Q.Find the sum of the vectors $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the median $AD$ of $\triangle ABC$ is bisected at the point $E$ and $BE$ is produced to meet the side $AC$ at $F$. Then the vector $\overline{BF} =$ (A) $\dfrac{3}{2}\, \overline{EF}$ (B) $2\, \overline{EF}$ (C) $3\, \overline{EF}$ (D) $4\, \overline{EF}$

   ›Reveal solutionSolution

   Placing $B$ at the origin and writing $D,E,F$ as combinations of $\vec A=\bar a,\vec C=\bar c$ shows $F$ divides $AC$ in ratio $1:2$ from $A$, and $BF=4EF$ along the same line.

   Concept and Intuition

   $E$ (midpoint of the median $AD$) is the point that, when joined to a vertex and extended, splits the opposite side in a fixed ratio — a standard "median of a median" vector construction, best handled by placing one vertex at the origin so position vectors of the other two act as a basis.

   Step-by-Step Solution

   1. Take $B$ as origin, so $\vec B=0$. Let $\vec A=\bar a$, $\vec C=\bar c$.
   2. $D$ = midpoint of $BC$ = $\bar c/2$.
   3. $E$ = midpoint of $AD$ = $\dfrac{\bar a+\bar c/2}{2}=\dfrac{\bar a}{2}+\dfrac{\bar c}{4}$.
   4. Since $B=0$, any point on line $BE$ is $\lambda E$ for some scalar $\lambda$. Write $F$ on $AC$ as $F=(1-t)\bar a+t\bar c$.
   5. Equate: $(1-t)\bar a+t\bar c=\lambda\left(\dfrac{\bar a}{2}+\dfrac{\bar c}{4}\right)$. Matching coefficients: $1-t=\lambda/2$, $t=\lambda/4$. Solving: $t=\tfrac13,\ \lambda=\tfrac43$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. If $\overline{OA} = \bar{a}, \overline{OD} = \bar{d}, \dfrac{OA}{CB} = 2$ and $\dfrac{OD}{AB} = \dfrac{1}{3}$, then $\overline{AD} + \overline{OC} + \overline{DC} =$ (A) $\bar{d} + \bar{a}$ (B) $5\bar{a} + 3\bar{d}$ (C) $6\bar{d}$ (D) $7\bar{a}$

   ›Reveal solutionSolution

   Using $O$ as the origin and the given parallel-side ratios to pin down $B$ and $C$ in terms of $\bar a,\bar d$, the required sum of vectors telescopes to $6\bar d$ (the $\bar a$ terms cancel).

   Concept and Intuition

   In a pentagon built from two pairs of parallel sides with known ratios, every vertex's position vector can be written in terms of the two "free" vectors $\bar a=\overline{OA}$ and $\bar d=\overline{OD}$. Once all vertices are pinned down, any combination of side vectors reduces to simple algebra.

   Step-by-Step Solution

   1. Let $O$ be the origin, so $A=\bar a$, $D=\bar d$.
   2. $\overline{OD}\parallel\overline{AB}$ with $\dfrac{OD}{AB}=\dfrac13$ means $\overline{AB}=3\,\overline{OD}=3\bar d$ (same sense), so $B=A+3\bar d=\bar a+3\bar d$.
   3. $\overline{OA}\parallel\overline{CB}$ with $\dfrac{OA}{CB}=2$ means $\overline{CB}=\tfrac12\overline{OA}=\tfrac12\bar a$, i.e. $B-C=\tfrac12\bar a\Rightarrow C=B-\tfrac12\bar a=\left(\bar a+3\bar d\right)-\tfrac12\bar a=\tfrac12\bar a+3\bar d$.
   4. Now compute each required vector:
      • $\overline{AD}=D-A=\bar d-\bar a$.
      • $\overline{OC}=C=\tfrac12\bar a+3\bar d$.
      • $\overline{DC}=C-D=\left(\tfrac12\bar a+3\bar d\right)-\bar d=\tfrac12\bar a+2\bar d$. …

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3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.In a regular hexagon $ABCDEF$, $\overline{AB} = \bar{a}$ and $\overline{BC} = \bar{b}$, then $\overline{FA} =$ (A) $\bar{a} - \bar{b}$ (B) $\bar{a} + \bar{b}$ (C) $\bar{b} - \bar{a}$ (D) $2\bar{b} - \bar{a}$

   ›Reveal solutionSolution

   In a regular hexagon $ABCDEF$ with $\overline{AB}=\bar a,\ \overline{BC}=\bar b$, the closing side $\overline{FA}=\bar a-\bar b$.

