Q.Find the scalar and vector products of two vectors $\vec{a} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} - 3\hat{k}$.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A 15 W machine used $(3i + 5j + 6k)$ N force for 6 seconds to lift a body through a distance of X m along x-axis. The value of X is (A) 20m (B) 25m (C) 30m (D) 35m

   ›Reveal solutionSolution

   Total work from power×time is 90 J; only the x-component of the force (3 N)

   acts along the x-axis displacement, giving X = 30 m. Answer: (C).

   Concept and Intuition

   Work done by a constant force over a displacement is $W = \vec F \cdot \vec d$. If the displacement is purely along one axis, only the force

   component along that axis contributes to the work — the other components

   are perpendicular to the motion and do zero work.

   Step-by-Step Solution

   1. Total work delivered by the 15 W machine in 6 s: $W = P t = 15\times 6 = 90$ J.
   2. The displacement is $X$ m purely along the x-axis, i.e. $\vec d = X\hat i$.
   3. Force is $\vec F = 3\hat i + 5\hat j + 6\hat k$ N; only its x-component ($F_x = 3$ N) does work on an x-axis displacement (the j and k components …

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2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Work done in moving a charge perpendicular to electric field $\vec{E}$ is (A) positive (B) negative (C) zero (D) infinity

   ›Reveal solutionSolution

   Displacement perpendicular to the electric field does zero work on the charge.

   Concept and Intuition

   Work done by a force is $W = \vec F \cdot \vec d = Fd\cos\theta$. The electrostatic force on a charge is $q\vec E$, which is parallel to $\vec E$. If the displacement is perpendicular to $\vec E$, the angle between force and displacement is $90°$, and the dot product vanishes — physically, this is exactly like moving along an equipotential surface, where no work is needed.

   Step-by-Step Solution

   1. Force on charge: $\vec F = q\vec E$, direction along $\vec E$. …

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3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If a force $\vec{F} = (3\hat{i} - 2\hat{j})$ N acting on a body displaces it from point (1 m, 2 m) to point (2 m, 0 m), then work done by the force is (A) 5 J (B) 6 J (C) 4 J (D) 7 J

   ›Reveal solutionSolution

   This tests the dot-product definition of work done by a constant force in vector form; the answer is (D) 7 J.

   Concept and Intuition

   Work done by a constant force $\vec{F}$ during a displacement $\vec{d}$ is $W = \vec{F}\cdot\vec{d}$, not simply force magnitude times distance travelled along some path — because a force does work only along the component of displacement parallel to it. So the first step is always to find the displacement vector, then take the dot product.

   Step-by-Step Solution

   1. Initial point: $(1, 2)$ m; final point: $(2, 0)$ m.
   2. Displacement vector: $\vec{d} = (2-1)\hat{i} + (0-2)\hat{j} = \hat{i} - 2\hat{j}$ m.
   3. Given force: $\vec{F} = 3\hat{i} - 2\hat{j}$ N. …

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4. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If a constant force of $(2\hat{i}+3\hat{j}+4\hat{k})\ \mathrm{N}$ acting on a body of mass 5 kg displaces it from $(3\hat{i}-4\hat{k})\ \mathrm{m}$ to $(2\hat{i}+2\hat{j}+3\hat{k})\ \mathrm{m}$, then the work done by the force on the body is (A) 32 J (B) 28 J (C) 36 J (D) 44 J

   ›Reveal solutionSolution

   Work done by a constant force is the dot product of force and displacement vectors; here it works out to 32 J.

   Concept and Intuition

   For a constant force, work is simply $W = \vec F\cdot\vec d$, where $\vec d$ is the displacement vector (final position minus initial position) — mass is irrelevant to this calculation since force and displacement are both given directly.

   Step-by-Step Solution

   1. Displacement $\vec d = \vec r_f - \vec r_i = (2\hat i+2\hat j+3\hat k) - (3\hat i+0\hat j-4\hat k) = (2-3)\hat i + (2-0)\hat j + (3-(-4))\hat k = -\hat i+2\hat j+7\hat k$ m.
   2. Force $\vec F = 2\hat i+3\hat j+4\hat k$ N. …

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5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.A vector perpendicular to the vector $(4\hat{i} - 3\hat{j})$ is (A) $4\hat{i} + 3\hat{j}$ (B) $6\hat{i}$ (C) $3\hat{i} - 4\hat{j}$ (D) $7\hat{k}$

   ›Reveal solutionSolution

   This tests the basic perpendicularity test via the dot product, plus the geometric fact that $\hat k$ is orthogonal to every vector confined to the $xy$-plane.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Since $\hat i,\hat j,\hat k$ are mutually perpendicular unit vectors, any vector lying purely along $\hat k$ is automatically perpendicular to any vector lying purely in the $\hat i$–$\hat j$ plane — no computation needed once you see this structure.

