Q.Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes 1 and 2 respectively and when $\vec{a}\cdot\vec{b}=1$.

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $\bar{a} = -4\bar{i} + 2\bar{j} + 4\bar{k}$, $\bar{b} = \sqrt{2}\bar{i} - \sqrt{2}\bar{j}$ are two vectors then angle between the vectors $2\bar{a}$ and $\dfrac{\bar{b}}{2}$ is (A) $30^\circ$ (B) $135^\circ$ (C) $90^\circ$ (D) $0^\circ$

   ›Reveal solutionSolution

   The angle between $2\bar a$ and $\bar b/2$ equals the angle between $\bar a$ and $\bar b$ (scalar multiples by positive numbers don't change direction); computing that angle gives $135^\circ$.

   Concept and Intuition

   Multiplying a vector by a positive scalar only changes its magnitude, not its direction. So $\theta(2\bar a,\ \bar b/2)=\theta(\bar a,\ \bar b)$, and we can use the original vectors directly in the cosine formula.

   Step-by-Step Solution

   1. $\bar a\cdot\bar b=(-4)(\sqrt2)+(2)(-\sqrt2)+(4)(0)=-4\sqrt2-2\sqrt2+0=-6\sqrt2$.
   2. $|\bar a|=\sqrt{(-4)^2+2^2+4^2}=\sqrt{16+4+16}=\sqrt{36}=6$.
   3. $|\bar b|=\sqrt{(\sqrt2)^2+(-\sqrt2)^2+0^2}=\sqrt{2+2}=\sqrt4=2$.
   4. $\cos\theta=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{-6\sqrt2}{6\times2}=\dfrac{-6\sqrt2}{12}=-\dfrac{\sqrt2}{2}=-\dfrac1{\sqrt2}$.
   5. $\theta=\cos^{-1}\left(-\dfrac1{\sqrt2}\right)=135^\circ$.

   Common Mistakes

   • Wasting time actually computing $2\bar a$ and $\bar b/2$ component-wise instead of recognizing the angle is scale-invariant.
   • Sign slip in the dot product from the negative components.

   ✓Final answer

   The correct option is (B) — $135^\circ$.

   ANSWER: B

2. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.If $\vec{a}$ and $\vec{b}$ are two vectors such that $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} < 0$ and $|\vec{a}\cdot\vec{b}| = |\vec{a}\times\vec{b}|$ then the angle between the vectors $\vec{a}$ and $\vec{b}$ is ________ (A) $\dfrac{\pi}{4}$ (B) $\mathrm{Sec}^{-1}(-\sqrt{2})$ (C) $\mathrm{Tan}^{-1}\left(\dfrac{-1}{2}\right)$ (D) $\mathrm{Sin}^{-1}\left(\dfrac{1}{2}\right)$

   ›Reveal solutionSolution

   The two conditions together force $\theta=135^\circ$, which is precisely $\sec^{-1}(-\sqrt2)$.

   Concept and Intuition

   $\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\cos\theta$, so a negative value means the angle is obtuse. The magnitude condition compares the dot and cross product magnitudes, which are $|\vec a||\vec b|\,|\cos\theta|$ and $|\vec a||\vec b|\,|\sin\theta|$ respectively.

   Step-by-Step Solution

   1. $\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}<0\Rightarrow\cos\theta<0\Rightarrow\theta$ is obtuse (between $90^\circ$ and $180^\circ$).
   2. $|\vec a\cdot\vec b|=|\vec a\times\vec b|\Rightarrow|\vec a||\vec b||\cos\theta|=|\vec a||\vec b||\sin\theta|\Rightarrow|\cos\theta|=|\sin\theta|\Rightarrow\tan\theta=\pm1$.
   3. Combined with $\theta$ obtuse, the only solution in $(90^\circ,180^\circ)$ is $\theta=135^\circ$.
   4. Checking option (B): $\sec^{-1}(-\sqrt2)$ means $\cos\theta=-\dfrac1{\sqrt2}\Rightarrow\theta=135^\circ$ — matches exactly.
   5. Options (A), (C), (D) give acute or non-matching angles ($45^\circ$, $\tan^{-1}(-1/2)$, $30^\circ$ or $150^\circ$ respectively — none is exactly $135^\circ$).