   Concept and Intuition

   In a regular hexagon, opposite sides are parallel and equal in magnitude but point in opposite directions, and consecutive side-vectors are related by a fixed $60^\circ$ rotation. Rather than track this abstractly, placing coordinates on a unit circle makes the vector relations immediate and safe from sign errors.

   Step-by-Step Solution

   1. Place the regular hexagon's vertices on a unit circle at angles $0^\circ,60^\circ,120^\circ,180^\circ,240^\circ,300^\circ$: $A=(1,0),\ B=(0.5,0.866),\ C=(-0.5,0.866),\ D=(-1,0),\ E=(-0.5,-0.866),\ F=(0.5,-0.866)$.
   2. Compute $\bar a=\overline{AB}=B-A=(-0.5,0.866)$ and $\bar b=\overline{BC}=C-B=(-1,0)$.
   3. Compute $\overline{FA}=A-F=(1-0.5,\,0-(-0.866))=(0.5,0.866)$. …

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4. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If L, M, N are the mid points of the sides PQ, QR and RP of $\triangle PQR$ respectively, then $\overline{QM} + \overline{LN} + \overline{ML} + \overline{RN} - \overline{MN} - \overline{QL} =$ (A) $\overline{PQ} + \overline{QR} + \overline{LM} + \overline{MN}$ (B) $\overline{LP} + \overline{PM} + \overline{MQ}$ (C) $\overline{PQ} + \overline{QR} - \overline{PR}$ (D) $\overline{LM} + \overline{MN} + \overline{NR}$

   ›Reveal solutionSolution

   The six-term vector sum on the left simplifies exactly to the zero vector; among the choices only $\overline{PQ}+\overline{QR}-\overline{PR}$ is identically zero, so that is the match.

   Concept and Intuition

   Midpoints of a triangle's sides are just averages of the vertex position vectors. Any vector joining two of the six points $P,Q,R,L,M,N$ can be written as a combination of $P,Q,R$, so a sum of several such vectors can always be reduced to a single combination of $P,Q,R$ and simplified.

   Step-by-Step Solution

   1. Let $P,Q,R$ be the position vectors of the vertices. Then $L=\dfrac{P+Q}{2},\; M=\dfrac{Q+R}{2},\; N=\dfrac{R+P}{2}$.
   2. Compute each vector needed: $\overline{QM}=M-Q=\dfrac{R-Q}{2}$, $\quad \overline{LN}=N-L=\dfrac{R-Q}{2}$, $\overline{ML}=L-M=\dfrac{P-R}{2}$, $\quad \overline{RN}=N-R=\dfrac{P-R}{2}$, $\overline{MN}=N-M=\dfrac{P-Q}{2}$, $\quad \overline{QL}=L-Q=\dfrac{P-Q}{2}$.
   3. Add them with the signs given: $\overline{QM}+\overline{LN}+\overline{ML}+\overline{RN}-\overline{MN}-\overline{QL} = (R-Q) + (P-R) - (P-Q) = 0$. …

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5. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $\overline{AB} = 2\bar{i} + 3\bar{j} - 6\bar{k}$; $\overline{BC} = 6\bar{i} - 2\bar{j} + 3\bar{k}$ are the vectors along two sides of a triangle ABC, then perimeter of triangle ABC is: (A) $21$ (B) $\sqrt{74} + 14$ (C) $\sqrt{74} + 19$ (D) $\sqrt{74} + 3$

   ›Reveal solutionSolution

   The third side is found from $\overline{CA}=-(\overline{AB}+\overline{BC})$ (vectors around a closed triangle sum to zero); the perimeter is $14+\sqrt{74}$.

