Q.A girl walks 4 km towards west, then she walks 3 km in a direction $30^\circ$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Showing the 12 most recent of 17 on this concept.

1. 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 $60°$.
   3. Let the three vectors have magnitudes $a,2a,3a$ along $\hat d_1,\hat d_2,\hat d_3$. Then $|\text{sum}|^2 = a^2+(2a)^2+(3a)^2 + 2(a)(2a)\cos60°+2(2a)(3a)\cos60°+2(a)(3a)\cos60°$.
   4. Compute: $a^2+4a^2+9a^2=14a^2$; cross terms $= \cos60°\times[2(2a^2)+2(6a^2)+2(3a^2)]=\tfrac12\times22a^2=11a^2$.
   5. Total: $14a^2+11a^2=25a^2$, so $|\text{sum}|=5a$.

   Common Mistakes

   • Assuming the face diagonals are mutually perpendicular (they are not — that's true for edges, not face diagonals from a shared vertex).
   • Forgetting the factor of $2$ in the cross terms of $|\vec u+\vec v+\vec w|^2$.

   ✓Final answer

   The correct option is (B) — $5a$.

   ANSWER: B

2. 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: $\vec{OA}+\vec{OB}+\vec{OC}=R\left[(\cos45^\circ+1+\cos45^\circ),\ (\sin45^\circ+0-\sin45^\circ)\right]=R\left[1+2\cos45^\circ,\ 0\right]$.
   5. $\cos45^\circ=\dfrac{\sqrt2}2$, so the $x$-component is $R\left(1+2\cdot\dfrac{\sqrt2}2\right)=R(1+\sqrt2)$.
   6. Magnitude of the resultant $=(\sqrt2+1)R$ (the $y$-component is zero).

   Common Mistakes

   • Placing $A$ and $C$ on the same side of $B$ instead of symmetric opposite sides, breaking the cancellation.
   • Forgetting that the resultant's magnitude is just the (positive) $x$-component once the $y$-components cancel.

   ✓Final answer

   The correct option is (B) — $(\sqrt{2}+1)R$.

   ANSWER: B

3. 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: $-1-1+1+\tfrac12+1-\tfrac12=0$.
   4. Sum the $y$-components: $-\sqrt3+0+0-\tfrac{\sqrt3}{2}-\sqrt3-\tfrac{\sqrt3}{2} = -3\sqrt3$.
   5. So the total sum is $(0,-3\sqrt3)$.
   6. $\vec{BF}=F-B=(0,-\sqrt3)$, and $3\vec{BF}=(0,-3\sqrt3)$ — exactly matches the sum.

   Common Mistakes

   • Sign slips when subtracting coordinates for each vector; double-check the direction (tail to head) each time.

   ✓Final answer

   The correct option is (D) — $3\vec{BF}$.

   ANSWER: D

4. 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: $t=\tfrac13,\ \lambda=\tfrac43$.
   6. So $\vec{BF}=F=\dfrac43E=\dfrac43\vec{BE}$, and $\vec{EF}=F-E=\dfrac13E=\dfrac13\vec{BE}$.
   7. Therefore $\vec{BF}=4\left(\dfrac13\vec{BE}\right)=4\,\vec{EF}$.

   Common Mistakes

   • Forgetting $B,E,F$ are collinear so $F$ must be a scalar multiple of $E$ once $B$ is the origin — solving without this shortcut leads to messy simultaneous equations.
   • Sign/ratio slip when solving the two linear equations for $t$ and $\lambda$.

   ✓Final answer

   The correct option is (D) — $4\,\overline{EF}$.

   ANSWER: D

5. 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, $\tfrac{\sqrt3}{2}p=-\sqrt3\Rightarrow p=-2$.
   5. Matching the $x$-component: $-\tfrac12(-2)-q=0\Rightarrow 1-q=0\Rightarrow q=1$.
   6. So $\vec{CE}=-2\vec a+\vec b=\vec b-2\vec a$.

   Common Mistakes

   • Confusing $\vec{CE}$ with $\vec{EC}$ (sign reversal) or misplacing which vertex vector corresponds to $\vec a$ vs $\vec b$.

   ✓Final answer

   The correct option is (C) — $\vec{b} - 2\vec{a}$.

   ANSWER: C

6. 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 $\overline{FA}=A-F=(1-0.5,\,0-(-0.866))=(0.5,0.866)$.
   4. Solve $\overline{FA}=p\bar a+q\bar b$: matching $y$-components, $0.866p=0.866\Rightarrow p=1$. Matching $x$-components, $-0.5(1)-q=0.5\Rightarrow q=-1$.
   5. So $\overline{FA}=\bar a-\bar b$, confirmed by direct substitution: $(-0.5,0.866)-(-1,0)=(0.5,0.866)$ ✓.

   Common Mistakes

   • Assuming $\overline{FA}$ is simply $-(\bar a+\bar b)$ or another combination without verifying via explicit coordinates.
   • Sign confusion about hexagon vertex traversal direction.

   ✓Final answer

   The correct option is (A) — $\bar a-\bar b$.

   ANSWER: A

7. 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: $\vec{BE}+\vec{AF} = \dfrac{(A+C-2B)+(B+C-2A)}{2} = \dfrac{2C-A-B}{2}$.
   6. Compare with $\vec{DC} = C - D = C-\dfrac{A+B}{2} = \dfrac{2C-A-B}{2}$. They match exactly.

