Q.If the vertices A, B, C of a triangle ABC are $(1, 2, 3), (-1, 0, 0), (0, 1, 2),$ respectively, then find $\angle ABC. [\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}].$

Showing the 12 most recent of 46 on this concept.

1. 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: $\vec{AB}\cdot\vec{AC}=(5)(7)+(-4)(20)+(15)(3)=35-80+45=0$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$ and $C(-1,1,2)$ are the vertices of a tetrahedron, then the acute angle between its face OAB and edge BC is (A) $\cos^{-1}\left(\dfrac{6\sqrt{2}}{5\sqrt{7}}\right)$ (B) $\sin^{-1}\left(\dfrac{6\sqrt{2}}{5\sqrt{7}}\right)$ (C) $\tan^{-1}\left(\dfrac{6\sqrt{2}}{5\sqrt{7}}\right)$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   The angle between a line and a plane is found from $\sin\phi=\dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$, which evaluates to $\sin^{-1}\!\left(\dfrac{6\sqrt2}{5\sqrt7}\right)$.

   Concept and Intuition

   The angle between a line and a plane is the complement of the angle between the line and the plane's normal. If $\psi$ is the angle between the line's direction $\vec d$ and the normal $\vec n$, then the line-plane angle is $\phi=90^\circ-\psi$, so $\sin\phi=\cos\psi=\dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$ — a sine formula, not a cosine formula, which is the key distinguishing feature from the line-normal or plane-plane angle formulas.

   Step-by-Step Solution

   1. Face $OAB$ contains $O,A(1,2,1),B(2,1,3)$; its normal is $\vec n=\vec{OA}\times\vec{OB}$.
   2. $\vec n=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&1\\2&1&3\end{vmatrix}=\mathbf i(6-1)-\mathbf j(3-2)+\mathbf k(1-4)=(5,-1,-3)$.
   3. Edge direction $\vec d=\vec{BC}=C-B=(-1-2,\,1-1,\,2-3)=(-3,0,-1)$.
   4. $\vec n\cdot\vec d=5(-3)+(-1)(0)+(-3)(-1)=-15+0+3=-12$.
   5. $|\vec n|=\sqrt{25+1+9}=\sqrt{35}$, $|\vec d|=\sqrt{9+0+1}=\sqrt{10}$.
   6. $\sin\phi=\dfrac{12}{\sqrt{35}\sqrt{10}}=\dfrac{12}{\sqrt{350}}=\dfrac{12}{5\sqrt{14}}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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. $=3^2+5^2-2(3)(5)\cos(\pi/3) = 9+25-30\left(\dfrac12\right)=34-15=19$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $|\bar a|=5,\ |\bar b|=8,\ |\bar c|=11$:

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

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

   🔒 Start your 14-day free trial to see the full solution →
6. 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. $\bar n_1\cdot\bar n_2=3(3)+0(-1)+(-3)(-2)=9+0+6=15$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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: $49 = 25+9+2\bar b\cdot\bar c \Rightarrow 2\bar b\cdot\bar c = 15 \Rightarrow \bar b\cdot\bar c=7.5$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $38+2\vec a\cdot\vec c=48\Rightarrow\vec a\cdot\vec c=5$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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: $\vec{OP}\cdot\vec{OQ}=(0)(4)+(1)(-2)+(2)(1)=0-2+2=0$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $\bar c.\bar a=15.5-3-7.5=5$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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 →
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. $\vec A\cdot\vec B=(2)(4)+(4)(2)+(4)(-4)=8+8-16=0$. …

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