   Concept and Intuition

   For any triangle traversed $A\to B\to C\to A$, the position-difference vectors satisfy $\overline{AB}+\overline{BC}+\overline{CA}=\bar 0$, since you return to your starting point. This lets us find the third side vector without knowing actual coordinates of $A,B,C$.

   Step-by-Step Solution

   1. $|\overline{AB}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7$.
   2. $|\overline{BC}|=\sqrt{6^2+(-2)^2+3^2}=\sqrt{36+4+9}=\sqrt{49}=7$.
   3. $\overline{AB}+\overline{BC}=(2+6,\ 3-2,\ -6+3)=(8,1,-3)$.
   4. Since $\overline{AB}+\overline{BC}+\overline{CA}=\bar0$, $\overline{CA}=-(8,1,-3)=(-8,-1,3)$.
   5. $|\overline{CA}|=\sqrt{(-8)^2+(-1)^2+3^2}=\sqrt{64+1+9}=\sqrt{74}$.
   6. Perimeter $=|AB|+|BC|+|CA|=7+7+\sqrt{74}=14+\sqrt{74}$. …

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6. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Three vectors of magnitudes $a, 2a, 3a$ are along the directions of the diagonals of 3 adjacent faces of a cube that meet in a point. Then the magnitude of the sum of those diagonals is (A) $4a$ (B) $5a$ (C) $6a$ (D) $8a$

   ›Reveal solutionSolution

   Three face-diagonal directions from one cube vertex are mutually inclined at $60°$; combining vectors of magnitude $a,2a,3a$ along them gives a resultant of magnitude $5a$.

   Concept and Intuition

   Place the cube vertex at the origin with edges along the coordinate axes. The face diagonals of the three faces meeting at that vertex point along directions like $(1,1,0)$, $(0,1,1)$, $(1,0,1)$ (up to scale). Computing the angle between any two of these using the dot product shows it is always $60°$ — a fixed geometric fact about a cube, independent of its size. This turns a 3D vector-addition problem into a plain application of the law of cosines (extended to three vectors) once the mutual angle is known.

   Step-by-Step Solution

   1. Take a unit cube with a vertex at the origin. The three face diagonals from that vertex (on the $xy$, $yz$, $zx$ faces) point along $\hat d_1=\frac{1}{\sqrt2}(1,1,0)$, $\hat d_2=\frac{1}{\sqrt2}(0,1,1)$, $\hat d_3=\frac{1}{\sqrt2}(1,0,1)$.
   2. Check the mutual angle: $\hat d_1\cdot\hat d_2=\frac12(0+1+0)=\frac12=\cos60°$. By symmetry all three pairs give the same $60°$. …

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7. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.In a $\triangle ABC$, $|\overline{CB}| = \bar{a}$, $|\overline{CA}| = \bar{b}$, $|\overline{AB}| = \bar{c}$ and CD is the median through the vertex C. Then $\overline{CA}.\overline{CD} =$ (A) $\dfrac{1}{4}(3a^2+b^2-c^2)$ (B) $\dfrac{1}{4}(a^2+3b^2-c^2)$ (C) $\dfrac{1}{4}(a^2+b^2-3c^2)$ (D) $\dfrac{1}{4}(-3a^2-b^2+c^2)$

   ›Reveal solutionSolution

   Writing the median vector as the average of the two side-vectors and using the law-of-cosines relation for $\vec a\cdot\vec b$ gives $\vec{CA}\cdot\vec{CD}=\frac14(a^2+3b^2-c^2)$.

   Concept and Intuition

   Vector dot products of triangle sides connect directly to the law of cosines: expanding $|\vec{AB}|^2=|\vec a-\vec b|^2$ produces the dot product $\vec a\cdot\vec b$ in terms of the side lengths $a,b,c$. The median to a side is naturally the average of the two vectors from the opposite vertex, since the midpoint's position vector is the average of the endpoints.

