Q.Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

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 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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3. 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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4. 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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5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the plane $x - y + z + 4 = 0$ divides the line joining the points $P(2,3,-1)$ and $Q(1,4,-2)$ in the ratio $l:m$, then $l+m$ is (A) $1$ (B) $3$ (C) $-1$ (D) $4$

   ›Reveal solutionSolution

   Parametrize the point dividing the segment, substitute into the plane equation to find the dividing fraction, then convert to the ratio $l:m$.

   Concept and Intuition

   If a point divides $PQ$ in ratio $l:m$ from $P$, it can be written as $P+\dfrac{l}{l+m}(Q-P)$. Substituting into the plane's equation and solving for the fraction $t=\dfrac{l}{l+m}$ gives the ratio directly.

   Step-by-Step Solution

   1. $Q-P = (1-2,4-3,-2+1) = (-1,1,-1)$.
   2. General point: $(2-t,\ 3+t,\ -1-t)$ where $t=\dfrac{l}{l+m}$.
   3. Substitute into $x-y+z+4=0$: $(2-t)-(3+t)+(-1-t)+4 = 2-3t = 0 \Rightarrow t=\dfrac23$.
   4. So $\dfrac{l}{l+m}=\dfrac23 \Rightarrow \dfrac{l}{m} = \dfrac{2/3}{1/3}=2$, i.e. $l:m=2:1$. …

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6. 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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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 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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9. 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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10. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Suppose P and Q lie on $3x + 4y - 4 = 0$ and $5x - y - 4 = 0$ respectively. If the midpoint of PQ is $(1, 5)$, then the slope of the line passing through P and Q is (A) $\dfrac{83}{35}$ (B) $\dfrac{63}{35}$ (C) $\dfrac{-3}{4}$ (D) $\dfrac{3}{4}$

    ›Reveal solutionSolution

    Use the midpoint condition to express $Q$ in terms of $P$, substitute into $Q$'s line, solve, and compute the slope; the answer is $\frac{83}{35}$.

    Concept and Intuition

    Given that $P$ lies on one line and $Q$ on another, and their midpoint is a known point, we can express $Q$'s coordinates in terms of $P$'s using the midpoint formula, then use $Q$'s own line equation to solve for $P$ directly (a system of two linear equations).

    Step-by-Step Solution

    1. Let $P=(p_1,p_2)$ with $3p_1+4p_2=4$.
    2. Midpoint of $PQ$ is $(1,5)$, so $Q=(2-p_1,\,10-p_2)$.
    3. $Q$ lies on $5x-y-4=0$: $5(2-p_1)-(10-p_2)-4=0 \Rightarrow 10-5p_1-10+p_2-4=0 \Rightarrow -5p_1+p_2=4$, i.e. $p_2=4+5p_1$.
    4. Substitute into $P$'s equation: $3p_1+4(4+5p_1)=4 \Rightarrow 3p_1+16+20p_1=4\Rightarrow 23p_1=-12\Rightarrow p_1=-\dfrac{12}{23}$.
    5. $p_2=4+5\left(-\dfrac{12}{23}\right)=\dfrac{92-60}{23}=\dfrac{32}{23}$. …

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11. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If the point $P$ which divides the line segment joining $A(1,1,1)$ and $B(2,2,2)$ in the ratio $1:m$ lies on the plane $x + 2y + 3z - 1 = 0$, then $m =$ (A) $-\dfrac{3}{2}$ (B) $\dfrac{4}{3}$ (C) $-\dfrac{11}{5}$ (D) $-\dfrac{1}{2}$

    ›Reveal solutionSolution

    Since $A$ and $B$ have identical $x=y=z$ patterns, $P$ also has equal coordinates $p$; substituting into the plane equation and solving for $m$ gives $-11/5$.

    Concept and Intuition

    The section formula for a point dividing $AB$ in ratio $l:m$ (from $A$) is $P=\dfrac{l\,B+m\,A}{l+m}$. Because $A=(1,1,1)$ and $B=(2,2,2)$ both have all three coordinates equal, $P$'s three coordinates are automatically equal too — this collapses the 3D problem to a single unknown $p$, which we then plug into the plane equation.

    Step-by-Step Solution

    1. With ratio $1:m$, $P=\dfrac{1\cdot(2,2,2)+m\cdot(1,1,1)}{1+m}=\left(\dfrac{2+m}{1+m},\dfrac{2+m}{1+m},\dfrac{2+m}{1+m}\right)$.
    2. Let $p=\dfrac{2+m}{1+m}$ (all coordinates of $P$).
    3. Plane: $x+2y+3z-1=0 \Rightarrow p+2p+3p-1=0 \Rightarrow 6p=1 \Rightarrow p=\dfrac16$. …

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12. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.Let $A = (-3, -2, 7)$ and $B = (3, 1, -2)$. Let a plane perpendicular to the line segment AB divide AB in the ratio $2 : 1$. Then the intercept made by the plane on y-axis is (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{3}$ (C) $2$ (D) $-1$

    ›Reveal solutionSolution

    This tests finding a plane perpendicular to a line segment at a given internal division point, then reading off an axis intercept.

    Concept and Intuition

    A plane perpendicular to segment AB has $\vec{AB}$ as its normal vector; if it also passes through the point dividing AB in a given ratio, that point (found via the section formula) pins down the plane completely.

    Step-by-Step Solution

    1. Section formula for the point dividing AB in ratio $2:1$ (from A to B): $\dfrac{2B+1\cdot A}{3}$.
    2. $2B=(6,2,-4)$, plus $A=(-3,-2,7)$: sum $=(3,0,3)$; divide by 3: point $=(1,0,1)$.
    3. Normal to the plane $=\vec{AB}=B-A=(6,3,-9)$, simplify to $(2,1,-3)$. …

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