   Step-by-Step Solution

   1. Given vector: $\vec A=4\hat i-3\hat j$ (no $\hat k$ component).
   2. Check (A) $4\hat i+3\hat j$: $\vec A\cdot(4\hat i+3\hat j)=16-9=7\ne0$.
   3. Check (B) $6\hat i$: $\vec A\cdot 6\hat i=24\ne0$.
   4. Check (C) $3\hat i-4\hat j$: $\vec A\cdot(3\hat i-4\hat j)=12+12=24\ne0$. …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If a force $(3\hat{i}+2\hat{j}+5\hat{k})$ N acting on a body displaces it through $(2\hat{i}+2\hat{j}+1\hat{k})$ m, then the work done by the force on the body is (A) 40 J (B) 20 J (C) 15 J (D) 25 J

   ›Reveal solutionSolution

   Work is the dot product of force and displacement vectors: $W=\vec F\cdot\vec d=15\ \mathrm J$.

   Concept and Intuition

   Work done by a constant force through a displacement is defined as the scalar (dot) product $W=\vec F\cdot\vec d=F_xd_x+F_yd_y+F_zd_z$ — only the component of force along the displacement direction contributes.

   Step-by-Step Solution

   1. $\vec F=3\hat i+2\hat j+5\hat k$ N, $\vec d=2\hat i+2\hat j+1\hat k$ m.
   2. $W=\vec F\cdot\vec d = (3)(2)+(2)(2)+(5)(1)$.
   3. $=6+4+5=15$.
   4. Units: N$\cdot$m = J. So $W=15\ \mathrm J$. …

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7. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.A force of $(4\vec{i} + 2\vec{j} + \vec{k})$ N is acting on a particle of mass 2 kg displaces the particle from a position of $(2\vec{i} + 2\vec{j} + \vec{k})$ m to a position of $(4\vec{i} + 3\vec{j} + 2\vec{k})$ m. The work done by the force on the particle in joules is (A) 21 J (B) 11 J (C) 14 J (D) 18 J

   ›Reveal solutionSolution

   Work done by a constant force is $\vec F\cdot\Delta\vec r$; computing this dot product directly gives 11 J, option (B).

   Concept and Intuition

   For a constant force, work depends only on the net displacement vector (not the path), so we just need the displacement between the two given position vectors and dot it with the force.

   Step-by-Step Solution

   1. Displacement $\Delta\vec r=(4\vec i+3\vec j+2\vec k)-(2\vec i+2\vec j+\vec k)=2\vec i+\vec j+\vec k$ (in metres).
   2. Force $\vec F=4\vec i+2\vec j+\vec k$ (in newtons).
   3. Work $W=\vec F\cdot\Delta\vec r=(4)(2)+(2)(1)+(1)(1)=8+2+1=11$ J. …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Two forces whose magnitudes are in the ratio 5:3 are acting at a point at an angle 60° simultaneously. If the resultant of the two forces is 35 N, then the magnitudes of two forces respectively are (A) 3N, 5N (B) 25 N, 9N (C) 25 N, 15 N (D) 12 N, 20 N

   ›Reveal solutionSolution

   Using the ratio to write the forces as $5k,3k$ and applying the resultant-of-two-vectors formula with $\cos 60^\circ=\frac12$ gives $k=5$, so the forces are 25 N and 15 N.

   Concept and Intuition

   When two forces act at a point at a known angle, their resultant magnitude is governed by the parallelogram/triangle law: $R^2=F_1^2+F_2^2+2F_1F_2\cos\theta$. Expressing forces in a given ratio as a common multiple $k$ turns this into a simple algebraic equation for $k$.

   Step-by-Step Solution

   1. Let the two forces be $F_1=5k$ and $F_2=3k$ (ratio $5:3$).
   2. Resultant formula: $R^2 = F_1^2+F_2^2+2F_1F_2\cos\theta$, with $\theta=60^\circ$, $\cos60^\circ=\tfrac12$.
   3. Substitute: $R^2 = 25k^2+9k^2+2(5k)(3k)\left(\tfrac12\right) = 25k^2+9k^2+15k^2=49k^2$.
   4. So $R=7k$. Given $R=35\,\text{N}$: $7k=35 \Rightarrow k=5$. …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.A body of mass 2 kg is moving with a constant acceleration of $(2\hat{i} + 3\hat{j} - \hat{k})$ ms$^{-2}$. If the displacement made by the body is $(3\hat{i} - \hat{j} + 2\hat{k})$ m then the work done is (A) 22 J (B) 2 J (C) 12 J (D) 10 J

   ›Reveal solutionSolution

   Work done by a constant force is the dot product of force and displacement; computing $\vec F=m\vec a$ and dotting with $\vec d$ gives 2 J.

   Concept and Intuition

   For a constant force, work done is simply $W = \vec F\cdot\vec d$ — a scalar (dot) product, so components along mutually perpendicular directions multiply and add, with no cross terms.

