Q.Find the angle between the vectors $2\hat{i}-\hat{j}+\hat{k}$ and $3\hat{i}+4\hat{j}-\hat{k}$.

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.The angle between two vectors $(\hat{i}+\hat{j})$ and $(\hat{j}+\hat{k})$ is (A) $60^{\circ}$ (B) $30^{\circ}$ (C) $45^{\circ}$ (D) $90^{\circ}$

   ›Reveal solutionSolution

   A direct application of the dot-product formula for the angle between two vectors. Answer: (A) $60^{\circ}$.

   Concept and Intuition

   The angle between two vectors can be found from $\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta$. Writing each vector in component form and computing the dot product and magnitudes directly gives $\cos\theta$, from which $\theta$ follows.

   Step-by-Step Solution

   1. Write $\vec{A} = (1,1,0)$ and $\vec{B} = (0,1,1)$.
   2. Compute the dot product: $\vec{A}\cdot\vec{B} = (1)(0)+(1)(1)+(0)(1) = 1$.
   3. Compute magnitudes: $|\vec{A}| = \sqrt{1^{2}+1^{2}+0^{2}} = \sqrt{2}$, similarly $|\vec{B}| = \sqrt{2}$.
   4. Apply the formula: $\cos\theta = \dfrac{1}{\sqrt{2}\cdot\sqrt{2}} = \dfrac{1}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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. $\vec A\cdot\vec B=(2)(4)+(4)(2)+(4)(-4)=8+8-16=0$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The angle made by the resultant vector of two vectors $2\bar{i}+3\bar{j}+4\bar{k}$ and $2\bar{i}-7\bar{j}-4\bar{k}$ with x-axis is (A) $60°$ (B) $45°$ (C) $90°$ (D) $120°$

   ›Reveal solutionSolution

   Add the vectors component-wise, then use the direction-cosine formula with the x-axis. Answer: $45°$.

   Concept and Intuition

   The angle a vector makes with the x-axis is found from its direction cosine $\cos\theta = \dfrac{R_x}{|\vec R|}$, where $R_x$ is the x-component of the resultant vector and $|\vec R|$ is its magnitude.

   Step-by-Step Solution

   1. Add the vectors: $(2\bar i+3\bar j+4\bar k) + (2\bar i-7\bar j-4\bar k) = 4\bar i - 4\bar j + 0\bar k$.
   2. Magnitude: $|\vec R| = \sqrt{4^2+(-4)^2+0^2} = \sqrt{32} = 4\sqrt2$.
   3. Direction cosine with x-axis: $\cos\theta = \dfrac{R_x}{|\vec R|} = \dfrac{4}{4\sqrt2} = \dfrac{1}{\sqrt2}$. …

   🔒 Start your 14-day free trial to see the full solution →
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 …

   🔒 Start your 14-day free trial to see the full solution →
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: $|n_1|=\sqrt{16+4+9}=\sqrt{29}$, $|n_2|=\sqrt{9+16+4}=\sqrt{29}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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

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

   🔒 Start your 14-day free trial to see the full solution →
8. 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. $|\bar b|=\sqrt{(\sqrt2)^2+(-\sqrt2)^2+0^2}=\sqrt{2+2}=\sqrt4=2$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $|\bar f+\bar g+\bar h|=a\sqrt3$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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: $6(\bar a\cdot\bar b)=3\Rightarrow \bar a\cdot\bar b = \dfrac12$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.Let $(\bar{a}, \bar{b})$ denote the angle between vectors $\bar{a}$ and $\bar{b}$. If $\bar{a} = 2\bar{i} + 3\bar{j} + 6\bar{k}$, $\bar{a}.\bar{b} = 4$ and $(\bar{a}, \bar{b}) = \cos^{-1}\left(\dfrac{4}{21}\right)$, then $\bar{a} + \bar{b} =$ (A) $3\bar{i} + \bar{j} + 8\bar{k}$ (B) $3\bar{i} + 5\bar{j} + 4\bar{k}$ (C) $3\bar{i} + 5\bar{j} + 8\bar{k}$ (D) $\bar{i} + \bar{j} + 8\bar{k}$

    ›Reveal solutionSolution

    Using $|\bar a+\bar b|^2=|\bar a|^2+2\bar a\cdot\bar b+|\bar b|^2$ pins down the magnitude of $\bar a+\bar b$, which uniquely identifies the matching option: $\bar i+\bar j+8\bar k$.

    Concept and Intuition

    Even without knowing $\bar b$ explicitly, the magnitude of $\bar a+\bar b$ can be computed purely from $|\bar a|$, $|\bar b|$, and $\bar a\cdot\bar b$ — this is the vector analogue of the law of cosines. Since only one answer option has that exact magnitude, it must be the correct vector sum (and it can be verified to be consistent with an actual vector $\bar b$).

    Step-by-Step Solution

    1. $\bar a=2\bar i+3\bar j+6\bar k\Rightarrow|\bar a|=\sqrt{4+9+36}=\sqrt{49}=7$.
    2. $\cos(\bar a,\bar b)=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{4}{21}$. Substituting $\bar a\cdot\bar b=4$: $\dfrac{4}{7|\bar b|}=\dfrac{4}{21}\Rightarrow|\bar b|=3$.
    3. $|\bar a+\bar b|^2=|\bar a|^2+2(\bar a\cdot\bar b)+|\bar b|^2=49+2(4)+9=66$.
    4. Compute $|\cdot|^2$ for each option: (A) $9+1+64=74$; (B) $9+25+16=50$; (C) $9+25+64=98$; (D) $1+1+64=66$.
    5. Only (D) matches 66. …

    🔒 Start your 14-day free trial to see the full solution →