   Step-by-Step Solution

   1. Take $C$ as the origin. Then $\vec{CA}=\vec b$ (magnitude $b$) and $\vec{CB}=\vec a$ (magnitude $a$).
   2. $D$, the midpoint of $AB$, has position vector $\vec{CD}=\dfrac{\vec a+\vec b}{2}$ (average of $A$ and $B$'s position vectors from $C$).
   3. $\vec{CA}\cdot\vec{CD} = \vec b\cdot\dfrac{\vec a+\vec b}{2} = \dfrac{\vec a\cdot\vec b}{2}+\dfrac{b^2}{2}$. …

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8. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If C is the midpoint of the line segment $\overline{AB}$ and P is any point outside the line AB, then (A) $\overline{PA}+\overline{PB}+2\overline{PC}=\bar{0}$ (B) $\overline{PA}+\overline{PB}+\overline{PC}=\bar{0}$ (C) $\overline{PA}+\overline{PB}=2\overline{PC}$ (D) $\overline{PA}+\overline{PB}=\overline{PC}$

   ›Reveal solutionSolution

   This tests the midpoint vector identity: for $C$ the midpoint of $AB$ and any external point $P$,

   $\overline{PA}+\overline{PB}=2\overline{PC}$.

   Concept and Intuition

   The position vector of the midpoint of a segment is the average of the endpoints' position

   vectors. Writing every vector $\overline{PX}$ as $X-P$ (position vector of $X$ minus that of $P$,

   taking any convenient origin) turns this geometric fact into simple algebra.

   Step-by-Step Solution

   1. $C$ is the midpoint of $\overline{AB}$, so (taking any origin) $C=\dfrac{A+B}{2}$, i.e. $A+B=2C$.
   2. $\overline{PA}=A-P$ and $\overline{PB}=B-P$.
   3. Adding: $\overline{PA}+\overline{PB}=(A+B)-2P = 2C-2P = 2(C-P)$.
   4. Since $\overline{PC}=C-P$, this is exactly $\overline{PA}+\overline{PB}=2\overline{PC}$. …

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9. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.OABCD is a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel. Also, it is given that $\dfrac{OA}{CB}=2, \dfrac{OD}{AB}=\dfrac{1}{3}$. If $\overline{OA}=\bar a, \overline{OD}=\bar d$, then $\overline{AD}+\overline{OC}+\overline{DC} =$ (A) $\bar d - \bar a$ (B) $\dfrac{1}{2}\bar a + 3\bar d$ (C) $\dfrac{1}{2}\bar a + 2\bar d$ (D) $6\bar d$

   ›Reveal solutionSolution

   Express every vertex of the pentagon in terms of $\bar a=\overline{OA}$ and $\bar d=\overline{OD}$ using the two given parallel/ratio conditions, then add the three requested vectors.

   Concept and Intuition

   A pentagon with two pairs of parallel sides can be fully built from two independent vectors ($\bar a$ and $\bar d$) plus the given ratios — exactly like building a trapezoid from its two parallel sides. Once every vertex's position vector (relative to $O$) is known, any requested vector sum is pure algebra.

   Step-by-Step Solution

   1. Take $O$ as the origin. $A=\bar a$, $D=\bar d$.
   2. $\overline{OD}/\overline{AB} = 1/3$ with $\overline{OD}\parallel\overline{AB}$ (same sense, as in a trapezoid's parallel sides) gives $\overline{AB}=3\bar d$, so $B = A+\overline{AB} = \bar a+3\bar d$.
   3. $\overline{OA}/\overline{CB}=2$ with $\overline{OA}\parallel\overline{CB}$ gives $\overline{CB}=\bar a/2$, i.e. $\overline{BC}=-\bar a/2$, so $C = B+\overline{BC} = \bar a+3\bar d-\bar a/2 = \bar a/2+3\bar d$. …

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10. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $D, E$ and $F$ are respectively mid points of $AB, AC$ and $BC$ in $\triangle ABC$, then $\vec{BE} + \vec{AF}$ is equal to (A) $\vec{DC}$ (B) $\dfrac{3}{2}\vec{BF}$ (C) $\dfrac{1}{2}\vec{BF}$ (D) $\dfrac{1}{2}\vec{DC}$

    ›Reveal solutionSolution

    Expressing every midpoint in terms of the triangle's position vectors and adding shows $\vec{BE}+\vec{AF}=\vec{DC}$.

