Q.The vectors $\vec{a}=3\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}-2\hat{k}$ are the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The angle between the diagonals of the parallelogram whose adjacent sides are $2i + 4j - 5k$, $i + 2j + 3k$ is (A) $\cos^{-1}\left(\dfrac{7}{\sqrt{69}}\right)$ (B) $\cos^{-1}\left(\dfrac{1}{7\sqrt{69}}\right)$ (C) $\cos^{-1}\left(\dfrac{1}{7}\right)$ (D) $\cos^{-1}\left(\dfrac{31}{7\sqrt{69}}\right)$

   ›Reveal solutionSolution

   The diagonals of a parallelogram built from adjacent side vectors $\vec a,\vec b$ are $\vec a+\vec b$ and $\vec a-\vec b$; the angle between them follows from the dot-product formula.

   Concept and Intuition

   If a parallelogram has adjacent sides $\vec a$ and $\vec b$ (both starting from the same vertex), the two diagonals are $\vec a+\vec b$ (the 'long' diagonal from that vertex to the opposite vertex) and $\vec a-\vec b$ (the other diagonal). The angle between them is just the angle between these two vectors.

   Step-by-Step Solution

   1. $\vec a = 2i+4j-5k,\ \vec b=i+2j+3k$.
   2. $\vec d_1=\vec a+\vec b = 3i+6j-2k$; $\vec d_2=\vec a-\vec b = i+2j-8k$.
   3. $\vec d_1\cdot\vec d_2 = 3(1)+6(2)+(-2)(-8) = 3+12+16=31$.
   4. $|\vec d_1| = \sqrt{9+36+4}=\sqrt{49}=7$.
   5. $|\vec d_2| = \sqrt{1+4+64}=\sqrt{69}$.
   6. $\cos\theta = \dfrac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|} = \dfrac{31}{7\sqrt{69}}$, so $\theta=\cos^{-1}\left(\dfrac{31}{7\sqrt{69}}\right)$. …

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2. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Let $|\bar a|=2, |\bar b|=3$ and the angle between $\bar a$ and $\bar b$ be $\dfrac{\pi}{3}$. If a parallelogram is constructed with adjacent sides $2\bar a+3\bar b$ and $\bar a-\bar b$, then its shorter diagonal is of length (A) 108 (B) 172 (C) $6\sqrt3$ (D) $2\sqrt{43}$

   ›Reveal solutionSolution

   The diagonals of a parallelogram with sides $\vec p,\vec q$ are $\vec p+\vec q$ and $\vec p-\vec q$; computing their magnitudes shows the shorter diagonal is $6\sqrt3$.

   Concept and Intuition

   For a parallelogram with adjacent sides $\vec p=2\vec a+3\vec b$ and $\vec q=\vec a-\vec b$, the diagonals are $\vec p+\vec q$ and $\vec p-\vec q$. Their lengths are found using $|\vec u|^2=\vec u\cdot\vec u$, expanded with $|\vec a|^2=4$, $|\vec b|^2=9$, and $\vec a\cdot\vec b = 2\cdot3\cdot\cos(\pi/3)=3$.

   Step-by-Step Solution

   1. $\vec p+\vec q = 3\vec a+2\vec b$. $|\vec p+\vec q|^2 = 9|\vec a|^2+12(\vec a\cdot\vec b)+4|\vec b|^2 = 9(4)+12(3)+4(9)=36+36+36=108$.
   2. $\vec p-\vec q = \vec a+4\vec b$. $|\vec p-\vec q|^2 = |\vec a|^2+8(\vec a\cdot\vec b)+16|\vec b|^2 = 4+24+144=172$. …

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3. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.If three consecutive vertices of a parallelogram are $A(4,3,5)$, $B(0,6,0)$, $C(-8,1,4)$ and $D$ is the fourth vertex, then the angle between $\overline{AC}$ and $\overline{BD}$ is (A) $\cos^{-1}\left(\dfrac{65}{\sqrt{149}\sqrt{161}}\right)$ (B) $\cos^{-1}\left(\dfrac{55}{\sqrt{149}\sqrt{161}}\right)$ (C) $\cos^{-1}\left(\dfrac{73}{\sqrt{149}\sqrt{161}}\right)$ (D) $\cos^{-1}\left(\dfrac{15}{\sqrt{149}\sqrt{161}}\right)$

   ›Reveal solutionSolution

   Find the fourth vertex using the parallelogram diagonal-bisection property, then compute the angle between the two diagonals via the dot product formula.

