Q.The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{a}\cdot\vec{b}=2\sqrt{3}$ is
(A) $\dfrac{\pi}{6}$
(B) $\dfrac{\pi}{3}$
(C) $\dfrac{\pi}{2}$
(D) $\dfrac{5\pi}{2}$

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$. If $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$ and $|\vec{a}+\vec{b}+\vec{c}|=4\sqrt3$, then the angle between $\vec{a}$ and $\vec{c}$ is (A) $\cos^{-1}\left(\dfrac{2}{5}\right)$ (B) $\dfrac{\pi}{3}$ (C) $\cos^{-1}\left(\dfrac{2}{3}\right)$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   Expand $|\vec a+\vec b+\vec c|^2$; the perpendicularity conditions kill two of the three cross terms, leaving $\vec a\cdot\vec c$ to solve for. Answer: (B).

   Concept and Intuition

   Squaring a vector sum brings out all pairwise dot products; when some pairs are given as perpendicular, those dot-product terms vanish, isolating the one unknown dot product — here $\vec a\cdot\vec c$, which directly gives the angle between $\vec a$ and $\vec c$.

   Step-by-Step Solution

   1. $|\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a\cdot\vec b+2\vec b\cdot\vec c+2\vec a\cdot\vec c$.
   2. Since $\vec a\perp\vec b$, $\vec a\cdot\vec b=0$; since $\vec b\perp\vec c$, $\vec b\cdot\vec c=0$.
   3. So $|\vec a+\vec b+\vec c|^2=4+9+25+2\vec a\cdot\vec c=38+2\vec a\cdot\vec c$.
   4. Given $|\vec a+\vec b+\vec c|=4\sqrt3\Rightarrow|\vec a+\vec b+\vec c|^2=48$. So $38+2\vec a\cdot\vec c=48\Rightarrow\vec a\cdot\vec c=5$. …

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2. 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: $6(\bar a\cdot\bar b)=3\Rightarrow \bar a\cdot\bar b = \dfrac12$. …

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3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\bar{a},\bar{b},\bar{c}$ are three vectors such that $|\bar{a}|=2$, $|\bar{b}|=3$, $|\bar{c}|=5$, $|\bar{a}+\bar{b}+\bar{c}|=\sqrt{69}$. If $(\bar{a},\bar{b})=(\bar{b},\bar{c})=\dfrac{\pi}{3}$ then $(\bar{c},\bar{a})=$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Expanding $|\bar a+\bar b+\bar c|^2$ and using the two known angles isolates $\bar c.\bar a$, giving the third angle as $\frac{\pi}{3}$.

   Concept and Intuition

   The squared magnitude of a vector sum expands into the sum of squared magnitudes plus twice the pairwise dot products. With two of the three pairwise angles already known, this single scalar equation is enough to solve for the third dot product — and hence the third angle.

   Step-by-Step Solution

   1. Expand: $|\bar a+\bar b+\bar c|^2=|\bar a|^2+|\bar b|^2+|\bar c|^2+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$.
   2. Substitute known magnitudes: $69 = 4+9+25+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a) = 38+2(\ldots)$.
   3. So $\bar a.\bar b+\bar b.\bar c+\bar c.\bar a=\dfrac{69-38}{2}=\dfrac{31}{2}=15.5$.
   4. Compute $\bar a.\bar b=|\bar a||\bar b|\cos\dfrac{\pi}{3}=2\cdot3\cdot\dfrac12=3$.
   5. Compute $\bar b.\bar c=|\bar b||\bar c|\cos\dfrac{\pi}{3}=3\cdot5\cdot\dfrac12=7.5$.
   6. So $\bar c.\bar a=15.5-3-7.5=5$. …

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4. 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 $\theta=135^\circ$. …

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5. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is ______ (A) $\dfrac{\pi}{3}$ (B) $\cos^{-1}\left(\dfrac{1}{6}\right)$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   Tests finding the angle between two planes via the angle between their normal vectors.

   Concept and Intuition

   The angle between two planes equals the angle between their normal vectors (up to supplementary ambiguity, resolved by taking the acute angle). If a plane is $Ax+By+Cz=D$, its normal vector is $(A,B,C)$, and the angle between two normals is found using the dot-product formula.

   Step-by-Step Solution

   1. Plane 1: $2x-y+z=6$, normal $n_1=(2,-1,1)$.
   2. Plane 2: $x+y+2z=3$, normal $n_2=(1,1,2)$.
   3. $n_1\cdot n_2 = 2(1)+(-1)(1)+1(2)=2-1+2=3$.
   4. $|n_1|=\sqrt{4+1+1}=\sqrt6$, $|n_2|=\sqrt{1+1+4}=\sqrt6$.
   5. $\cos\theta=\dfrac{3}{\sqrt6\cdot\sqrt6}=\dfrac{3}{6}=\dfrac12$.
   6. So $\theta=\cos^{-1}\left(\dfrac12\right)=\dfrac{\pi}{3}$.

   Common Mistakes …

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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $\bar{f}, \bar{g}, \bar{h}$ be mutually orthogonal vectors of equal magnitudes, then the angle between the vectors $\bar{f} + \bar{g} + \bar{h}$ and $\bar{h}$ is (A) $\cos^{-1}\left(\dfrac{\sqrt{3}}{4}\right)$ (B) $\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$ (C) $\pi - \cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$ (D) $\pi - \cos^{-1}\left(\dfrac{\sqrt{3}}{4}\right)$

   ›Reveal solutionSolution

   Uses orthogonality to kill cross dot products; answer is $\cos^{-1}(1/\sqrt3)$.

