Q.The angle between $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$ is

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $\vec{a} = 4\vec{i} - \vec{j} + \alpha\vec{k}$ and $\vec{b} = \vec{i} + \alpha\vec{j} - 4\vec{k}$ be two vectors. If $\alpha_1, \alpha_2$ ($\alpha_1 < \alpha_2$) are two different values of $\alpha$ such that $(\vec{a}, \vec{b}) = \cos^{-1}\left(-\dfrac{2}{7}\right)$, then $\alpha_1 + 2\alpha_2 =$ (A) $15$ (B) $24$ (C) $33$ (D) $52$

   ›Reveal solutionSolution

   Setting up $\cos\theta=\vec a\cdot\vec b/(|\vec a||\vec b|)=-2/7$ gives a quadratic in $\alpha$ with roots $2$ and $15.5$; then $\alpha_1+2\alpha_2=33$.

   Concept and Intuition

   Both vectors have the same magnitude expression in $\alpha$ (a nice simplification to notice first), which keeps the resulting equation a clean single-variable quadratic instead of something messier.

   Step-by-Step Solution

   1. $\vec a\cdot\vec b=4(1)+(-1)(\alpha)+\alpha(-4)=4-\alpha-4\alpha=4-5\alpha$.
   2. $|\vec a|=\sqrt{16+1+\alpha^2}=\sqrt{17+\alpha^2}$ and $|\vec b|=\sqrt{1+\alpha^2+16}=\sqrt{17+\alpha^2}$ — identical.
   3. So $\cos\theta=\dfrac{4-5\alpha}{17+\alpha^2}=-\dfrac27$.
   4. Cross-multiply: $7(4-5\alpha)=-2(17+\alpha^2)\Rightarrow28-35\alpha=-34-2\alpha^2\Rightarrow2\alpha^2-35\alpha+62=0$.
   5. Discriminant $=35^2-4(2)(62)=1225-496=729=27^2$. Roots: $\alpha=\dfrac{35\pm27}{4}$, i.e. $\alpha=2$ or $\alpha=15.5$.
   6. Since $\alpha_1<\alpha_2$: $\alpha_1=2,\ \alpha_2=15.5$. Then $\alpha_1+2\alpha_2=2+2(15.5)=2+31=33$.

   Common Mistakes

   • Not noticing $|\vec a|=|\vec b|$ and computing both magnitudes the long way (more error-prone but same result).
   • Sign error cross-multiplying the negative cosine value.

   ✓Final answer

   The correct option is (C) — $33$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The resultant magnitude of two vectors is equal to magnitude of both the vectors separately, then the angle between the two vectors is (A) $60°$ (B) $80°$ (C) $120°$ (D) $45°$

   ›Reveal solutionSolution

   Setting the resultant's magnitude equal to each vector's own magnitude in the law of cosines forces $\cos\theta=-1/2$, i.e. $\theta=120°$ — option (C).

   Concept and Intuition

   For two vectors of equal magnitude $A=B=R$ (and the resultant also equal to $R$), the vector triangle formed is actually an equilateral triangle (all three sides equal), whose interior angle is $60°$ — but the angle between the vectors (the angle you'd measure if you place them tail-to-tail) is the exterior supplement, $180°-60°=120°$. The law of cosines derivation below confirms this directly.

   Step-by-Step Solution

   1. Law of cosines for resultant: $R^2=A^2+B^2+2AB\cos\theta$.
   2. Given $A=B=R$ (resultant magnitude equals magnitude of each vector): $R^2=R^2+R^2+2R^2\cos\theta$.
   3. Divide throughout by $R^2$: $1=1+1+2\cos\theta \Rightarrow 1=2+2\cos\theta$.
   4. So $2\cos\theta=-1 \Rightarrow \cos\theta=-\dfrac12$.
   5. $\theta=\cos^{-1}(-1/2)=120°$.

   Common Mistakes

   • Confusing the angle between the vectors with the internal angle of the vector triangle (which would be $60°$) — these are supplementary, not equal.
   • Sign error when moving terms, giving $\cos\theta=+1/2$ (i.e. $60°$) instead of $-1/2$.

   ✓Final answer

   The correct option is (C) — $120°$.

   ANSWER: C

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. $=3^2+5^2-2(3)(5)\cos(\pi/3) = 9+25-30\left(\dfrac12\right)=34-15=19$.
   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

4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $A(1, 0, 1)$, $B(0, 1, -1)$, $C(-1, 1, 0)$ are the vertices of a triangle $ABC$, then $\cos^2 A + \cos^2 B =$ (A) $\dfrac{1}{2\sqrt{3}}$ (B) $\dfrac{1}{3\sqrt{2}}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{7}{9}$

   ›Reveal solutionSolution

   Use position vectors to get the two enclosed sides at each vertex, apply the vector dot-product formula for the cosine of the included angle, then combine the squares.