   Common Mistakes

   • Picking the acute solution $\theta=45^\circ$ from $\tan\theta=\pm1$ without checking the sign condition $\cos\theta<0$, which rules it out.

   ✓Final answer

   The correct option is (B) — $\mathrm{Sec}^{-1}(-\sqrt{2})$.

   ANSWER: B

3. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If the angle between two unit vectors $\vec{A}$ and $\vec{B}$ is $\theta$, then $|\vec{A}+\vec{B}|$ is (A) $2\cos\dfrac{\theta}{2}$ (B) $2\sin\dfrac{\theta}{2}$ (C) $0$ (D) $\cos\dfrac{\theta}{2}$

   ›Reveal solutionSolution

   Expand $|\vec A+\vec B|^2$ using the dot product and the half-angle identity $1+\cos\theta=2\cos^2(\theta/2)$.

   Concept and Intuition

   For two unit vectors, the parallelogram-law expansion directly gives the magnitude of the sum in terms of the angle between them.

   Step-by-Step Solution

   1. $|\vec A+\vec B|^2=\vec A\cdot\vec A+2\vec A\cdot\vec B+\vec B\cdot\vec B=|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta$.
   2. Since $|\vec A|=|\vec B|=1$: $|\vec A+\vec B|^2=1+1+2\cos\theta=2+2\cos\theta$.
   3. Use $1+\cos\theta=2\cos^2(\theta/2)$: $2+2\cos\theta=4\cos^2(\theta/2)$.
   4. $|\vec A+\vec B|=\sqrt{4\cos^2(\theta/2)}=2\left|\cos(\theta/2)\right|=2\cos(\theta/2)$ since $\theta\in[0,\pi]$ makes $\theta/2\in[0,\pi/2]$, where cosine is non-negative.

   Common Mistakes

   • Using $1-\cos\theta=2\sin^2(\theta/2)$ (the identity for $|\vec A-\vec B|$) instead of the correct one for the sum.
   • Forgetting the absolute value / sign consideration when taking the square root.

   ✓Final answer

   The correct option is (A) — $2\cos\dfrac{\theta}{2}$.

   ANSWER: A

4. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The value of $\dfrac{(\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2}{2(\vec{a})^2(\vec{b})^2}$ is (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   The identity $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2$ makes the given ratio collapse instantly to $\frac12$, independent of the actual vectors.

   Concept and Intuition

   The cross-product magnitude captures the $\sin\theta$ part of the angle between two vectors, while the dot product captures the $\cos\theta$ part. Squaring and adding them recovers $a^2b^2(\sin^2\theta+\cos^2\theta)=a^2b^2$ — a clean Pythagorean-style identity that eliminates the angle entirely.

   Step-by-Step Solution

   1. Recall $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ and $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$, where $\theta$ is the angle between $\vec a$ and $\vec b$.
   2. Square both: $(\vec a\times\vec b)^2 = a^2b^2\sin^2\theta$ and $(\vec a\cdot\vec b)^2=a^2b^2\cos^2\theta$.
   3. Add: $(\vec a\times\vec b)^2+(\vec a\cdot\vec b)^2 = a^2b^2(\sin^2\theta+\cos^2\theta) = a^2b^2$.
   4. Substitute into the given expression: $\dfrac{a^2b^2}{2a^2b^2} = \dfrac12$.

   Common Mistakes

   • Trying to compute this for specific vectors instead of recognizing the general Pythagorean-style identity.
   • Confusing $(\vec a\times\vec b)^2$ (meaning $|\vec a\times\vec b|^2$, a scalar) with an actual vector square.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{2}$.

   ANSWER: C

5. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.Find the angle between the vectors $\vec{A} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\vec{B} = 4\hat{i} + 2\hat{j} - 4\hat{k}$. (A) $0^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

   ›Reveal solutionSolution

   The dot product of the two vectors is exactly zero, so the angle between them is $90^\circ$.

   Concept and Intuition

   The angle between two vectors is found from $\cos\theta=\dfrac{\vec A\cdot\vec B}{|\vec A||\vec B|}$; a zero dot product directly signals perpendicularity without needing the magnitudes.

   Step-by-Step Solution

   1. $\vec A\cdot\vec B=(2)(4)+(4)(2)+(4)(-4)=8+8-16=0$.
   2. Since $|\vec A|,|\vec B|\ne0$, $\cos\theta=0\Rightarrow\theta=90^\circ$.

