left(\dfrac{12}5,-\dfrac{12}5,-\dfrac95\right)$. …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.PQRS is a quadrilateral and $\overline{PQ}=\bar{a}$, $\overline{QR}=\bar{b}$, $\overline{SP}=\bar{a}-\bar{b}$, M is the midpoint of QR and X is a point on $\overline{SM}$ such that $\overline{SX}=\dfrac{4}{5}\overline{SM}$. If $\overline{SM}=m(4\bar{a}-\bar{b})$ and $\overline{SX}=n(4\bar{a}-\bar{b})$, then $m+n=$ (A) $9/10$ (B) $10/9$ (C) $11/9$ (D) $4/3$

    ›Reveal solutionSolution

    This tests position-vector manipulation in a quadrilateral; working everything out from a chosen

    origin gives $m=1/2$, $n=2/5$, so $m+n=9/10$.

    Concept and Intuition

    When only relative vectors between points are given (like $\overline{PQ}, \overline{QR}, \overline{SP}$), the cleanest approach is to fix one point as the origin and express every other

    point's position vector in terms of the given vectors $\bar a,\bar b$. Midpoints and points dividing

    a segment in a given ratio then follow from simple averaging/ratio formulas.

    Step-by-Step Solution

    1. Choose $P$ as the origin, so $P=\bar 0$.
    2. $\overline{PQ}=\bar a \Rightarrow Q=\bar a$.
    3. $\overline{QR}=\bar b \Rightarrow R = Q+\bar b = \bar a+\bar b$.
    4. $\overline{SP}=\bar a-\bar b \Rightarrow P - S = \bar a-\bar b \Rightarrow S = P-(\bar a-\bar b)=\bar b-\bar a$.
    5. $M$ = midpoint of $QR$: $M=\dfrac{Q+R}{2}=\dfrac{\bar a+(\bar a+\bar b)}{2}=\bar a+\dfrac{\bar b}{2}$.
    6. $\overline{SM}=M-S=\left(\bar a+\dfrac{\bar b}{2}\right)-(\bar b-\bar a)=2\bar a-\dfrac{\bar b}{2} =\dfrac12(4\bar a-\bar b)$. Comparing with $\overline{SM}=m(4\bar a-\bar b)$: $m=\dfrac12$. …

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