   Concept and Intuition

   In parallelogram $ABCD$ (vertices in order), the diagonals $AC$ and $BD$ bisect each other, i.e. their midpoints coincide: $A+C=B+D$. This directly gives the fourth vertex $D$ once $A,B,C$ are known.

   Step-by-Step Solution

   1. $D = A+C-B = (4-8-0,\,3+1-6,\,5+4-0) = (-4,-2,9)$.
   2. $\overrightarrow{AC} = C-A = (-8-4,\,1-3,\,4-5) = (-12,-2,-1)$.
   3. $\overrightarrow{BD} = D-B = (-4-0,\,-2-6,\,9-0) = (-4,-8,9)$.
   4. Dot product: $\overrightarrow{AC}\cdot\overrightarrow{BD} = (-12)(-4)+(-2)(-8)+(-1)(9) = 48+16-9=55$.
   5. $|\overrightarrow{AC}| = \sqrt{144+4+1}=\sqrt{149}$; $|\overrightarrow{BD}| = \sqrt{16+64+81}=\sqrt{161}$. …

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4. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.$\bar{a}, \bar{b}$ are non-collinear vectors, $|\bar{a}|=2\sqrt{2}$, $|\bar{b}|=3$ and the angle between $\bar{a}$ and $\bar{b}$ is $45^0$. Then the lengths of the diagonals of the parallelogram whose adjacent sides are represented by the vectors $5\bar{a}+2\bar{b}$ and $\bar{a}-3\bar{b}$ are (A) $15, 593$ (B) $15, \sqrt{593}$ (C) $225, \sqrt{593}$ (D) $225, 593$

   ›Reveal solutionSolution

   This tests using the parallelogram-diagonal sum/difference of two side vectors together with the

   dot product; the diagonal lengths come out to $15$ and $\sqrt{593}$.

   Concept and Intuition

   If a parallelogram has adjacent sides represented by vectors $\bar u,\bar v$ from a common vertex,

   its two diagonals are $\bar u+\bar v$ and $\bar u-\bar v$ (sum and difference). Their lengths are

   found by expanding $|\bar u\pm\bar v|^2=|\bar u|^2+|\bar v|^2\pm2\,\bar u\cdot\bar v$, which needs

   only $|\bar u|,|\bar v|$ and the angle between them.

   Step-by-Step Solution

   1. Here $\bar u=5\bar a+2\bar b$ and $\bar v=\bar a-3\bar b$.
   2. First find $\bar a\cdot\bar b = |\bar a||\bar b|\cos45^\circ = 2\sqrt2\times3\times\dfrac{\sqrt2}{2} = 6$. Also $|\bar a|^2=8,\ |\bar b|^2=9$.
   3. Diagonal 1: $\bar u+\bar v = (5\bar a+2\bar b)+(\bar a-3\bar b)=6\bar a-\bar b$.

   $|6\bar a-\bar b|^2 = 36|\bar a|^2 - 12(\bar a\cdot\bar b)+|\bar b|^2 = 36(8)-12(6)+9 = 288-72+9=225.$

   So this diagonal has length $\sqrt{225}=15$.

   4. Diagonal 2: $\bar u-\bar v = (5\bar a+2\bar b)-(\bar a-3\bar b)=4\bar a+5\bar b$. …

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5. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let ABCD be a parallelogram and $2\vec{i}+\vec{j}$, $4\vec{i}+5\vec{j}+4\vec{k}$ and $-\vec{i}-4\vec{j}-3\vec{k}$ be the position vectors of the vertices A, B, D respectively. Then the position vector of one of the point of trisection of the diagonal AC is (A) $\dfrac{1}{3}(5\vec{i}+2\vec{j}-\vec{k})$ (B) $\dfrac{1}{3}(5\vec{i}+2\vec{j}+\vec{k})$ (C) $\dfrac{1}{3}(5\vec{i}+4\vec{j}+\vec{k})$ (D) $\dfrac{1}{3}(3\vec{i}+2\vec{j}+\vec{k})$

   ›Reveal solutionSolution

   Use the parallelogram diagonal-bisection property to find $C$, then compute the point one-third of the way from $A$ to $C$. Answer: (B).

   Concept and Intuition

   In a parallelogram $ABCD$ (vertices in order), the diagonals $AC$ and $BD$ bisect each other, so their midpoints coincide: $\dfrac{A+C}{2}=\dfrac{B+D}{2}\Rightarrow C=B+D-A$. Once $C$ is known, the two trisection points of segment $AC$ are just the points at parameters $1/3$ and $2/3$ along it.