   Concept and Intuition

   When three vectors are mutually perpendicular and of equal magnitude, their sum is the space-diagonal of a cube built on them. The angle any diagonal makes with an edge is a classic $\cos^{-1}(1/\sqrt3)$ result.

   Step-by-Step Solution

   1. Let $|\bar f|=|\bar g|=|\bar h|=a$, and $\bar f\cdot\bar g=\bar g\cdot\bar h=\bar h\cdot\bar f=0$.
   2. $(\bar f+\bar g+\bar h)\cdot\bar h = \bar f\cdot\bar h + \bar g\cdot\bar h + \bar h\cdot\bar h = 0+0+a^2 = a^2$.
   3. $|\bar f+\bar g+\bar h|^2 = |\bar f|^2+|\bar g|^2+|\bar h|^2 + 2(\bar f\cdot\bar g+\bar g\cdot\bar h+\bar h\cdot\bar f) = 3a^2$, so $|\bar f+\bar g+\bar h|=a\sqrt3$. …

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7. 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. $|\bar b|=\sqrt{(\sqrt2)^2+(-\sqrt2)^2+0^2}=\sqrt{2+2}=\sqrt4=2$. …

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8. 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: $\vec{OP}\cdot\vec{OQ}=(0)(4)+(1)(-2)+(2)(1)=0-2+2=0$. …

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

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

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10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.Angle between the planes $\bar{r}.(12\bar{i}+4\bar{j}-3\bar{k}) = 5$ and $\bar{r}.(5\bar{i}+3\bar{j}+4\bar{k}) = 7$ is (A) $\cos^{-1}\left(\dfrac{12}{13}\right)$ (B) $\cos^{-1}\left(\dfrac{6\sqrt{2}}{13}\right)$ (C) $\cos^{-1}\left(\dfrac{3\sqrt{2}}{13}\right)$ (D) $\cos^{-1}\left(\dfrac{6}{13}\right)$

    ›Reveal solutionSolution

    The angle between two planes equals the angle between their normal vectors.

    Using the dot product formula, the cosine of the angle is $\frac{6\sqrt{2}}{13}$, so the correct option is (B).

    The key idea: the angle between two planes is defined as the angle between their normal vectors.

    Given plane equations in vector form $\bar{r} \cdot \bar{n} = d$, the normal vectors are simply the coefficients of $\bar{i}, \bar{j}, \bar{k}$.

    1. Identify the normal vectors

       For the first plane: $\bar{n}_1 = 12\bar{i} + 4\bar{j} - 3\bar{k}$

       For the second plane: $\bar{n}_2 = 5\bar{i} + 3\bar{j} + 4\bar{k}$

    2. Compute the dot product

       $\bar{n}_1 \cdot \bar{n}_2 = (12)(5) + (4)(3) + (-3)(4) = 60 + 12 - 12 = 60$

    3. Compute the magnitudes

       $|\bar{n}_1| = \sqrt{12^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$

       $|\bar{n}_2| = \sqrt{5^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5\sqrt{2}$

    4. Apply the dot product formula for the angle

       $\cos\theta = \frac{\bar{n}_1 \cdot \bar{n}_2}{|\bar{n}_1|\,|\bar{n}_2|} = \frac{60}{13 \cdot 5\sqrt{2}} = \frac{60}{65\sqrt{2}} = \frac{12}{13\sqrt{2}}$ …

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11. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.The vectors $3\bar{a} - 5\bar{b}$ and $2\bar{a} + \bar{b}$ are mutually perpendicular and the vectors $\bar{a} + 4\bar{b}$ and $-\bar{a} + \bar{b}$ are also mutually perpendicular then the acute angle between $\bar{a}$ and $\bar{b}$ is (A) $\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$ (B) $\cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)$ (C) $\pi - \cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$ (D) $\pi - \cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)$

    ›Reveal solutionSolution

    This tests translating two perpendicularity (dot product = 0) conditions into linear equations relating $|\bar a|^2$, $|\bar b|^2$, and $\bar a\cdot\bar b$, then solving for the angle. The acute angle is $\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$.

    Concept and Intuition

    Two vectors are perpendicular exactly when their dot product is zero. Expanding each given perpendicularity condition using distributivity of the dot product yields a linear relation among $A=\bar a\cdot\bar a$, $B=\bar b\cdot\bar b$, and $M=\bar a\cdot\bar b$. Two such conditions give two equations in three unknowns, but since we only need the RATIO $\cos\theta = M/\sqrt{AB}$, we can express everything in terms of $M$ and solve.

    Step-by-Step Solution

    1. $(3\bar a-5\bar b)\cdot(2\bar a+\bar b)=0$: expand $\Rightarrow 6A+3M-10M-5B=0 \Rightarrow 6A-5B-7M=0$ … (i)
    2. $(\bar a+4\bar b)\cdot(-\bar a+\bar b)=0$: expand $\Rightarrow -A+M-4M+4B=0 \Rightarrow 4B-A-3M=0 \Rightarrow A=4B-3M$ … (ii)
    3. Substitute (ii) into (i): $6(4B-3M)-5B-7M=0 \Rightarrow 24B-18M-5B-7M=0 \Rightarrow 19B-25M=0 \Rightarrow B=\dfrac{25M}{19}$.
    4. From (ii): $A=4\left(\dfrac{25M}{19}\right)-3M=\dfrac{100M}{19}-\dfrac{57M}{19}=\dfrac{43M}{19}$.
    5. Since $A=|\bar a|^2>0$ and $B=|\bar b|^2>0$, $M$ must be positive. …

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12. 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. $\vec A\cdot\vec B=(2)(4)+(4)(2)+(4)(-4)=8+8-16=0$. …

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