Q.The position vector of the point which divides the join of points $2\vec{a}-3\vec{b}$ and $\vec{a}+\vec{b}$ in the ratio 3 : 1 is
(A) $\dfrac{3\vec{a}-2\vec{b}}{2}$
(B) $\dfrac{7\vec{a}-8\vec{b}}{4}$
(C) $\dfrac{3\vec{a}}{4}$
(D) $\dfrac{5\vec{a}}{4}$

Showing the 12 most recent of 62 on this concept.

1. 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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2. 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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3. 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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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 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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6. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If a point $C$ divides the line segment joining the points with the position vectors $2\bar{i}-3\bar{j}+2\bar{k}$ and $3\bar{i}-\bar{j}-2\bar{k}$ in the ratio $2:3$, then the distance of $C$ from the point with position vector $2\bar{i}-\bar{j}+\bar{k}$ is (A) $\dfrac{7}{5}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   The section formula gives $C=(12/5,-11/5,2/5)$; the distance from $C$ to $(2,-1,1)$ works out to exactly $7/5$.

   Concept and Intuition

   A point $C$ dividing segment $AB$ internally in ratio $m:n$ (i.e. $AC:CB=m:n$) has position vector $C=\dfrac{nA+mB}{m+n}$ — the "near" endpoint gets the larger weight. Once $C$'s coordinates are found, distance to another given point is just the standard 3D distance formula.

   Step-by-Step Solution

   1. $A=(2,-3,2)$, $B=(3,-1,-2)$, ratio $AC:CB=2:3$, so $m=2,n=3$: $C=\dfrac{3A+2B}{2+3}=\dfrac{3A+2B}5$.
   2. $3A=(6,-9,6)$; $2B=(6,-2,-4)$; sum $=(12,-11,2)$; divide by 5: $C=\left(\dfrac{12}5,-\dfrac{11}5,\dfrac25\right)$.
   3. Target point $P=(2,-1,1)=\left(\dfrac{10}5,-\dfrac55,\dfrac55\right)$ (converting to fifths for easy subtraction).
   4. $C-P = \left(\dfrac{12-10}5,\ \dfrac{-11+5}5,\ \dfrac{2-5}5\right) = \left(\dfrac25,-\dfrac65,-\dfrac35\right)$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\bar i + \bar j - \bar k$, $7\bar i - 2\bar j - 3\bar k$ and $-5\bar i - 2\bar j + 5\bar k$ are the position vectors of the points $A, B, C$ respectively, then the position vector of the point of intersection of the bisector of $\angle BAC$ and side $BC$ is (A) $\dfrac{1}{16}(27\bar i - 32\bar j + 2\bar k)$ (B) $\dfrac{1}{4}(7\bar i - 8\bar j + 2\bar k)$ (C) $\dfrac{1}{4}(7\bar i + 8\bar j + 2\bar k)$ (D) $\dfrac{1}{16}(28\bar i - 32\bar j + 2\bar k)$

   ›Reveal solutionSolution

   The angle bisector from a vertex divides the opposite side in the ratio of the two adjacent sides ($BD:DC=AB:AC$); applying the section formula gives $D=\dfrac14(7\bar i-8\bar j+2\bar k)$.

   Concept and Intuition

   The internal angle-bisector theorem says the bisector of $\angle A$ meets side $BC$ at a point $D$ such that $BD:DC=AB:AC$. Once that ratio is known, the section formula (weighted average of the endpoints, weighted inversely to the adjacent segment) locates $D$ exactly.

   Step-by-Step Solution

   1. $A=\bar i+\bar j-\bar k,\ B=7\bar i-2\bar j-3\bar k,\ C=-5\bar i-2\bar j+5\bar k$.
   2. $\overrightarrow{AB}=B-A=(6,-3,-2)$, so $AB=\sqrt{36+9+4}=\sqrt{49}=7$.
   3. $\overrightarrow{AC}=C-A=(-6,-3,6)$, so $AC=\sqrt{36+9+36}=\sqrt{81}=9$.
   4. By the angle-bisector theorem, $BD:DC=AB:AC=7:9$.
   5. By the section formula for internal division in ratio $m:n=7:9$: $D=\dfrac{n\cdot B+m\cdot C}{m+n}=\dfrac{9B+7C}{16}$. …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.On a line with direction cosines $l, m, n$, $A(x_1, y_1, z_1)$ is a fixed point. If $B = (x_1 + 4kl, y_1 + 4km, z_1 + 4kn)$ and $C = (x_1 + kl, y_1 + km, z_1 + kn)$ $(k>0)$ then the ratio in which the point B divides the line segment joining A and C is (A) 1 : 2 (B) 1 : -4 (C) 4 : -3 (D) 4 : 3

   ›Reveal solutionSolution

   Since $A,B,C$ are collinear points on a line with common direction cosines, the ratio in which one divides the segment joining the other two can be read off directly from the scalar multiples of the direction vector — here $B$ lies beyond $C$ (as seen from $A$), so it divides $AC$ externally, in ratio $4:-3$.

   Concept and Intuition

   All three points lie on the same line through direction $(l,m,n)$, so we can parametrize every point on the line by a single scalar $t$ via $A + t\,k(l,m,n)$. Comparing the multiples of $k(l,m,n)$ used for each point tells us instantly how far along the line each one sits, and hence the ratio of division — no need for full 3D distance/section-formula algebra.

   Step-by-Step Solution

   1. Write each point as $A + t\cdot k(l,m,n)$:

   • $A$ corresponds to $t=0$
   • $C = A+k(l,m,n)$ corresponds to $t=1$
   • $B = A+4k(l,m,n)$ corresponds to $t=4$

   2. We want the ratio in which $B$ divides the segment joining $A$ and $C$. If $B$ divides $AC$ in ratio $m:n$ (using the section formula $B=\dfrac{mC+nA}{m+n}$), then in terms of the parameter $t$:

   $\frac{m}{m+n} = \frac{t_B-t_A}{t_C-t_A} \text{ is not quite it — instead use: } m(t_C-t_B)=n(t_B-t_A)$

   More directly: $\dfrac{m}{n}=\dfrac{t_B-t_A}{t_C-t_B} = \dfrac{4-0}{1-4}=\dfrac{4}{-3}$. …

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10. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The coordinates of the point which divides the line joining the points $(2, 3, 4)$ and $(3, -4, 7)$ in the ratio $2 : 4$ externally is (A) $(10, 1, 1)$ (B) $(1, 10, 1)$ (C) $(10, -10, 10)$ (D) $(1, 1, 10)$

    ›Reveal solutionSolution

    Apply the external section formula with $A(2,3,4)$, $B(3,-4,7)$, ratio $2:4$. Answer: (B).

    Concept and Intuition

    External division extends the line segment $AB$ beyond one of its endpoints; the formula $P=\dfrac{mB-nA}{m-n}$ (note the minus sign in the denominator, unlike internal division's $m+n$) locates that point.

    Step-by-Step Solution

    1. $A=(2,3,4)$, $B=(3,-4,7)$, ratio $m:n=2:4$.
    2. External section formula: $P=\dfrac{mB-nA}{m-n}=\dfrac{2(3,-4,7)-4(2,3,4)}{2-4}$.
    3. Numerator: $2(3,-4,7)=(6,-8,14)$; $4(2,3,4)=(8,12,16)$; difference $=(6-8,\ -8-12,\ 14-16)=(-2,-20,-2)$. …

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11. 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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12. 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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