Q.In triangle $ABC$ (Fig 10.18), which of the following is not true: (A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$ (B) $\vec{AB} + \vec{BC} - \vec{AC} = \vec{0}$ (C) $\vec{AB} + \vec{BC} - \vec{CA} = \vec{0}$ (D) $\vec{AB} - \vec{CB} + \vec{CA} = \vec{0}$

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If '$C$' is the midpoint of line segment $AB$ and '$P$' is any point not on the line $AB$, then (A) $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=0$ (B) $\overrightarrow{PA}+\overrightarrow{PB}+2\overrightarrow{PC}=0$ (C) $\overrightarrow{PA}-\overrightarrow{PC}=\overrightarrow{PC}-\overrightarrow{PB}$ (D) $\overrightarrow{PA}+\overrightarrow{PB}-\overrightarrow{PC}=0$

   ›Reveal solutionSolution

   The midpoint formula in vector form, $\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}$, rearranges exactly to option (C).

   Concept and Intuition

   For any point P (on or off the line AB) and C the midpoint of AB, the position vector of C relative to P is the average of the position vectors of A and B relative to P — this is the vector form of the midpoint formula, and it holds regardless of where P is.

   Step-by-Step Solution

   1. Since C is the midpoint of segment AB, $\overrightarrow{PC}=\overrightarrow{PA}+\overrightarrow{AC}=\overrightarrow{PA}+\tfrac12\overrightarrow{AB}$.
   2. Also $\overrightarrow{AB}=\overrightarrow{PB}-\overrightarrow{PA}$, so $\overrightarrow{PC}=\overrightarrow{PA}+\tfrac12(\overrightarrow{PB}-\overrightarrow{PA})=\tfrac12(\overrightarrow{PA}+\overrightarrow{PB})$.
   3. So $2\overrightarrow{PC}=\overrightarrow{PA}+\overrightarrow{PB}$, i.e. $\overrightarrow{PA}+\overrightarrow{PB}-2\overrightarrow{PC}=0$. …

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2. 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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3. 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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4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$ then $\vec{CE}$ equals (A) $\vec{b} - \vec{a}$ (B) $-\vec{b}$ (C) $\vec{b} - 2\vec{a}$ (D) $\vec{a} - 2\vec{b}$

   ›Reveal solutionSolution

   Expressing all hexagon vertices via coordinates and solving for $\vec{CE}$ in terms of $\vec a=\vec{AB}$ and $\vec b=\vec{BC}$ gives $\vec b-2\vec a$.

   Concept and Intuition

   Once two consecutive side vectors of a regular hexagon are known, every other side or diagonal can be written as an integer combination of them, because the hexagon's vertices are all fixed linear combinations of the first two edges.

   Step-by-Step Solution

   1. Place $A=(1,0)$, $B=(\tfrac12,\tfrac{\sqrt3}{2})$, $C=(-\tfrac12,\tfrac{\sqrt3}{2})$, $E=(-\tfrac12,-\tfrac{\sqrt3}{2})$.
   2. Then $\vec a=\vec{AB}=(-\tfrac12,\tfrac{\sqrt3}{2})$ and $\vec b=\vec{BC}=(-1,0)$.
   3. $\vec{CE}=E-C=(0,-\sqrt3)$.
   4. Write $\vec{CE}=p\vec a+q\vec b$: matching the $y$-component, $\tfrac{\sqrt3}{2}p=-\sqrt3\Rightarrow p=-2$. …

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5. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$ABCDEF$ is a regular hexagon. The sum of the vectors $\vec{BE}, \vec{BC}, \vec{EF}, \vec{BA}, \vec{CF}, \vec{AF}$ (A) $\vec{BF}$ (B) $2\vec{BF}$ (C) $\vec{FB}$ (D) $3\vec{BF}$

   ›Reveal solutionSolution

   Direct vector addition using hexagon coordinates shows the sum of the six vectors is exactly $3\vec{BF}$.

   Concept and Intuition

   For a regular hexagon, placing coordinates at the vertices (center at origin, circumradius 1) turns any vector-sum identity into simple coordinate arithmetic, avoiding error-prone geometric reasoning.

   Step-by-Step Solution

   1. Place $A=(1,0)$, $B=(\tfrac12,\tfrac{\sqrt3}{2})$, $C=(-\tfrac12,\tfrac{\sqrt3}{2})$, $D=(-1,0)$, $E=(-\tfrac12,-\tfrac{\sqrt3}{2})$, $F=(\tfrac12,-\tfrac{\sqrt3}{2})$.
   2. Compute each vector: $\vec{BE}=(-1,-\sqrt3)$, $\vec{BC}=(-1,0)$, $\vec{EF}=(1,0)$, $\vec{BA}=(\tfrac12,-\tfrac{\sqrt3}{2})$, $\vec{CF}=(1,-\sqrt3)$, $\vec{AF}=(-\tfrac12,-\tfrac{\sqrt3}{2})$.
   3. Sum the $x$-components: $-1-1+1+\tfrac12+1-\tfrac12=0$. …

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6. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Let $ABCDEF$ be a regular hexagon with the vertices $A, B, C, D, E, F$ counterclockwise. Then the vector $\overrightarrow{AB} + \overrightarrow{BC}$ is equal/parallel to (A) $\overrightarrow{BC} + \overrightarrow{CD}$ (B) $\overrightarrow{CD} + \overrightarrow{DE}$ (C) $\overrightarrow{AF} + \overrightarrow{FE}$ (D) $\overrightarrow{FE} + \overrightarrow{ED}$

   ›Reveal solutionSolution

   $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$, a fixed vector. Checking all four options against a coordinate model of the hexagon shows only $\overrightarrow{FE}+\overrightarrow{ED}=\overrightarrow{FD}$ equals $\overrightarrow{AC}$ exactly.

   Concept and Intuition

   In any polygon, consecutive edge vectors telescope: $\overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR}$. So each option is really just a "skip one vertex" displacement vector between two hexagon vertices two apart. The question becomes: which of these skip-vectors equals $\overrightarrow{AC}$?

   Step-by-Step Solution

   1. Place the regular hexagon (counterclockwise) on a unit circle centered at $O$: $A=(1,0)$, $B=(0.5,0.866)$, $C=(-0.5,0.866)$, $D=(-1,0)$, $E=(-0.5,-0.866)$, $F=(0.5,-0.866)$.
   2. $\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC} = C-A = (-1.5,\ 0.866)$.
   3. Option (A): $\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{BD}=D-B=(-1.5,-0.866)$ — not equal (mirrored in $y$).
   4. Option (B): $\overrightarrow{CD}+\overrightarrow{DE}=\overrightarrow{CE}=E-C=(0,-1.732)$ — not equal.
   5. Option (C): $\overrightarrow{AF}+\overrightarrow{FE}=\overrightarrow{AE}=E-A=(-1.5,-0.866)$ — not equal. …

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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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