   Step-by-Step Solution

   1. $A=2\vec i+\vec j=(2,1,0)$, $B=(4,5,4)$, $D=(-1,-4,-3)$.
   2. $C=B+D-A=(4-1-2,\ 5-4-1,\ 4-3-0)=(1,0,1)=\vec i+\vec k$.
   3. $C-A=(1-2,\ 0-1,\ 1-0)=(-1,-1,1)$.
   4. Trisection point nearer $A$: $A+\tfrac13(C-A)=(2,1,0)+\tfrac13(-1,-1,1)=\left(\tfrac53,\tfrac23,\tfrac13\right)=\tfrac13(5\vec i+2\vec j+\vec k)$. …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}$, $2\bar{i}-\bar{j}$, $\bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively, then the quadrilateral $ABCD$ is a (A) square (B) rectangle (C) rhombus (D) parallelogram

   ›Reveal solutionSolution

   This tests recognizing a quadrilateral's type from vertex position vectors by comparing side vectors and lengths. $ABCD$ is a rhombus.

   Concept and Intuition

   For a quadrilateral $ABCD$ given by position vectors, first check $\vec{AB}=\vec{DC}$ (opposite sides equal and parallel) to confirm it's a parallelogram. Then compare adjacent side lengths $|AB|$ and $|BC|$: if equal, all four sides are equal (rhombus); if additionally perpendicular, it's a square.

   Step-by-Step Solution

   1. $A=(1,2,2)$, $B=(2,-1,0)$, $C=(1,1,3)$, $D=(0,4,5)$.
   2. $\vec{AB}=B-A=(1,-3,-2)$.
   3. $\vec{DC}=C-D=(1,-3,-2)$.
   4. Since $\vec{AB}=\vec{DC}$, $ABCD$ is a parallelogram.
   5. $\vec{BC}=C-B=(-1,2,3)$, so $|AB|=\sqrt{1+9+4}=\sqrt{14}$ and $|BC|=\sqrt{1+4+9}=\sqrt{14}$ — equal adjacent sides.
   6. Since a parallelogram with equal adjacent sides has all sides equal, $ABCD$ is a rhombus (or better, a square if adjacent sides are also perpendicular). …

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7. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.Let $\vec a = \hat i + \hat j + \hat k$, $\vec b = \hat i + 3\hat j + 5\hat k$ and $\vec c = 7\hat i + 9\hat j + 11\hat k$ then the area of parallelogram having diagonals $\vec a + \vec b$ and $\vec b + \vec c$ is (A) $4\sqrt6$ sq. units (B) $2\sqrt6$ sq. units (C) $\sqrt6$ sq. units (D) $6\sqrt6$ sq. units

   ›Reveal solutionSolution

   Area of a parallelogram from its diagonals $\vec d_1,\vec d_2$ is $\tfrac12|\vec d_1\times\vec d_2|$; here it comes out to $4\sqrt6$.

   Concept and Intuition

   If a parallelogram has diagonal vectors $\vec d_1$ and $\vec d_2$, its area is $\tfrac12|\vec d_1\times\vec d_2|$ — this follows because the diagonals split the parallelogram into triangles whose combined area works out to half the cross-product magnitude of the diagonals.

   Step-by-Step Solution

   1. $\vec a+\vec b = (1+1)\hat i+(1+3)\hat j+(1+5)\hat k = 2\hat i+4\hat j+6\hat k$.
   2. $\vec b+\vec c = (1+7)\hat i+(3+9)\hat j+(5+11)\hat k = 8\hat i+12\hat j+16\hat k$.
   3. Compute $(\vec a+\vec b)\times(\vec b+\vec c)$ using components $(2,4,6)$ and $(8,12,16)$:
      • $i$-component: $4\cdot16-6\cdot12 = 64-72=-8$
      • $j$-component: $-(2\cdot16-6\cdot8) = -(32-48)=16$
      • $k$-component: $2\cdot12-4\cdot8=24-32=-8$ …

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8. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $\bar a, \bar b$ be two non collinear vectors and the vector $\bar a + \bar b$ bisects the angle between $\bar a$ and $\bar b$, then (A) $|\bar a| = |\bar b|$ (B) angle between $\bar a, \bar b$ is $0^0$ (or) $\pi$ (C) $\bar a, \bar b$ always form adjacent sides of a square. (D) $\bar a, \bar b$ always form adjacent sides of a rectangle.