   Step-by-Step Solution

   1. Given: $m=2\,\text{kg}$, $\vec a = (2\hat i+3\hat j-\hat k)\,\text{ms}^{-2}$, $\vec d = (3\hat i-\hat j+2\hat k)\,\text{m}$.
   2. Force: $\vec F = m\vec a = 2(2\hat i+3\hat j-\hat k) = 4\hat i+6\hat j-2\hat k\,\text{N}$. …

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10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.A force, $\vec{F}=(4\hat{i}+3\hat{j}-5\hat{k})\ \mathrm{N}$ is acting on a body making an angle $\theta$ with the horizontal. Then the angle '$\theta$' is (A) $\cos^{-1}\left(\dfrac{2\sqrt{2}}{5}\right)$ (B) $\cos^{-1}\left(\dfrac{\sqrt{2}}{5}\right)$ (C) $\cos^{-1}\left(\dfrac{5\sqrt{2}}{9}\right)$ (D) $\cos^{-1}\left(\dfrac{3}{5\sqrt{2}}\right)$

    ›Reveal solutionSolution

    This tests resolving a 3-D force vector into a component along a reference (horizontal) direction and using $\cos\theta=$ (component)/(magnitude).

    Concept and Intuition

    The angle a vector makes with a given reference direction is obtained by projecting the vector onto that direction. If $\hat i$ represents the horizontal reference axis, then the horizontal component of $\vec F$ is simply $F_x$, and $\cos\theta$ between $\vec F$ and $\hat i$ is that component divided by the full magnitude of $\vec F$ — exactly like resolving a force along an incline or a reference line in 2-D, extended to 3-D.

    Step-by-Step Solution

    1. Given $\vec F = 4\hat i + 3\hat j - 5\hat k\ \mathrm N$.
    2. Magnitude: $|\vec F| = \sqrt{4^2+3^2+(-5)^2} = \sqrt{16+9+25} = \sqrt{50} = 5\sqrt2\ \mathrm N$.
    3. The horizontal reference direction is $\hat i$, so the component of $\vec F$ along it is $F_x = 4\ \mathrm N$.
    4. $\cos\theta = \dfrac{F_x}{|\vec F|} = \dfrac{4}{5\sqrt2}$.
    5. Rationalising: $\dfrac{4}{5\sqrt2}\times\dfrac{\sqrt2}{\sqrt2} = \dfrac{4\sqrt2}{10} = \dfrac{2\sqrt2}{5}$.
    6. So $\theta = \cos^{-1}\left(\dfrac{2\sqrt2}{5}\right)$.

    Common Mistakes …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.A motor boat is moving in a river with velocity $\vec{v} = 7\hat{i}+2\hat{j}-5\hat{k}\ \mathrm{ms^{-1}}$. If the flow of water offers resistive force $\vec{F} = 9\hat{i}+3\hat{j}-3\hat{k}\ \mathrm{N}$, then the power of the boat is (A) 13 W (B) 69 W (C) 12 W (D) 84 W

    ›Reveal solutionSolution

    Power associated with a force acting on a moving object is the dot product $\vec F\cdot\vec v$; computing the components directly gives $84$ W.

    Concept and Intuition

    Instantaneous power delivered by (or against) a force is $P=\vec F\cdot\vec v$ — only the component of force along the velocity direction contributes, which is exactly what the dot product captures.

    Step-by-Step Solution

    1. $\vec v=7\hat i+2\hat j-5\hat k\,\mathrm{ms^{-1}}$, $\vec F=9\hat i+3\hat j-3\hat k\,\mathrm N$.
    2. $P=\vec F\cdot\vec v=(9)(7)+(3)(2)+(-3)(-5)$.
    3. $=63+6+15=84\,\mathrm W$.

    Common Mistakes …

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12. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.The dot product of $\vec{A} = (\hat{i} + \hat{j} + \hat{k})$ and the unit vector parallel to $(\hat{i} - \hat{j} + \hat{k})$ is (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{2}{\sqrt{3}}$ (C) $\sqrt{3}$ (D) $3$

    ›Reveal solutionSolution

    Straightforward vector dot-product problem; the result is $1/\sqrt3$ after normalizing the second vector.

    Concept and Intuition

    To find the component of $\vec A$ along a given direction, we dot $\vec A$ with the unit vector in that direction (not the un-normalized vector). This requires first normalizing $(\hat i-\hat j+\hat k)$ by dividing by its own magnitude.

    Step-by-Step Solution

    1. Magnitude of $(\hat i - \hat j+\hat k)$: $\sqrt{1^2+(-1)^2+1^2} = \sqrt3$.
    2. Unit vector: $\hat n = \dfrac{\hat i - \hat j+\hat k}{\sqrt3}$.
    3. Dot product: $\vec A \cdot \hat n = (\hat i+\hat j+\hat k)\cdot \dfrac{\hat i-\hat j+\hat k}{\sqrt3} = \dfrac{(1)(1)+(1)(-1)+(1)(1)}{\sqrt3} = \dfrac{1-1+1}{\sqrt3} = \dfrac{1}{\sqrt3}$. …

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