   Common Mistakes

   • Sign slip while multiplying the $k$-components ($4\times(-4)=-16$, not $+16$).

   ✓Final answer

   The correct option is (D) — $90^\circ$.

   ANSWER: D

6. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Two vectors of same magnitude act at a point. Twice the product of the magnitudes of two vectors is equal to the square of the magnitude of their resultant. The angle between the two vectors is (A) $60°$ (B) $30°$ (C) $90°$ (D) $120°$

   ›Reveal solutionSolution

   Setting $2A^2$ (twice the product of equal magnitudes) equal to $R^2$ for two equal vectors directly forces $\cos\theta=0$, so $\theta=90°$.

   Concept and Intuition

   For two vectors of equal magnitude $A$ at angle $\theta$, the parallelogram law gives $R^2=2A^2(1+\cos\theta)$. The given condition ("twice the product of magnitudes equals square of resultant") is simply an equation relating $A$ and $R$ that pins down $\cos\theta$.

   Step-by-Step Solution

   1. Let both vectors have magnitude $A$. Resultant: $R^2 = A^2+A^2+2A\cdot A\cos\theta = 2A^2(1+\cos\theta)$.
   2. Given condition: $2(A)(A) = R^2$, i.e. $2A^2 = R^2$.
   3. Substitute: $2A^2 = 2A^2(1+\cos\theta)$.
   4. Divide both sides by $2A^2$ (nonzero): $1 = 1+\cos\theta \Rightarrow \cos\theta = 0$.
   5. $\theta = 90°$.

   Common Mistakes

   • Misreading "twice the product of the magnitudes" as $2A$ instead of $2A^2$ (product of two equal magnitudes is $A\times A=A^2$).
   • Sign error in the resultant formula (using $-2A^2\cos\theta$ instead of $+2A^2\cos\theta$).

   ✓Final answer

   The correct option is (C) — $90°$.

   ANSWER: C

7. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $\bar{a}, \bar{b}, \bar{c}$ are 3 vectors such that $|\bar{a}|=5, |\bar{b}|=8, |\bar{c}|=11$ and $\bar{a}+\bar{b}+\bar{c}=\bar{0}$ then the angle between the vectors $\bar{a}$ and $\bar{b}$ is (A) $\cos^{-1}\dfrac{2}{5}$ (B) $\cos^{-1}\dfrac{10}{11}$ (C) $\cos^{-1}\dfrac{41}{55}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   From $\bar c=-(\bar a+\bar b)$, $|\bar c|^2=|\bar a|^2+|\bar b|^2+2\,\bar a\!\cdot\!\bar b$ gives $\cos\theta=\tfrac25$.

   Since $\bar a+\bar b+\bar c=\bar 0$, we have $\bar c=-(\bar a+\bar b)$, so

   $|\bar c|^2=|\bar a+\bar b|^2=|\bar a|^2+|\bar b|^2+2|\bar a||\bar b|\cos\theta,$

   where $\theta$ is the angle between $\bar a$ and $\bar b$.

   Substituting $|\bar a|=5,\ |\bar b|=8,\ |\bar c|=11$:

   $121 = 25 + 64 + 2(5)(8)\cos\theta = 89 + 80\cos\theta.$

   $80\cos\theta = 32 \ \Rightarrow\ \cos\theta=\frac{32}{80}=\frac{2}{5}.$

   Hence $\theta=\cos^{-1}\dfrac{2}{5}$.

   ✓Final answer

   The angle between $\bar a$ and $\bar b$ is $\cos^{-1}\dfrac{2}{5}$ — option (A).

8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Let $\bar{a}, \bar{b}$ be two unit vector. If $\bar{c} = \bar{a} + 2\bar{b}$ and $\bar{d} = 5\bar{a} - 4\bar{b}$ are perpendicular to each other, then the angle between $\bar{a}$ and $\bar{b}$ is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   Expand the perpendicularity condition $\bar c\cdot\bar d=0$ to isolate $\bar a\cdot\bar b$.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Expanding the dot product of linear combinations of unit vectors reduces everything to the single unknown $\bar a\cdot\bar b=\cos\theta$.