   Concept and Intuition

   The angle of a triangle at a vertex is the angle between the two vectors formed by that vertex to the other two vertices. Vector algebra gives this directly via $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$, with no need to first find all three side lengths and invoke the cosine rule separately for each angle.

   Step-by-Step Solution

   1. Compute the vectors from $A$: $\vec{AB}=B-A=(-1,1,-2)$, $\vec{AC}=C-A=(-2,1,-1)$.
   2. $|\vec{AB}|=\sqrt{1+1+4}=\sqrt6$, $|\vec{AC}|=\sqrt{4+1+1}=\sqrt6$, and $\vec{AB}\cdot\vec{AC}=(-1)(-2)+(1)(1)+(-2)(-1)=2+1+2=5$.
   3. So $\cos A=\dfrac{5}{6}$, giving $\cos^2A=\dfrac{25}{36}$.
   4. Compute the vectors from $B$: $\vec{BA}=A-B=(1,-1,2)$, $\vec{BC}=C-B=(-1,0,1)$.
   5. $|\vec{BA}|=\sqrt6$, $|\vec{BC}|=\sqrt{1+0+1}=\sqrt2$, and $\vec{BA}\cdot\vec{BC}=(1)(-1)+(-1)(0)+(2)(1)=1$.
   6. So $\cos B=\dfrac{1}{\sqrt6\cdot\sqrt2}=\dfrac{1}{2\sqrt3}$, giving $\cos^2B=\dfrac{1}{12}$.
   7. Add: $\cos^2A+\cos^2B=\dfrac{25}{36}+\dfrac{3}{36}=\dfrac{28}{36}=\dfrac{7}{9}$.

   Common Mistakes

   • Using $\vec{AB}$ and $\vec{BA}$ inconsistently for the two angles (each vertex's angle uses the vectors emanating from that vertex).
   • Forgetting to square the cosines before adding (the question asks for $\cos^2A+\cos^2B$, not $\cos A+\cos B$).

   ✓Final answer

   The correct option is (D) — $\dfrac{7}{9}$.

   ANSWER: D

5. 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 $\bar c.\bar a=15.5-3-7.5=5$.
   7. Also $\bar c.\bar a=|\bar c||\bar a|\cos\theta=5\cdot2\cdot\cos\theta=10\cos\theta$, so $10\cos\theta=5 \Rightarrow \cos\theta=\dfrac12$.
   8. Therefore $\theta=\dfrac{\pi}{3}$.

   Common Mistakes

   • Forgetting the factor of 2 in the cross-term expansion of $|\bar a+\bar b+\bar c|^2$.
   • Mixing up which pair of vectors' angle is asked (here $(\bar c,\bar a)$, not $(\bar a, \bar c)$ reversed with a different sign — dot product is symmetric so this doesn't actually matter, but it's easy to substitute the wrong known angle by mistake).

   ✓Final answer

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

   ANSWER: C

6. 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

7. 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. $\sin\phi=\dfrac{12}{\sqrt{35}\sqrt{10}}=\dfrac{12}{\sqrt{350}}=\dfrac{12}{5\sqrt{14}}$.
   7. Rationalize to compare with the options: $\dfrac{12}{5\sqrt{14}}=\dfrac{12\sqrt{14}}{70}=\dfrac{6\sqrt{14}}{35}$, and $\dfrac{6\sqrt2}{5\sqrt7}=\dfrac{6\sqrt2\cdot\sqrt7}{5\cdot7}=\dfrac{6\sqrt{14}}{35}$ — identical. So $\phi=\sin^{-1}\!\left(\dfrac{6\sqrt2}{5\sqrt7}\right)$.

   Common Mistakes

   • Using $\cos\phi=\dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$ (correct for line-vs-normal, wrong for line-vs-plane) — this is exactly what separates options (A)/(C) from the correct (B).
   • Taking the normal as $\vec{AB}\times\vec{AO}$ or similar without checking it still spans the same plane (any two independent vectors in the face work, direction/sign doesn't matter since we take absolute value).

   ✓Final answer

   The correct option is (B) — $\sin^{-1}\left(\dfrac{6\sqrt{2}}{5\sqrt{7}}\right)$.

   ANSWER: B

8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Angle between a diagonal of a cube and a diagonal of its face which are coterminus is (A) $\dfrac{\pi}{2}$ (B) $\cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)$ (C) $\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$ (D) $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

   ›Reveal solutionSolution

   Placing a cube vertex at the origin and writing down the coterminous space diagonal and face diagonal as vectors, the dot-product formula gives $\cos\theta=\sqrt{2/3}$.

   Concept and Intuition

   "Coterminus" here means both diagonals start from the same vertex. Vector geometry makes 3D angle problems in a cube routine: assign coordinates to the cube's vertices, write each diagonal as a displacement vector from the shared vertex, and use $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$.

