Physics · System Of Particles And Rotational Motion · NEET
Write three rows. Top row is always the unit vectors i, j, k. Second row is the components of the FIRST vector A: (Ax, Ay, Az). Third row is the components of the SECOND vector B: (Bx, By, Bz). Order matters: A goes above B because A x B is not the same as B x A (they differ by a minus sign). So the setup is: row1 = i j k, row2 = Ax Ay Az, row3 = Bx By Bz.
When you expand a 3x3 determinant along the top row, the signs follow the pattern + - + . So i keeps its plus, j gets a minus, and k keeps its plus. This is not physics - it is just the rule for expanding any determinant. Forgetting this minus is the single most common mistake. Formula: A x B = i(Ay*Bz - Az*By) - j(Ax*Bz - Az*Bx) + k(Ax*By - Ay*Bx).
For the i term, cover column 1 (the i column) and take the remaining 2x2 block: (Ay*Bz - Az*By). For the j term, cover column 2 and take (Ax*Bz - Az*Bx), then put a minus in front. For the k term, cover column 3 and take (Ax*By - Ay*Bx). Each minor is 'main diagonal product minus off-diagonal product'.
Going in the cyclic order i -> j -> k -> i gives plus: i x j = k, j x k = i, k x i = j. Going backwards gives minus: j x i = -k, k x j = -i, i x k = -j. Any vector crossed with itself is zero: i x i = 0, j x j = 0, k x k = 0. These fall out of the determinant automatically, so you do not have to memorise them, but they are a fast shortcut for simple cases like the NEET 2020 torque question.
The cross product A x B is always perpendicular to BOTH A and B. Point the fingers of your right hand along A, curl them toward B, and your thumb points along A x B (right-hand rule). The determinant already gives you the correct signed components, so the direction is built into the answer - the right-hand rule is just a physical check.
The torque about the origin when a force 3j N acts on a particle whose position vector is 2k m is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. They are equal in size but point in opposite directions: A x B = -(B x A). This is why the first vector must go in the second row of the determinant and the second vector in the third row.
Yes. Torque tau = r x F and angular momentum L = r x p are both cross products. Put r in the second row and F (or p) in the third row, then expand exactly the same way. Most NEET rotation questions are solved with this one skill.
Just fill the missing components with zeros. So 2k becomes (0, 0, 2). The determinant handles the zeros automatically and many terms vanish, making the maths short.
|A x B| = |A||B| sin(theta), where theta is the angle between the two vectors. It is largest when they are perpendicular (sin 90 = 1) and zero when they are parallel (sin 0 = 0). The determinant gives both the magnitude and the direction in one step.
Because torque, angular momentum, and the direction of angular velocity all use it. Setting up and expanding the determinant correctly is a repeatable, high-yield skill that shows up almost every year in the rotational motion unit.