   Step-by-Step Solution

   1. $\bar c\cdot\bar d = (\bar a+2\bar b)\cdot(5\bar a-4\bar b) = 5(\bar a\cdot\bar a) -4(\bar a\cdot\bar b) + 10(\bar b\cdot\bar a) -8(\bar b\cdot\bar b)$.
   2. Since $|\bar a|=|\bar b|=1$: $=5(1) + 6(\bar a\cdot\bar b) - 8(1) = 6(\bar a\cdot\bar b) - 3$.
   3. Set to zero: $6(\bar a\cdot\bar b)=3\Rightarrow \bar a\cdot\bar b = \dfrac12$.
   4. Since both are unit vectors, $\bar a\cdot\bar b=\cos\theta = \dfrac12 \Rightarrow \theta = \dfrac{\pi}{3}$.

   Common Mistakes

   • Sign errors when combining the $-4$ and $+10$ cross terms (they add, not cancel).
   • Forgetting $|\bar a|=|\bar b|=1$ so $\bar a\cdot\bar a=\bar b\cdot\bar b=1$.

   ✓Final answer

   The correct option is (C) — $\dfrac{\pi}{3}$.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $P = (0, 1, 2), Q = (4, -2, 1)$ and $O = (0, 0, 0)$ then $\angle POQ =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Compute the dot product of the position vectors of $P$ and $Q$ from the origin; a zero dot product means the angle between them is a right angle.

   Concept and Intuition

   For any two vectors $\vec u,\vec v$, $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$. If the dot product is zero, $\cos\theta=0$ regardless of the magnitudes, so $\theta=\dfrac{\pi}{2}$ — no need to even compute the lengths.

   Step-by-Step Solution

   1. Since $O$ is the origin, $\vec{OP}=P-O=(0,1,2)$ and $\vec{OQ}=Q-O=(4,-2,1)$.
   2. Dot product: $\vec{OP}\cdot\vec{OQ}=(0)(4)+(1)(-2)+(2)(1)=0-2+2=0$.
   3. A zero dot product means $\vec{OP}\perp\vec{OQ}$, so $\angle POQ=\dfrac{\pi}{2}$.

   Common Mistakes

   • Sign slip while multiplying corresponding components.
   • Unnecessarily computing $|\vec{OP}|,|\vec{OQ}|$ when the zero dot product alone already settles the angle.

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi}{2}$.

   ANSWER: D

10. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.Let $\bar a = a_1\bar i + a_2\bar j + a_3\bar k$ where $a_1, a_2, a_3$ and $|\bar a|$ are rational numbers. If $\bar a$ makes an angle $45^0$ with $\bar b$ and $\bar b = \sqrt2\bar i + 3\sqrt2\bar j + 4\bar k$ then $\bar a$ lies in (A) XY - plane (B) YZ - plane (C) XZ - plane (D) along the bisector of the angle between $\bar k$ and $-\bar b$

    ›Reveal solutionSolution

    This tests separating rational and irrational parts of a dot-product equation to pin down a vector's component.

    Concept and Intuition

    $\bar b = \sqrt2\bar i+3\sqrt2\bar j+4\bar k$ has magnitude $|\bar b|=\sqrt{2+18+16}=6$. The angle condition between $\bar a$ and $\bar b$ mixes rational coefficients ($a_1,a_2,a_3,|\bar a|$) with $\bar b$'s irrational ($\sqrt2$-scaled) and rational (the $k$-component, 4) parts — since $a_1,a_2,a_3,|\bar a|$ are all rational, the equation must balance separately in its rational and $\sqrt2$-irrational parts.

    Step-by-Step Solution

    1. $\cos45° = \dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}$, with $|\bar b|=6$ and $\cos45°=\tfrac{\sqrt2}{2}$.
    2. $\bar a\cdot\bar b = a_1\sqrt2+3a_2\sqrt2+4a_3 = \sqrt2(a_1+3a_2)+4a_3$.
    3. Setting this equal to $|\bar a|\cdot6\cdot\tfrac{\sqrt2}{2}=3\sqrt2\,|\bar a|$: $\sqrt2(a_1+3a_2)+4a_3 = 3\sqrt2\,|\bar a|$.
    4. Since $a_1,a_2,a_3,|\bar a|$ are rational, the equation splits into a rational part ($4a_3$ on the left, $0$ on the right since the right side is purely a $\sqrt2$ multiple) and an irrational $\sqrt2$ part.
    5. Rational part: $4a_3=0 \Rightarrow a_3=0$.
    6. $a_3=0$ means $\bar a$ has no $\bar k$-component, so it lies entirely in the XY-plane.