   ›Reveal solutionSolution

   The sum vector bisecting the angle between two non-collinear vectors is the classic rhombus-diagonal property, giving $|\bar a|=|\bar b|$.

   Concept and Intuition

   Geometrically, $\bar a$ and $\bar b$ are adjacent sides of a parallelogram with diagonal $\bar a+\bar b$ from the common vertex. That diagonal bisects the vertex angle exactly when the parallelogram is a rhombus (equal adjacent sides) — this is a standard fact about parallelograms, not specific to squares or rectangles.

   Step-by-Step Solution

   1. Let $\theta$ be the angle between $\bar a$ and $\bar b$, and let $\bar c=\bar a+\bar b$ bisect it.
   2. The angle bisector condition (via the formula for the internal bisector direction of two vectors) is: the bisector direction is $\hat a+\hat b$ (unit vectors), i.e. proportional to $\dfrac{\bar a}{|\bar a|}+\dfrac{\bar b}{|\bar b|}$. …

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9. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.Let $\overline{OA} = -4\bar{i} + 3\bar{k}$, $\overline{OB} = 14\bar{i} + 2\bar{j} - 5\bar{k}$. $\overline{OD}$ bisects $\angle AOB$ and $|\overline{OD}| = \sqrt{6}$, then $\overline{OD} =$ (A) $\pm(\bar{i}+\bar{j}+2\bar{k})$ (B) $\pm(\bar{i}+2\bar{j}+\bar{k})$ (C) $\pm(2\bar{i}+\bar{j}+\bar{k})$ (D) $\pm\frac{1}{\sqrt{2}}(2\bar{i}+\bar{j}+\sqrt{7}\bar{k})$

   ›Reveal solutionSolution

   The internal angle bisector direction of two vectors is along the sum of their unit vectors; scale this direction to have the given magnitude to get $\overline{OD}$.

   Concept and Intuition

   For two vectors $\vec a,\vec b$ from a common point $O$, the direction that bisects the angle between them is $\hat a+\hat b$ (sum of unit vectors) — this is a standard vector-geometry fact, since it "averages" the two directions with equal weight regardless of their original magnitudes.

   Step-by-Step Solution

   1. $\overline{OA}=-4\bar i+3\bar k$, $|\overline{OA}|=\sqrt{16+0+9}=5$. Unit vector $\hat a=\left(-\dfrac45,0,\dfrac35\right)$.
   2. $\overline{OB}=14\bar i+2\bar j-5\bar k$, $|\overline{OB}|=\sqrt{196+4+25}=\sqrt{225}=15$. Unit vector $\hat b=\left(\dfrac{14}{15},\dfrac2{15},-\dfrac13\right)$.
   3. Bisector direction $\hat a+\hat b$: $x$: $-\dfrac45+\dfrac{14}{15}=-\dfrac{12}{15}+\dfrac{14}{15}=\dfrac2{15}$; $y$: $0+\dfrac2{15}=\dfrac2{15}$; $z$: $\dfrac35-\dfrac13=\dfrac9{15}-\dfrac5{15}=\dfrac4{15}$.
   4. So $\hat a+\hat b=\dfrac2{15}(1,1,2)$, i.e., direction $\propto(1,1,2)$, with $|(1,1,2)|=\sqrt{1+1+4}=\sqrt6$. …

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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.The figure formed by the four points $(\hat{i}+\hat{j}-\hat{k})$, $(2\hat{i}+3\hat{j})$, $(5\hat{j}-2\hat{k})$ and $(\hat{k}-\hat{j})$ is (A) Trapezium (B) Rectangle (C) Parallelogram (D) Quadrilateral

    ›Reveal solutionSolution

    Testing the four side vectors for parallelism shows neither pair of opposite sides is parallel, ruling out trapezium/parallelogram/rectangle, leaving a general quadrilateral.

    Concept and Intuition

    To classify a quadrilateral from its vertices, compute the four side vectors (in the given cyclic order) and check whether opposite sides are parallel (trapezium/parallelogram) and/or equal in length with right angles (rectangle) — if none of these hold, it is simply a general quadrilateral.

