Q.Find the sine of the angle between the vectors $\vec{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}-2\hat{j}+4\hat{k}$.

Showing the 12 most recent of 46 on this concept.

1. 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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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 $\theta=135^\circ$. …

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3. 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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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 $(\vec a\cdot\vec b)^2=a^2b^2\cos^2\theta$. …

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5. 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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6. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Let $\pi_1$ be the plane determined by the vectors $\bar{i}+2\bar{j}$ and $3\bar{j}-2\bar{k}$. Let $\pi_2$ be the plane determined by the vectors $\bar{j}+2\bar{k}$ and $3\bar{k}-2\bar{i}$. If $\theta$ is the angle between $\pi_1$ and $\pi_2$, then $\cos\theta=$ (A) $\dfrac{7}{26}$ (B) $-\dfrac{14}{29}$ (C) $-\dfrac{32}{5\sqrt{2}}$ (D) $\dfrac{23}{38}$

   ›Reveal solutionSolution

   The angle between two planes equals the angle between their normal vectors, found here via cross products of the given spanning vectors, giving $\cos\theta=-\dfrac{14}{29}$.

   Concept and Intuition

   A plane spanned by two vectors has a normal vector equal to their cross product. Once both planes' normals are known, the angle between the planes is the angle between these normals (up to a sign ambiguity, which the options resolve for us).

   Step-by-Step Solution

   1. $\pi_1$ is spanned by $\bar i+2\bar j=(1,2,0)$ and $3\bar j-2\bar k=(0,3,-2)$. Normal $n_1 = (1,2,0)\times(0,3,-2)$: $n_1 = \left(2(-2)-0(3),\ -(1(-2)-0(0)),\ 1(3)-2(0)\right) = (-4,\ 2,\ 3)$.
   2. $\pi_2$ is spanned by $\bar j+2\bar k=(0,1,2)$ and $3\bar k-2\bar i=(-2,0,3)$. Normal $n_2 = (0,1,2)\times(-2,0,3)$: $n_2 = \left(1(3)-2(0),\ -(0(3)-2(-2)),\ 0(0)-1(-2)\right) = (3,\ -4,\ 2)$.
   3. Dot product: $n_1\cdot n_2 = (-4)(3)+(2)(-4)+(3)(2) = -12-8+6 = -14$.
   4. Magnitudes: $|n_1|=\sqrt{16+4+9}=\sqrt{29}$, $|n_2|=\sqrt{9+16+4}=\sqrt{29}$. …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $\theta$ is the angle between $\overline{f} = i + 2j - 3k$ and $\overline{g} = 2i - 3j + ak$ and $\sin\theta = \sqrt{\dfrac{24}{28}}$ then $7a^2 + 24a =$ (A) $10$ (B) $12$ (C) $36$ (D) $15$

   ›Reveal solutionSolution

   This tests using $\cos\theta=\dfrac{\overline f\cdot\overline g}{|\overline f||\overline g|}$ together with $\sin^2\theta+\cos^2\theta=1$ to build an equation in the unknown $a$. Answer: $7a^2+24a=10$.

   Concept and Intuition

   Given $\sin\theta$ between two vectors, first get $\cos^2\theta$ from the Pythagorean identity, then equate $\cos^2\theta$ to $\left(\dfrac{\overline f\cdot\overline g}{|\overline f||\overline g|}\right)^2$ — squaring avoids sign ambiguity and leaves a clean polynomial equation in $a$.

   Step-by-Step Solution

   1. $\overline f=i+2j-3k$, $\overline g=2i-3j+ak$.
   2. $\overline f\cdot\overline g=(1)(2)+(2)(-3)+(-3)(a)=2-6-3a=-(4+3a)$.
   3. $|\overline f|^2=1+4+9=14$. $|\overline g|^2=4+9+a^2=13+a^2$.
   4. Given $\sin^2\theta=\dfrac{24}{28}=\dfrac67$, so $\cos^2\theta=1-\dfrac67=\dfrac17$. …

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8. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Let L be the line passing through the points $\bar{i}-9\bar{k}$ and $7\bar{j}+\bar{k}$ and $\pi$ be the plane passing through the point $6\bar{i}+\bar{j}$ and perpendicular to the vector $\bar{i}+\bar{j}+\bar{k}$. If $\theta$ is the angle between L and $\pi$, then $\sin\theta =$ (A) $\dfrac{8\sqrt{2}}{15}$ (B) $\dfrac{3\sqrt{3}}{8}$ (C) $\dfrac{7}{13}$ (D) $\dfrac{24}{25}$

   ›Reveal solutionSolution

   This tests the line–plane angle formula $\sin\theta=\dfrac{|\vec d\cdot\bar n|}{|\vec d||\bar n|}$ using $L$'s direction vector and $\pi$'s normal; the answer is $\dfrac{8\sqrt2}{15}$.

