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Torque problems in two dimensions can be done with force × perpendicular distance. As soon as a question gives position and force as vectors, or asks for torque about a point that is not the origin, you need τ = r × F, and that is where sign and position errors creep in. This post shows how to set up r, compute the cross product reliably, check the result, and get torque about an axis.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "moment of a force, torque" under Rotational Motion, and its mathematics units include scalar and vector products. The NEET (UG) 2026 syllabus lists torque too. The NCERT Class 11 chapter Systems of Particles and Rotational Motion (2026-27 reprint) defines the vector product in Section 6.5, defines torque as τ = r × F in Section 6.7.1, and works an example in Example 6.5. Check the current year's syllabus.
What exactly is r?
NCERT defines the moment of force about the origin as τ = r × F, where r is the position vector of the particle the force acts on. For any other reference point O, r is the vector from O to the point where the force acts:
r = (position of the point of application) − (position of O)
The magnitude is rF sin θ, which equals F times the perpendicular distance from O to the force's line of action. The direction is perpendicular to both r and F, given by the right-hand rule.
Worked example 1: same force, two reference points
A force F = 3î + 4ĵ N acts at the point P = (2, 1, 0) m.
About the origin. r = 2î + ĵ. In the xy-plane only the z-component survives: τ_z = r_xF_y − r_yF_x = 2 × 4 − 1 × 3 = 5. So τ = 5k̂ N m, anticlockwise seen from +z.
Check with the perpendicular distance: |F| = 5 N, so the distance from the origin to the line of action is 5/5 = 1 m.
About O′ = (1, −1, 0). Now r = P − O′ = î + 2ĵ. τ_z = 1 × 4 − 2 × 3 = −2. So τ = −2k̂ N m, clockwise. The same force gives a different torque, even a different sense of rotation, about a different point.
Worked example 2: a full three-dimensional cross product
A force F = 2î − ĵ + 3k̂ N acts at P = (1, 2, 0) m. Find the torque about O = (0, 0, 1) m.
Step 1: r = P − O = î + 2ĵ − k̂.
Step 2: use τ = (r_yF_z − r_zF_y, r_zF_x − r_xF_z, r_xF_y − r_yF_x).
x-component: 2 × 3 − (−1)(−1) = 6 − 1 = 5
y-component: (−1)(2) − 1 × 3 = −2 − 3 = −5
z-component: 1 × (−1) − 2 × 2 = −1 − 4 = −5
Step 3: τ = 5î − 5ĵ − 5k̂ N m, with magnitude 5√3 ≈ 8.66 N m.
Check 1. τ must be perpendicular to both r and F. r · τ = 5 − 10 + 5 = 0 and F · τ = 10 + 5 − 15 = 0. Both zero.
Check 2. Slide the point of application along the line of action, to P + F = (3, 1, 3). Then r = 3î + ĵ + 2k̂ and r × F = (1 × 3 − 2 × (−1), 2 × 2 − 3 × 3, 3 × (−1) − 1 × 2) = (5, −5, −5). The same torque, as it must be: only the line of action matters, not where along it the force is applied.
How do you get torque about an axis?
Torque about a point is a vector; torque about an axis is its component along that axis. For an axis through O parallel to z, the torque in Example 2 is the z-component, −5 N m. For a general axis with unit vector n̂ through O, take n̂ · τ.
Why does a couple give the same torque everywhere?
A couple is two equal and opposite forces with different lines of action. NCERT's Example 6.7 shows its moment does not depend on the point you take moments about. With F at A and −F at B, the total torque is (r_A − r_B) × F, and the reference point has cancelled out. For F = 5ĵ N at (1, 0, 0) m and −5ĵ N at (−1, 0, 0) m, the torque is 10k̂ N m about the origin, and also 10k̂ N m about (3, 4, 1).
Why do students find rotation hard?
Rimoldini and Singh (2005) studied introductory students and physics juniors on rotational and rolling motion. All groups showed similar difficulties, higher mathematical sophistication did not seem to help, and some difficulties came from related problems with linear motion. Fluent cross products are not enough on their own; you also need to know which r you are crossing.
Which mistakes should you avoid?
Using the position of P when the pivot is not the origin. Subtract O first.
Computing F × r. That is −τ.
Dropping the minus in the middle component of the determinant. The component formula above avoids it.
Skipping the perpendicularity check.
For cross-product drills, see vector algebra for JEE maths, and for torques in equilibrium, torque and equilibrium: ladders and beams.
A practice routine
Redo NCERT Example 6.5, then repeat it about a point of your choice other than the origin.
For three forces, compute τ about two points and run both checks.
Find the torque about a given axis with n̂ · τ.
Build a couple and confirm its torque about three points.
If the idea of a vector product is still shaky, the Fermi Concept Library covers every concept in physics and mathematics, so you can revise it on its own before returning to torque.
Try this now: find the torque of F = î + ĵ N acting at (0, 3, 0) m about the point (1, 1, 0) m.