   Common Mistakes

   • Sign errors when subtracting position vectors to get a vector between two points (must be head minus tail).
   • Assuming the answer must involve $F$ because $F$ appears in the given vectors — here it cancels out.

   ✓Final answer

   The correct option is (A) — $\vec{DC}$.

   ANSWER: A

8. 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.
   6. Option (D): $\overrightarrow{FE}+\overrightarrow{ED}=\overrightarrow{FD}=D-F=(-1.5,\ 0.866)$ — exactly matches $\overrightarrow{AC}$.

   Common Mistakes

   • Assuming symmetry means any "two-step" vector matches — the hexagon has 6-fold rotational symmetry, so only the skip-vector that is a rotation by a multiple of $60°$ aligned correctly (here, effectively the same vector, not just parallel) will match; a quick coordinate check avoids guessing.

   ✓Final answer

   The correct option is (D) — $\overrightarrow{FE} + \overrightarrow{ED}$.

   ANSWER: D

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 $C = B+\overline{BC} = \bar a+3\bar d-\bar a/2 = \bar a/2+3\bar d$.
   4. Pentagon closure: $\overline{OA}+\overline{AB}+\overline{BC}+\overline{CD}+\overline{DO}=\bar 0$. We have $\overline{DO}=-\bar d$, and substituting the known vectors: $\bar a+3\bar d-\bar a/2+\overline{CD}-\bar d=\bar 0 \Rightarrow \overline{CD} = -\bar a/2-2\bar d$, so $\overline{DC}=\bar a/2+2\bar d$.
   5. $\overline{AD} = D - A = \bar d-\bar a$.
   6. Sum: $\overline{AD}+\overline{OC}+\overline{DC} = (\bar d-\bar a)+(\bar a/2+3\bar d)+(\bar a/2+2\bar d) = (-\bar a+\bar a/2+\bar a/2) + (\bar d+3\bar d+2\bar d) = 0+6\bar d = 6\bar d$.

   Common Mistakes

   • Getting the direction sense of the parallel-side ratios backwards (which flips signs) — cross-checking with a concrete coordinate example (e.g. $\bar a=(2,0)$, $\bar d=(0,1)$) confirms the sum comes out to $(0,6)=6\bar d$.

   ✓Final answer

   The correct option is (D) — $6\bar d$.

   ANSWER: D

10. 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:
       • $\overline{AD}=D-A=\bar d-\bar a$.
       • $\overline{OC}=C=\tfrac12\bar a+3\bar d$.
       • $\overline{DC}=C-D=\left(\tfrac12\bar a+3\bar d\right)-\bar d=\tfrac12\bar a+2\bar d$.
    5. Sum: $\overline{AD}+\overline{OC}+\overline{DC}=(\bar d-\bar a)+\left(\tfrac12\bar a+3\bar d\right)+\left(\tfrac12\bar a+2\bar d\right)$.
    6. The $\bar a$ coefficients: $-1+\tfrac12+\tfrac12=0$. The $\bar d$ coefficients: $1+3+2=6$. So the sum is $6\bar d$.

    Common Mistakes

    • Getting the direction of a parallel-side ratio backwards (e.g. using $\overline{BC}$ instead of $\overline{CB}$, which flips a sign).
    • Losing track of which vertex is the "free" origin-anchored one when building up $B$ and $C$.

    ✓Final answer

    The correct option is (C) — $6\bar d$.

    ANSWER: C

11. 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 $=|AB|+|BC|+|CA|=7+7+\sqrt{74}=14+\sqrt{74}$.

    Common Mistakes

    • Adding instead of taking the negative sum for $\overline{CA}$ (sign error in the closed-triangle vector relation).
    • Forgetting $\overline{CA}$, not $\overline{AC}$, is what's needed to close the loop $A\to B\to C\to A$ (they only differ by sign, but it must be tracked correctly through the loop equation).

    ✓Final answer

    The correct option is (B) — $\sqrt{74}+14$.

    ANSWER: B

12. 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. $\vec{CA}\cdot\vec{CD} = \vec b\cdot\dfrac{\vec a+\vec b}{2} = \dfrac{\vec a\cdot\vec b}{2}+\dfrac{b^2}{2}$.
    4. Since $\vec{AB}=\vec a-\vec b$ has magnitude $c$: $c^2=a^2+b^2-2\vec a\cdot\vec b \Rightarrow \vec a\cdot\vec b = \dfrac{a^2+b^2-c^2}{2}$.
    5. Substitute: $\vec{CA}\cdot\vec{CD} = \dfrac{1}{2}\cdot\dfrac{a^2+b^2-c^2}{2} + \dfrac{b^2}{2} = \dfrac{a^2+b^2-c^2+2b^2}{4} = \dfrac{a^2+3b^2-c^2}{4}$.

    Common Mistakes

    • Mixing up which side vector corresponds to $a$, $b$, $c$ (standard convention: $a=BC$, $b=CA$, $c=AB$).
    • Forgetting the median's position vector is the average, not the sum, of the two vertex vectors.

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{4}(a^2+3b^2-c^2)$.

    ANSWER: B