    Concept and Intuition

    Midpoint vector problems become simple bookkeeping once every named point is written as a position vector combination of the triangle's vertices. Vector addition of two such combinations is just adding coefficients.

    Step-by-Step Solution

    1. Let $A,B,C$ denote the position vectors of the vertices.
    2. $D$ = midpoint of $AB$: $D=\dfrac{A+B}{2}$. $E$ = midpoint of $AC$: $E=\dfrac{A+C}{2}$. $F$ = midpoint of $BC$: $F=\dfrac{B+C}{2}$.
    3. $\vec{BE} = E - B = \dfrac{A+C}{2}-B = \dfrac{A+C-2B}{2}$.
    4. $\vec{AF} = F - A = \dfrac{B+C}{2}-A = \dfrac{B+C-2A}{2}$.
    5. Sum: $\vec{BE}+\vec{AF} = \dfrac{(A+C-2B)+(B+C-2A)}{2} = \dfrac{2C-A-B}{2}$. …

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11. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Let $A,B,C$ be three points on a circle of radius $R$. if $O$ is the centre of the circle and $\angle AOB=45^\circ$, $\angle BOC=45^\circ$ then the resultant of $\overrightarrow{OA}, \overrightarrow{OB}$ and $\overrightarrow{OC}$ has magnitude (A) $2R$ (B) $(\sqrt{2}+1)R$ (C) $2\sqrt{2}\,R$ (D) $4\sqrt{2}\,R$

    ›Reveal solutionSolution

    Placing the three points symmetrically about $B$ makes the $y$-components of $\vec{OA}$ and $\vec{OC}$ cancel, leaving a resultant of magnitude $(\sqrt2+1)R$.

    Concept and Intuition

    With equal central angles on either side of $B$, the configuration is symmetric — the perpendicular components of $\vec{OA}$ and $\vec{OC}$ cancel, and only the components along $\vec{OB}$'s direction add up.

    Step-by-Step Solution

    1. Set up coordinates with $O$ at the origin and $B$ along the angle $0^\circ$: $\vec{OB}=R(1,0)$.
    2. Since $\angle AOB=45^\circ$, place $A$ at angle $45^\circ$: $\vec{OA}=R(\cos45^\circ,\sin45^\circ)$.
    3. Since $\angle BOC=45^\circ$, place $C$ at angle $-45^\circ$: $\vec{OC}=R(\cos45^\circ,-\sin45^\circ)$.
    4. Sum: $\vec{OA}+\vec{OB}+\vec{OC}=R\left[(\cos45^\circ+1+\cos45^\circ),\ (\sin45^\circ+0-\sin45^\circ)\right]=R\left[1+2\cos45^\circ,\ 0\right]$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.In $\triangle OAC$, if $B$ is the midpoint of side $AC$ and $\overrightarrow{OA} = \vec{a}, \overrightarrow{OB} = \vec{b}$ then $\overrightarrow{OC} =$ (A) $2\vec{b} - \vec{a}$ (B) $\vec{b} - 2\vec{a}$ (C) $\vec{a} - 2\vec{b}$ (D) $\vec{a} - \vec{b}$

    ›Reveal solutionSolution

    The midpoint formula for position vectors directly gives $\overrightarrow{OC}=2\vec b-\vec a$.

    Concept and Intuition

    If $B$ is the midpoint of segment $AC$, then the position vector of $B$ (from any common origin $O$) is the average of the position vectors of $A$ and $C$: $\vec{OB} = \dfrac{\vec{OA}+\vec{OC}}{2}$. This is just the vector form of "midpoint = average of endpoints".

    Step-by-Step Solution

    1. Given $\overrightarrow{OA}=\vec a$, $\overrightarrow{OB}=\vec b$, and $B$ the midpoint of $AC$.
    2. Midpoint formula: $\vec b = \dfrac{\vec a + \overrightarrow{OC}}{2}$.
    3. Multiply both sides by 2: $2\vec b = \vec a + \overrightarrow{OC}$. …

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