Q.If $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$ and $\hat{i}-6\hat{j}-\hat{k}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

Showing the 12 most recent of 46 on this concept.

1. 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

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

3. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.A (1, 2, 1), B (2, 3, 2), C (3, 1, 3) and D (2, 1, 3) are the vertices of a tetrahedron. If $\theta$ is the angle between the faces ABC and ABD then $\cos\theta =$ (A) $\dfrac{5}{\sqrt{14}}$ (B) $\dfrac{15}{8\sqrt{7}}$ (C) $\dfrac{3}{\sqrt{14}}$ (D) $\dfrac{5}{2\sqrt{7}}$

   ›Reveal solutionSolution

   This tests finding the dihedral angle between two faces of a tetrahedron via their normal vectors; the answer is $\cos\theta=\dfrac{5}{2\sqrt7}$.

   Concept and Intuition

   The angle between two planes equals the angle between their normal vectors (up to supplement). Each face's normal is found as the cross product of two edge vectors lying in that face, both measured from the shared vertex $A$ and $B$ common to both faces.

   Step-by-Step Solution

   1. $\overrightarrow{AB}=(1,1,1)$, $\overrightarrow{AC}=(2,-1,2)$, $\overrightarrow{AD}=(1,-1,2)$.
   2. Normal to face $ABC$: $\bar n_1=\overrightarrow{AB}\times\overrightarrow{AC}=\big(1(2)-1(-1),\ -(1(2)-1(2)),\ 1(-1)-1(2)\big)=(3,0,-3)$.
   3. Normal to face $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD}=\big(1(2)-1(-1),\ -(1(2)-1(1)),\ 1(-1)-1(1)\big)=(3,-1,-2)$.
   4. $\bar n_1\cdot\bar n_2=3(3)+0(-1)+(-3)(-2)=9+0+6=15$.
   5. $|\bar n_1|=\sqrt{9+0+9}=3\sqrt2$, $|\bar n_2|=\sqrt{9+1+4}=\sqrt{14}$.
   6. $\cos\theta=\dfrac{15}{3\sqrt2\cdot\sqrt{14}}=\dfrac{15}{3\sqrt{28}}=\dfrac{15}{6\sqrt7}=\dfrac{5}{2\sqrt7}$.

   Common Mistakes

   • Using non-common vertex edges (e.g. mixing up which two edges belong to which face) — always pick edges from the shared vertex.
   • Arithmetic slip simplifying $\sqrt{28}=2\sqrt7$.

   ✓Final answer

   The correct option is (D) — $\dfrac{5}{2\sqrt{7}}$.

   ANSWER: D

4. 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

5. 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

6. 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).

7. 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

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 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

10. 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 $38+2\vec a\cdot\vec c=48\Rightarrow\vec a\cdot\vec c=5$.
    5. $\cos\theta=\dfrac{\vec a\cdot\vec c}{|\vec a||\vec c|}=\dfrac{5}{2\times5}=\dfrac12\Rightarrow\theta=\dfrac{\pi}{3}$.

    Common Mistakes

    • Forgetting to double the cross terms when expanding $|\vec a+\vec b+\vec c|^2$.
    • Assuming $\vec a\perp\vec c$ as well just because both are perpendicular to $\vec b$ — that's not implied in 3D.

    ✓Final answer

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

    ANSWER: B

11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.The position vectors of the vertices A and B of a triangle $ABC$ are $\bar{i}+3\bar{j}+4\bar{k}$ and $2\bar{i}+\bar{j}+2\bar{k}$ respectively. If $|\overline{AC}| = 5$ and angle $A = \pi/3$, then $|\overline{BC}| =$ (A) $\sqrt{26}$ (B) $3\sqrt{19}$ (C) $3\sqrt{26}$ (D) $\sqrt{19}$

    ›Reveal solutionSolution

    Find $|AB|$ from the position vectors, then apply the law of cosines at the known angle $A$.

    Concept and Intuition

    Once we know two sides meeting at a vertex ($AB$ and $AC$) and the included angle there ($A$), the third side $BC$ is fixed by the law of cosines — position vectors are just a way of encoding the side length $AB$.

    Step-by-Step Solution

    1. $A=(1,3,4)$, $B=(2,1,2)$, so $\overline{AB}=B-A=(1,-2,-2)$, giving $|AB|=\sqrt{1^2+(-2)^2+(-2)^2}=\sqrt9=3$.
    2. We are given $|AC|=5$ and $\angle A = \pi/3$ (the angle between $AB$ and $AC$ at vertex $A$).
    3. By the law of cosines in $\triangle ABC$: $BC^2 = AB^2+AC^2-2\cdot AB\cdot AC\cos A$.
    4. $=3^2+5^2-2(3)(5)\cos(\pi/3) = 9+25-30\left(\dfrac12\right)=34-15=19$.
    5. $BC=\sqrt{19}$.

    Common Mistakes

    • Computing $|AB|$ from the wrong difference (e.g. $A-B$ instead of $B-A$ — though the magnitude is the same, sign slips elsewhere are common).
    • Using $\cos(\pi/3)=1$ or a wrong standard value instead of $1/2$.

    ✓Final answer

    The correct option is (D) — $\sqrt{19}$.

    ANSWER: D

12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If A = (0, 4, -3), B = (5, 0, 12) and C = (7, 24, 0), then $\angle BAC =$ (A) $60°$ (B) $\mathrm{Cos}^{-1}\left(\frac{16}{\sqrt{13}}\right)$ (C) $\mathrm{Cos}^{-1}\left(\frac{13}{38}\right)$ (D) $90°$

    ›Reveal solutionSolution

    Form the two vectors from $A$ and dot them — the dot product vanishes, so the angle is a right angle. Answer: $90°$.

    Concept and Intuition

    The angle at vertex $A$ between rays $AB$ and $AC$ is found from $\cos(\angle BAC)=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}$. If the numerator (the dot product) is zero, the angle is exactly $90°$ regardless of the vector magnitudes — so it's worth checking the dot product first before computing any magnitudes.

    Step-by-Step Solution

    1. $\vec{AB}=B-A=(5-0,\,0-4,\,12-(-3))=(5,-4,15)$.
    2. $\vec{AC}=C-A=(7-0,\,24-4,\,0-(-3))=(7,20,3)$.
    3. Dot product: $\vec{AB}\cdot\vec{AC}=(5)(7)+(-4)(20)+(15)(3)=35-80+45=0$.
    4. A zero dot product between two nonzero vectors means they are perpendicular, so $\angle BAC=90°$.

    Common Mistakes

    • Sign errors when subtracting coordinates to form the vectors (especially with the negative $z$-coordinate of $A$).
    • Computing the full $\cos\theta$ expression (magnitudes and all) when the zero dot product already settles the answer.

    ✓Final answer

    The correct option is (D) — $90°$.

    ANSWER: D