Q.Find angle $\theta$ between the vectors $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.

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: $\cos\theta = \dfrac{1}{\sqrt{2}\cdot\sqrt{2}} = \dfrac{1}{2}$.
   5. So $\theta = \cos^{-1}(1/2) = 60^{\circ}$.

   Common Mistakes

   • Miscomputing the dot product by forgetting that $\hat{i},\hat{j},\hat{k}$ are mutually orthogonal (only matching components contribute).
   • Arithmetic slip converting $\cos\theta = 1/2$ into an angle other than $60^{\circ}$.

   ✓Final answer

   The correct option is (A) — $60^{\circ}$.

   ANSWER: A

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

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

4. 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: $|n_1|=\sqrt{16+4+9}=\sqrt{29}$, $|n_2|=\sqrt{9+16+4}=\sqrt{29}$.
   5. $\cos\theta = \dfrac{-14}{\sqrt{29}\cdot\sqrt{29}} = \dfrac{-14}{29}$.

   Common Mistakes

   • Sign or component errors in the cross-product determinant expansion (a very common source of error in these vector-geometry problems).
   • Forgetting that the angle between planes uses the angle between normals directly (with the sign convention matching the given options).

   ✓Final answer

   The correct option is (B) — $-\dfrac{14}{29}$.

   ANSWER: B

5. 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 $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$.
   5. $|\bar n_1|=5$, $|\bar n_2|=\sqrt{16+9+4}=\sqrt{29}$.
   6. $\bar n_1.\bar n_2 = 5(-4)+0(3)+0(2)=-20$.
   7. $\cos\theta=\dfrac{-20}{5\sqrt{29}}=\dfrac{-4}{\sqrt{29}}$, so $\theta=\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

   Common Mistakes

   • Sign errors in the cross-product cofactor expansion (especially the middle term's negative sign).
   • Using edge vectors not sharing a common vertex — always build both normals from vectors emanating from the shared edge's endpoint (here $A$) to keep the computation clean.

   ✓Final answer

   The correct option is (A) — $\mathrm{Cos}^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

   ANSWER: A

6. 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

7. 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

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 $|\bar n|=\sqrt{3}$.
   5. $\sin\theta = \dfrac{16}{5\sqrt6\cdot\sqrt3}=\dfrac{16}{5\sqrt{18}}=\dfrac{16}{15\sqrt2}=\dfrac{16\sqrt2}{30}=\dfrac{8\sqrt2}{15}$.

   Common Mistakes

   • Using $\cos\theta=\dfrac{|\vec d\cdot\bar n|}{|\vec d||\bar n|}$ (that formula is for the angle between two lines or two planes, not a line and a plane).
   • Wasting time trying to use the given points to build the plane's Cartesian equation — only the normal is needed here.

   ✓Final answer

   The correct option is (A) — $\dfrac{8\sqrt2}{15}$.

   ANSWER: A

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

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

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

   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).

10. 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

    • Forgetting to normalize (divide by the magnitudes), landing on the wrong cosine value.
    • Not recognizing $\cos^{-1}(1/2)=\pi/3$ is a standard angle, and instead leaving the answer in an unsimplified inverse-cosine form (which happens to also appear as a distractor option (B) computed for different vectors).

    ✓Final answer

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

    ANSWER: A

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.
    6. $\cos\theta = \dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|} = \dfrac{M}{\sqrt{AB}} = \dfrac{M}{\sqrt{\frac{43M}{19}\cdot\frac{25M}{19}}} = \dfrac{M}{\frac{M}{19}\sqrt{43\times25}} = \dfrac{19}{\sqrt{1075}} = \dfrac{19}{5\sqrt{43}}$ (since $\sqrt{1075}=\sqrt{25\times43}=5\sqrt{43}$).
    7. This cosine value is positive, so $\theta=\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$ is already the acute angle.

    Common Mistakes

    • Sign errors expanding the dot products, especially the cross terms.
    • Forgetting to simplify $\sqrt{1075}$ to $5\sqrt{43}$, leaving the answer in an unmatched form.

    ✓Final answer

    The correct option is (A) — $\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$.

    ANSWER: A

12. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $\bar{a}+\bar{b}+\bar{c}=0$ and $|\bar{a}|=7, |\bar{b}|=5, |\bar{c}|=3$ then the angle between $\bar{b}$ and $\bar{c}$ is (A) $30^{0}$ (B) $45^{0}$ (C) $60^{0}$ (D) $90^{0}$

    ›Reveal solutionSolution

    Using $\bar a+\bar b+\bar c=0$ to eliminate $\bar a$ and squaring magnitudes gives the dot product $\bar b\cdot\bar c$, from which the angle between $\bar b$ and $\bar c$ is $60^\circ$.

    Concept and Intuition

    When three vectors sum to zero, they form a closed triangle, and squaring one vector expressed as the negative sum of the other two directly yields the dot product (hence the angle) between those two.

    Step-by-Step Solution

    1. From $\bar a+\bar b+\bar c=0$, we get $\bar a = -(\bar b+\bar c)$.
    2. Take magnitudes squared: $|\bar a|^2 = |\bar b+\bar c|^2 = |\bar b|^2+|\bar c|^2+2\bar b\cdot\bar c$.
    3. Substitute: $49 = 25+9+2\bar b\cdot\bar c \Rightarrow 2\bar b\cdot\bar c = 15 \Rightarrow \bar b\cdot\bar c=7.5$.
    4. $\cos\theta = \frac{\bar b\cdot\bar c}{|\bar b||\bar c|} = \frac{7.5}{5\times3} = \frac{7.5}{15}=0.5$.
    5. So $\theta = \cos^{-1}(0.5) = 60^\circ$.

    Common Mistakes

    • Forgetting the factor of 2 in the expansion of $|\bar b+\bar c|^2$.
    • Computing the angle between $\bar a$ and one of the others instead of between $\bar b$ and $\bar c$ as asked.

    ✓Final answer

    The correct option is (C) — $60^{0}$.

    ANSWER: C