Work and Moment as Vector Products
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Direct answer
Two physical quantities carry nearly every JEE vector application: work and moment. Work is the dot product W = F·d = |F||d| cos θ, a scalar that vanishes whenever the displacement is perpendicular to the force — the reason uniform circular motion does no work. Moment (torque) is the cross product M = r × F, where r runs from the chosen reference point to the point of application of the force; its magnitude |r||F| sin θ equals force times perpendicular distance, and its direction comes from the right-hand screw rule. The moment about a specific axis is the component along that axis, computed as the scalar triple product â·(r × F). The two products interlock through Lagrange's identity |a × b|^2 + (a·b)^2 = |a|^2 |b|^2.
What you must remember
- Work formula: W = F·d = |F||d| cos θ; only the force component along the displacement contributes, so motion perpendicular to F costs nothing.
- Moment about a point: M = r × F (never F × r, which flips the sign); magnitude = force × perpendicular distance from the reference point to the line of action.
- Moment about an axis: M_axis = â·(r × F), the scalar triple product with the unit vector along the axis; only the component spinning around that axis survives.
- Couple: two equal, opposite, parallel forces; its moment = magnitude of one force × arm length, and it is independent of the choice of reference point.
- Rigid-body equilibrium: ΣF = 0 and ΣM = 0 must both hold; either alone permits rotation or translation.
- Lagrange's identity: |a × b|^2 + (a·b)^2 = |a|^2 |b|^2 — the geometric bridge between the two products, occasionally asked directly.
- Coplanar force system: forces whose lines of action keep a scalar triple product of zero with any connecting position vectors cannot twist the body out of plane.
Computing a moment and a work together
Let F = 2i + j - k act at the point P(1, 2, 3). The moment about the origin uses r = i + 2j + 3k: expand r × F to get i(2 × (-1) - 3 × 1) - j(1 × (-1) - 3 × 2) + k(1 × 1 - 2 × 2) = -5i + 7j - 3k, with magnitude √(25 + 49 + 9) = √83 ≈ 9.1 units. The moment about the z-axis is just the k-component, -3 — viewed from above (looking down the positive z-axis), the negative sign flags a clockwise twist. Same force, different question: if the point of application moves from A(1, 2, 3) to B(2, 3, 5), the displacement is d = i + j + 2k and the work is W = 2 + 1 - 2 = 1 unit. Notice how one scenario reads components out of a cross product while the other collapses a displacement to a single scalar — the exam's way of checking whether you know which product the physics demands.
The examiner's lens
JEE Main asks for direct evaluations: a cross product for a moment, a dot product for work, magnitudes from both — often dressed as physics-adjacent statements in the vector chapter. Advanced likes moments about a shifted origin (translate r and watch the moment change, unlike a couple's), moments about an arbitrary axis via the triple product, and equilibrium problems where the vanishing of ΣM yields equations per component. The signature errors are three: writing F × r and losing the direction; using the position vector of some point on the line of action other than the point of application (any point on the line of action actually gives the same moment about a fixed point — the slip is using a point off that line); and reporting work as negative when the angle exceeds 90°, which is legitimate physics but often mishandled in the arithmetic of signs. Remember also that moment depends on the reference point while a couple does not — a true/false favourite in both papers.
Frequently asked questions
Why is work a dot product and not a cross product?
Because work is energy, a scalar, and only the force component along the displacement contributes: W = |F||d| cos θ, which is exactly the dot product.
How is the moment of a force about a point computed?
As M = r × F with r drawn from the reference point to the point of application; its magnitude equals the force multiplied by the perpendicular distance to its line of action.
What does the scalar triple product â·(r × F) represent?
The moment of F about the axis with unit vector â — the component of the moment that actually rotates the body around that axis.
When is the work done by a force zero?
When displacement is zero or perpendicular to the force (cos θ = 0), as in uniform circular motion under a radial force.
Why does a couple's moment not depend on the reference point?
Because the two forces' translational effects cancel (ΣF = 0), leaving only the pure rotational pair whose moment F × arm is fixed by the forces themselves.