Q.If $\vec{a}$ and $\vec{b}$ are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that $BC = 1.5\,BA$.

Showing the 12 most recent of 62 on this concept.

1. 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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2. 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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3. 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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4. 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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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 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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7. 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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8. 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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9. 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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10. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.The position vectors of the points 'A' and 'B' with respect to 'O' are $2\hat{i} + 2\hat{j} + \hat{k}$ and $2\hat{i} + 4\hat{j} + 4\hat{k}$. The length of the internal bisector of $\angle BOA$ of $\triangle AOB$ is ________ (take proportionality Constant is 2) (A) $\dfrac{\sqrt{136}}{9}$ (B) $\dfrac{\sqrt{136}}{3}$ (C) $\dfrac{20}{3}$ (D) $\dfrac{25}{3}$

    ›Reveal solutionSolution

    The internal bisector from $O$ meets $AB$ at a point determined by the angle-bisector ratio $OA:OB=3:6=1:2$; its distance from $O$ works out to $\dfrac{\sqrt{136}}{3}$.

    Concept and Intuition

    The "length of the internal bisector" of an angle of a triangle conventionally means the length of the cevian from that vertex to the point where it meets the opposite side. The Angle Bisector Theorem tells us this point divides the opposite side in the ratio of the two adjacent sides, letting us locate it exactly with the section formula, then just measure the distance.

    Step-by-Step Solution

    1. Compute $|OA|=\sqrt{2^2+2^2+1^2}=\sqrt9=3$ and $|OB|=\sqrt{2^2+4^2+4^2}=\sqrt{36}=6$.
    2. The internal bisector of $\angle AOB$ meets side $AB$ at point $P$ with $AP:PB=OA:OB=3:6=1:2$.
    3. By the section formula, $P=\dfrac{2\vec A+1\cdot\vec B}{1+2}=\dfrac{2(2,2,1)+(2,4,4)}{3}=\dfrac{(4,4,2)+(2,4,4)}{3}=\dfrac{(6,8,6)}{3}=(2,\tfrac83,2)$. …

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11. 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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12. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.If C is a point on the straight line joining the points $A(-2+i)$ and $B(3-4i)$ in the Argand plane and $\dfrac{AC}{CB} = \dfrac{1}{2}$, then the argument of C is (A) $\operatorname{Tan}^{-1}3$ (B) $\operatorname{Tan}^{-1}2 - \pi$ (C) $\operatorname{Tan}^{-1}2$ (D) $\pi - \operatorname{Tan}^{-1}3$

    ›Reveal solutionSolution

    Section formula gives $C=\frac{-1-2i}{3}$, a third-quadrant point whose principal argument is $\tan^{-1}2-\pi$.

    Concept and Intuition

    In the Argand plane, "$C$ on segment $AB$ with $AC:CB=m:n$" is exactly the section-formula problem: $C=\dfrac{nA+mB}{m+n}$ (internal division). Once we have $C$'s coordinates, the argument is found from its quadrant using $\arctan(y/x)$, adjusted for quadrant.

    Step-by-Step Solution

    1. $A=-2+i,\ B=3-4i$, and $AC:CB=1:2$ means $C=\dfrac{2\cdot A+1\cdot B}{1+2}$ (larger weight to the farther point).
    2. $2A=-4+2i$. Adding $B=3-4i$: $2A+B=-1-2i$. Dividing by 3: $C=-\dfrac13-\dfrac23 i$.
    3. $C$ has both real and imaginary parts negative $\Rightarrow$ third quadrant. …

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