Concept understanding — Applications of Dot and Cross Product
The dot product a⋅b=∣a∣∣b∣cosθ and cross product a×b (magnitude ∣a∣∣b∣sinθ, direction perpendicular to both) are not just formulas — placed carefully, they reprove classical geometry theorems and compute two real mechanical quantities.
Vector proofs in geometry and trigonometry. The standard technique: put a convenient point (a triangle's vertex, a circle's centre) at the origin, write every other point as a position vector, and express every needed segment as a difference of position vectors. Then:
Perpendicularity is proved by showing a dot product is 0. E.g. if O is a circle's centre and M is the midpoint of chord AB (position vectors a,b with ∣a∣=∣b∣=r), then OM=2a+b and AB=b−a, and OM⋅AB=21(∣b∣2−∣a∣2)=0.
Equal lengths / rectangles come from expanding ∣p±q∣2=∣p∣2±2p⋅q+∣q∣2 and comparing.
Areas come from area of a parallelogram=∣p×q∣ for adjacent sides p,q; a triangle is half that, and a general quadrilateral with diagonals d1,d2 has area 21∣d1×d2∣ (proved by splitting along one diagonal, since the two triangles on either side add — same-sense cross products — instead of subtracting).
Compound-angle identities (cos(α∓β), sin(α±β)) drop out of dotting or crossing two unit vectors i^cosα+j^sinα and i^cosβ±j^sinβ placed at angles α,±β to the x-axis.
Work done. For a constant force F producing a displacement d, w=F⋅d. With several simultaneous forces, first add them to get the resultant force (and, similarly, use the actual displacement between the two given points as d), then take one dot product.
Torque (moment). For a force F applied at a point with position vector rrelative to the pivot, τ=r×F — a vector, whose magnitude is the turning strength and whose direction (via its direction cosines) is the rotation axis. As with work, combine multiple forces into a resultant first if several forces act at the same point.
Note
Work is a scalar (dot product); torque is a vector (cross product) — mixing the two up is the single most common slip in this topic.
With centre O as origin and diameter endpoints A(a),B(−a), any third point C(c) on the circle gives CA⋅CB=0.
CA=a−c, CB=−a−c, and ∣a∣=∣c∣=r.
✓Final answer
CA⋅CB=∣c∣2−∣a∣2=0, so ∠ACB=90∘. ■
Placing the centre at the origin makes the diameter's endpoints negatives of each other, and expanding CA⋅CB collapses to a difference of two equal radii-squared.
Step 1. Set up. Let O be the centre of a circle of radius r, and let AB be a diameter, so the position vectors of A,B are a,−a with ∣a∣=r. Let C be any other point on the circle, position vector c, ∣c∣=r.
Step 2. Form the two chords from C.CA=a−c,CB=−a−c.
Step 3. Dot them.
CA⋅CB=(a−c)⋅(−a−c)=−a⋅a−a⋅c+c⋅a+c⋅c=∣c∣2−∣a∣2.
Step 4. Use ∣a∣=∣c∣=r.CA⋅CB=r2−r2=0.
Step 5. Conclude.CA⊥CB, i.e. ∠ACB=90∘ — the angle subtended by a diameter at any point of the semicircle is a right angle.
✓Final answer
CA⋅CB=∣c∣2−∣a∣2=0 since both equal r2; hence ∠ACB=90∘. ■
Origin at the centre; write the diameter's endpoints as ±a
Forgetting B's position vector is −a, not an independent vector
Mixing up which two chords (CA,CB) must be dotted — it's the two from the point ON the circle to the two ends of the diameter