    Common Mistakes

    • Trying to solve for the full vector numerically instead of recognizing the rational/irrational split is the key shortcut.

    ✓Final answer

    The correct option is (A) — XY-plane.

    ANSWER: A

11. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.Let $\vec{a} = 4\vec{i} - \vec{j} + \alpha\vec{k}$ and $\vec{b} = \vec{i} + \alpha\vec{j} - 4\vec{k}$ be two vectors. If $\alpha_1, \alpha_2$ ($\alpha_1 < \alpha_2$) are two different values of $\alpha$ such that $(\vec{a}, \vec{b}) = \cos^{-1}\left(-\dfrac{2}{7}\right)$, then $\alpha_1 + 2\alpha_2 =$ (A) $15$ (B) $24$ (C) $33$ (D) $52$

    ›Reveal solutionSolution

    Setting up $\cos\theta=\vec a\cdot\vec b/(|\vec a||\vec b|)=-2/7$ gives a quadratic in $\alpha$ with roots $2$ and $15.5$; then $\alpha_1+2\alpha_2=33$.

    Concept and Intuition

    Both vectors have the same magnitude expression in $\alpha$ (a nice simplification to notice first), which keeps the resulting equation a clean single-variable quadratic instead of something messier.

    Step-by-Step Solution

    1. $\vec a\cdot\vec b=4(1)+(-1)(\alpha)+\alpha(-4)=4-\alpha-4\alpha=4-5\alpha$.
    2. $|\vec a|=\sqrt{16+1+\alpha^2}=\sqrt{17+\alpha^2}$ and $|\vec b|=\sqrt{1+\alpha^2+16}=\sqrt{17+\alpha^2}$ — identical.
    3. So $\cos\theta=\dfrac{4-5\alpha}{17+\alpha^2}=-\dfrac27$.
    4. Cross-multiply: $7(4-5\alpha)=-2(17+\alpha^2)\Rightarrow28-35\alpha=-34-2\alpha^2\Rightarrow2\alpha^2-35\alpha+62=0$.
    5. Discriminant $=35^2-4(2)(62)=1225-496=729=27^2$. Roots: $\alpha=\dfrac{35\pm27}{4}$, i.e. $\alpha=2$ or $\alpha=15.5$.
    6. Since $\alpha_1<\alpha_2$: $\alpha_1=2,\ \alpha_2=15.5$. Then $\alpha_1+2\alpha_2=2+2(15.5)=2+31=33$.

    Common Mistakes

    • Not noticing $|\vec a|=|\vec b|$ and computing both magnitudes the long way (more error-prone but same result).
    • Sign error cross-multiplying the negative cosine value.

    ✓Final answer

    The correct option is (C) — $33$.

    ANSWER: C

12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.For the resultant of two vectors $\vec{A}$ and $\vec{B}$ to maximum, the angle between them should be ____ (A) $180^\circ$ (B) $0^\circ$ (C) $90^\circ$ (D) $60^\circ$

    ›Reveal solutionSolution

    The resultant magnitude formula shows $R$ increases as $\cos\theta$ increases, so $R$ is maximum when $\theta=0^\circ$.

    Concept and Intuition

    Two vectors add most constructively when they point in exactly the same direction — there's no cancellation at all, so their magnitudes add directly ($R=A+B$), which is the largest possible resultant.

    Step-by-Step Solution

    1. Resultant magnitude: $R=\sqrt{A^2+B^2+2AB\cos\theta}$.
    2. $R$ is maximized when $\cos\theta$ is maximized, i.e. $\cos\theta=1 \Rightarrow \theta=0^\circ$.
    3. At $\theta=0^\circ$: $R=\sqrt{A^2+B^2+2AB}=\sqrt{(A+B)^2}=A+B$, the largest possible value.

    Common Mistakes

    • Confusing this with the condition for MINIMUM resultant, which occurs at $\theta=180^\circ$ (vectors anti-parallel), giving $R=|A-B|$.

    ✓Final answer

    The correct option is (B) — $0^\circ$.

    ANSWER: B