   Concept and Intuition

   The angle between a line and a plane is measured from the line to its projection on the plane, so it uses sine, not cosine — because the plane's normal is perpendicular to the plane itself. If $\phi$ is the angle between the line's direction $\vec d$ and the normal $\bar n$, then $\theta = 90^\circ-\phi$, so $\sin\theta=\cos\phi=\dfrac{|\vec d\cdot\bar n|}{|\vec d||\bar n|}$.

   Step-by-Step Solution

   1. Direction of $L$: $\vec d = (7\bar j+\bar k)-(\bar i-9\bar k) = -\bar i+7\bar j+10\bar k$.
   2. The plane is perpendicular to $\bar i+\bar j+\bar k$, so this vector IS the plane's normal $\bar n$ — the point $6\bar i+\bar j$ is not needed for the angle.
   3. $\vec d\cdot\bar n = (-1)(1)+(7)(1)+(10)(1)=16$.
   4. $|\vec d|=\sqrt{(-1)^2+7^2+10^2}=\sqrt{150}=5\sqrt6$, and $|\bar n|=\sqrt{3}$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the points A, B, C, D with position vectors $\bar{i}+\bar{j}-\bar{k}$, $\bar{i}-\bar{j}+2\bar{k}$, $\bar{i}-2\bar{j}+\bar{k}$, $2\bar{i}+\bar{j}+\bar{k}$ respectively form a tetrahedron, then the angle between the faces ABC and ABD of the tetrahedron is (A) $\mathrm{Cos}^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$ (B) $\mathrm{Cos}^{-1}\left(\dfrac{-4}{5}\right)$ (C) $\mathrm{Cos}^{-1}\left(\dfrac{3}{5}\right)$ (D) $\mathrm{Cos}^{-1}\left(\dfrac{\sqrt{29}}{\sqrt{33}}\right)$

   ›Reveal solutionSolution

   The angle between the two triangular faces sharing edge AB equals the angle between their normal vectors, computed via a pair of cross products as $\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

   Concept and Intuition

   The dihedral angle between two planes meeting along a common edge can be found from the angle between their normal vectors (normals are perpendicular to their respective planes, so the angle between normals directly reflects the angle between the planes, up to sign conventions).

   Step-by-Step Solution

   1. Position vectors: $A=(1,1,-1)$, $B=(1,-1,2)$, $C=(1,-2,1)$, $D=(2,1,1)$.
   2. Compute edge vectors from $A$: $\overrightarrow{AB}=B-A=(0,-2,3)$, $\overrightarrow{AC}=C-A=(0,-3,2)$, $\overrightarrow{AD}=D-A=(1,0,2)$.
   3. Normal to face $ABC$: $\bar n_1=\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\0&-3&2\end{vmatrix} = \bar i[(-2)(2)-(3)(-3)] - \bar j[(0)(2)-(3)(0)]+\bar k[(0)(-3)-(-2)(0)] = \bar i(-4+9) - \bar j(0) + \bar k(0) = (5,0,0)$.
   4. Normal to face $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$. …

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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.The dot product of unit vectors $\hat{n}_1$ and $\hat{n}_2$ that are parallel to $5\hat{i} + 12\hat{j}$ and $3\hat{i} + 4\hat{j}$ respectively is (A) $\frac{63}{65}$ (B) $63$ (C) $\frac{63}{4225}$ (D) $\frac{63}{845}$

    ›Reveal solutionSolution

    This tests unit-vector construction and the dot product formula $\hat a\cdot\hat b=\cos\theta$; the answer is $\frac{63}{65}$.

    Concept and Intuition

    A unit vector just points in the same direction as the original vector but has magnitude 1 — you get

    it by dividing each component by the vector's own magnitude. Once both vectors are unit vectors,

    their dot product is simply $\cos\theta$ between them, computed the usual way: sum of the products

    of corresponding components.

    Step-by-Step Solution

    1. Magnitude of $5\hat i+12\hat j$: $\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$. So $\hat n_1 = \dfrac{5}{13}\hat i + \dfrac{12}{13}\hat j$.
    2. Magnitude of $3\hat i+4\hat j$: $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. So $\hat n_2 = \dfrac{3}{5}\hat i + \dfrac{4}{5}\hat j$. …

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11. 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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12. 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 $1+\cos\theta=2\cos^2(\theta/2)$: $2+2\cos\theta=4\cos^2(\theta/2)$. …

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