   Step-by-Step Solution

   1. Take a unit cube with one vertex at the origin $O=(0,0,0)$ and edges along the axes, so the opposite vertex is $(1,1,1)$.
   2. The space (body) diagonal from $O$ is the vector $\vec d_1=(1,1,1)$, with $|\vec d_1|=\sqrt{3}$.
   3. A face diagonal from the same vertex $O$, lying in the face $z=0$, goes to $(1,1,0)$: $\vec d_2=(1,1,0)$, with $|\vec d_2|=\sqrt{2}$.
   4. Dot product: $\vec d_1\cdot\vec d_2=1(1)+1(1)+1(0)=2$.
   5. $\cos\theta=\dfrac{2}{\sqrt3\cdot\sqrt2}=\dfrac{2}{\sqrt6}=\dfrac{2}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}=\dfrac{2\sqrt6}{6}=\dfrac{\sqrt6}{3}$.
   6. Check this equals $\sqrt{2/3}$: $\left(\dfrac{\sqrt6}{3}\right)^2=\dfrac{6}{9}=\dfrac23$. Yes, so $\cos\theta=\sqrt{2/3}$.
   7. $\theta=\cos^{-1}\sqrt{2/3}$.

   Common Mistakes

   • Picking a face diagonal that does NOT share the same starting vertex as the space diagonal (not "coterminus"), which gives a different, wrong angle.
   • Arithmetic slip simplifying $\dfrac{2}{\sqrt6}$ to $\sqrt{2/3}$ — always square back to check.

   ✓Final answer

   The correct option is (B) — $\cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)$.

   ANSWER: B

9. 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

10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.The set of all real values of $c$ so that the angle between the vectors $\bar{a}=cx\bar{i}-6\bar{j}+3\bar{k}$ and $\bar{b}=x\bar{i}+2\bar{j}+2cx\bar{k}$ is an obtuse angle for all real $x$ is (A) $\left(0, \dfrac{4}{3}\right]$ (B) $\left(0, \dfrac{2}{3}\right]$ (C) $\left(-\dfrac{2}{3}, 0\right)$ (D) $\left(-\dfrac{4}{3}, 0\right]$

    ›Reveal solutionSolution

    This tests translating "angle is obtuse for all $x$" into "dot product is negative for all $x$" and then solving the resulting quadratic-in-$x$ inequality for the parameter $c$. The answer is $c\in\left(-\tfrac43,0\right]$.

    Concept and Intuition

    The angle between two vectors is obtuse exactly when their dot product is negative (and they aren't anti-parallel making it exactly $180^\circ$, which doesn't arise as a boundary issue here). So "obtuse for all real $x$" becomes: the expression $\bar a\cdot\bar b$, viewed as a quadratic in $x$ with $c$ as parameter, is negative for every $x$. That requires the parabola (in $x$) to open downward with no real roots — or be a negative constant.

    Step-by-Step Solution

    1. $\bar a\cdot\bar b=(cx)(x)+(-6)(2)+(3)(2cx)=cx^2-12+6cx=cx^2+6cx-12$.
    2. Need $f(x)=cx^2+6cx-12<0$ for all real $x$.
    3. Case $c=0$: $f(x)=-12<0$ always — satisfies the condition.
    4. Case $c\ne0$: for a quadratic to be negative for all $x$, it must open downward ($c<0$) and have no real roots (discriminant $<0$).
    5. Discriminant $=(6c)^2-4c(-12)=36c^2+48c<0\Rightarrow 12c(3c+4)<0\Rightarrow c(3c+4)<0$.
    6. This holds for $-\tfrac43<c<0$, which is already consistent with $c<0$.
    7. Combining both cases: $c\in\left(-\tfrac43,0\right)\cup\{0\}=\left(-\tfrac43,0\right]$.

    Common Mistakes

    • Forgetting to separately check $c=0$ (the quadratic degenerates to a constant) and thus excluding it from the answer.
    • Sign errors when factoring the discriminant inequality $12c(3c+4)<0$.

    ✓Final answer

    The correct option is (D) — $\left(-\dfrac43,0\right]$.

    ANSWER: D

11. 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 $|\bar f+\bar g+\bar h|=a\sqrt3$.
    4. $\cos\theta = \dfrac{(\bar f+\bar g+\bar h)\cdot\bar h}{|\bar f+\bar g+\bar h|\,|\bar h|} = \dfrac{a^2}{a\sqrt3\cdot a}=\dfrac{1}{\sqrt3}$.
    5. So $\theta = \cos^{-1}(1/\sqrt3)$.

    Common Mistakes

    • Forgetting the cross terms vanish because the vectors are orthogonal, not just non-collinear.
    • Mixing up $\cos^{-1}(\sqrt3/4)$ with $\cos^{-1}(1/\sqrt3)$ — the magnitude of the sum vector is $a\sqrt3$, not $2a$.

    ✓Final answer

    The correct option is (B) — $\cos^{-1}\left(\dfrac{1}{\sqrt3}\right)$.

    ANSWER: B

12. 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