Physics · Work, Energy And Power · NEET
It is a scalar (just a number, with a sign but no direction). Even though A and B each have a direction, A.B = AB cos(theta) is a product of magnitudes and cos(theta), which are all scalars. This is why work (W = F.d) and power (P = F.v) are scalar quantities in NEET, even though force, displacement and velocity are vectors.
The unit vectors i, j, k point along three mutually perpendicular axes. For i.i the angle is 0 degrees, so i.i = (1)(1)cos(0) = 1. For i.j the angle is 90 degrees, so i.j = (1)(1)cos(90) = 0. Rule to remember: same-name pairs (i.i, j.j, k.k) equal 1, and different-name pairs (i.j, j.k, k.i) equal 0. These rules are what collapse the full expansion into A.B = Ax Bx + Ay By + Az Bz.
Multiply the x-components together, the y-components together, the z-components together, then add. So if A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k, then A.B = Ax Bx + Ay By + Az Bz. Example: (3i + 4j - 5k).(5i + 4j + 3k) = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16. This component method is faster than finding the angle first, so use it whenever vectors are given in i, j, k form.
Zero. If theta = 90 degrees then cos(90) = 0, so A.B = 0. This works both ways: if you compute A.B in component form and get 0 (and neither vector is zero), the two vectors are perpendicular. NEET often uses this as a quick perpendicularity test, for example checking that velocity is perpendicular to position in circular motion.
Yes to both. Commutative: A.B = B.A, because cos(theta) is the same for either order. Distributive over addition: A.(B + C) = A.B + A.C. Also A.(lambda B) = lambda(A.B) for any real number lambda. But note the dot product of three vectors like A.B.C is meaningless, because A.B is already a scalar and you cannot dot a scalar with a vector.
A.A = A squared, the square of the magnitude. Since the angle of a vector with itself is 0, A.A = A A cos(0) = A^2. In components, A.A = Ax^2 + Ay^2 + Az^2, which is why the magnitude is A = sqrt(Ax^2 + Ay^2 + Az^2). This is a handy way to get a vector's length without a separate formula.
A particle moves from a point (-2i + 5j) to (4j + 3k) when a force of (4i + 3j) N is applied. How much work is done by the force?
A particle moves with a velocity (5i - 3j + 6k) m/s under the action of a constant force (10i + 10j + 20k) N. The instantaneous power supplied to the particle is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A.B = AB cos(theta), where A and B are the magnitudes and theta is the angle between the two vectors. Equivalently, in component form A.B = Ax Bx + Ay By + Az Bz.
i.i = j.j = k.k = 1 (same axis, angle 0) and i.j = j.k = k.i = 0 (perpendicular axes, angle 90 degrees). These rules turn the full expansion into A.B = Ax Bx + Ay By + Az Bz.
Work done by a constant force is W = F.d, and instantaneous power is P = F.v. Both are dot products, so both give scalar answers in joules and watts respectively. This is the main reason NEET asks you to master the dot product in this chapter.
Yes. When theta is between 90 and 180 degrees, cos(theta) is negative, so A.B is negative. In physics this shows up as negative work, for example when friction acts opposite to displacement.
The dot product gives a scalar (A.B = AB cos(theta)) and the cross product gives a vector (magnitude AB sin(theta), direction by the right-hand rule). Work and power use the dot product; torque and magnetic force use the cross product.