    Step-by-Step Solution

    1. Convert the position vectors to coordinates: $A=\hat i+\hat j-\hat k=(1,1,-1)$, $B=2\hat i+3\hat j=(2,3,0)$, $C=5\hat j-2\hat k=(0,5,-2)$, $D=\hat k-\hat j=(0,-1,1)$.
    2. Side vectors: $\vec{AB}=B-A=(1,2,1)$, $\vec{BC}=C-B=(-2,2,-2)$, $\vec{CD}=D-C=(0,-6,3)$, $\vec{DA}=A-D=(1,2,-2)$.
    3. Check closure: $\vec{AB}+\vec{BC}+\vec{CD}+\vec{DA}=(1-2+0+1,\,2+2-6+2,\,1-2+3-2)=(0,0,0)$ — confirms a valid closed quadrilateral.
    4. Check $\vec{AB}\parallel\vec{CD}$: ratios of components $1/0$ (undefined), $2/(-6)$, $1/3$ — not a consistent scalar, so NOT parallel. …

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11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The diagonals AC and BD of a rhombus ABCD intersect at the point (3, 4). If BD = $2\sqrt{2}$, A = (1,2), B = ($\alpha,\beta$), D = ($\gamma,\delta$) and $\alpha<\delta<\gamma<\beta$, then $\beta+\gamma-\delta =$ (A) 0 (B) $\alpha$ (C) $2\alpha$ (D) $3\alpha$

    ›Reveal solutionSolution

    Using the rhombus diagonal-bisection and perpendicularity properties, $B=(2,5)$ and $D=(4,3)$ (the ordering that satisfies $\alpha<\delta<\gamma<\beta$), giving $\beta+\gamma-\delta=3\alpha$.

    Concept and Intuition

    In a rhombus the diagonals bisect each other at right angles. Knowing one diagonal's midpoint and one endpoint ($A$) lets us find the opposite vertex $C$ by point reflection. The other diagonal, being perpendicular to $AC$ through the same midpoint, is then determined up to a $\pm$ swap of its two endpoints — the extra ordering condition ($\alpha<\delta<\gamma<\beta$) tells us which swap is intended.

    Step-by-Step Solution

    1. Midpoint of both diagonals is $(3,4)$. Since $A=(1,2)$ and $C$ is the reflection of $A$ through this midpoint: $C=2(3,4)-(1,2)=(6,8)-(1,2)=(5,6)$.
    2. Direction of $AC$: $C-A=(4,4)$, i.e. along $(1,1)$, unit vector $\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$.
    3. $BD\perp AC$, so $BD$'s direction is $\left(\frac{1}{\sqrt2},-\frac1{\sqrt2}\right)$ (or its negative).
    4. $BD=2\sqrt2$, so each of $B,D$ is at distance $\sqrt2$ from the midpoint $(3,4)$ along this direction: $(3,4)\pm\sqrt2\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)=(3,4)\pm(1,-1)$.
    5. This gives the two candidate points $(4,3)$ and $(2,5)$ for $\{B,D\}$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.M and N are the mid points of the sides BC and CD of a parallelogram ABCD respectively then $\overline{AM}+\overline{AN} =$ (A) $\frac{1}{3}\overline{AC}$ (B) $\frac{2}{3}\overline{AC}$ (C) $\frac{3}{4}\overline{AC}$ (D) $\frac{3}{2}\overline{AC}$

    ›Reveal solutionSolution

    Assign position vectors with $A$ as origin, express $M$ and $N$ as midpoints, and add the resulting vectors — the answer falls out as a clean multiple of $\overline{AC}$.

    Concept and Intuition

    In a parallelogram $ABCD$, placing $A$ at the origin makes $\overline{AC}=\overline{AB}+\overline{AD}$ (diagonal = sum of adjacent sides). Since $M,N$ are midpoints of $BC,CD$, their position vectors are simple averages, and summing $\overline{AM}+\overline{AN}$ naturally collects into a multiple of $\overline{AB}+\overline{AD}=\overline{AC}$.

    Step-by-Step Solution

    1. Let $A$ be the origin, $\overline{AB}=\vec b$, $\overline{AD}=\vec d$. Since $ABCD$ is a parallelogram, $C=B+D-A=\vec b+\vec d$ (as vectors from $A$).
    2. $M$ = midpoint of $BC$: $\overline{AM}=\dfrac{\vec b+(\vec b+\vec d)}2=\vec b+\dfrac{\vec d}2$.
    3. $N$ = midpoint of $CD$: $\overline{AN}=\dfrac{(\vec b+\vec d)+\vec d}2=\dfrac{\vec b}2+\vec d$. …

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