Skip to content

Physics · Ch 2 — Mathematical Methods

Scalar Product (Dot Product)

2.5.1

Scalar Product (Dot Product)

The scalar product (or dot product) of two non-zero vectors P⃗\vec P and Q⃗\vec Q is defined as the product of their magnitudes and the cosine of the angle θ\theta between them:

P⃗⋅Q⃗=PQcos⁡θ— (2.17)\vec P\cdot\vec Q = PQ\cos\theta \qquad \text{--- (2.17)}

The result is a SCALAR (a pure number with units), not a vector.

Geometric meaning (projection). P⃗⋅Q⃗=P(Qcos⁡θ)\vec P\cdot\vec Q = P(Q\cos\theta), i.e. the magnitude of P⃗\vec P times the component of Q⃗\vec Q along the direction of P⃗\vec P (its 'projection' onto P⃗\vec P). Equivalently, P⃗⋅Q⃗=Q(Pcos⁡θ)\vec P\cdot\vec Q = Q(P\cos\theta), the magnitude of Q⃗\vec Q times the component of P⃗\vec P along Q⃗\vec Q.

Properties.

  1. Commutative: P⃗⋅Q⃗=PQcos⁡θ=QPcos⁡θ=Q⃗⋅P⃗\vec P\cdot\vec Q = PQ\cos\theta = QP\cos\theta = \vec Q\cdot\vec P.
  2. Distributive: P⃗⋅(Q⃗+R⃗)=P⃗⋅Q⃗+P⃗⋅R⃗\vec P\cdot(\vec Q+\vec R) = \vec P\cdot\vec Q+\vec P\cdot\vec R.
  3. Special angles: if θ=0\theta=0 (parallel vectors), P⃗⋅Q⃗=PQ\vec P\cdot\vec Q = PQ — in particular i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1; if θ=90∘\theta=90^\circ (perpendicular vectors), P⃗⋅Q⃗=0\vec P\cdot\vec Q=0 — in particular i^⋅j^=j^⋅k^=k^⋅i^=0\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0; if θ=180∘\theta=180^\circ (anti-parallel), P⃗⋅Q⃗=−PQ\vec P\cdot\vec Q=-PQ.
  4. If P⃗=Q⃗\vec P=\vec Q, then P⃗⋅Q⃗=P2=Q2\vec P\cdot\vec Q = P^2 = Q^2.

Component formula. For P⃗=Pxi^+Pyj^+Pzk^\vec P = P_x\hat i+P_y\hat j+P_z\hat k and Q⃗=Qxi^+Qyj^+Qzk^\vec Q = Q_x\hat i+Q_y\hat j+Q_z\hat k,

P⃗⋅Q⃗=PxQx+PyQy+PzQz\vec P\cdot\vec Q = P_xQ_x+P_yQ_y+P_zQ_z

This follows by expanding P⃗⋅Q⃗\vec P\cdot\vec Q term by term and using i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1 while i^⋅j^=j^⋅k^=k^⋅i^=0\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0 — every cross term between different unit vectors vanishes, leaving only the three matching-axis products.

A caution. If a⃗⋅b⃗=a⃗⋅c⃗\vec a\cdot\vec b = \vec a\cdot\vec c with a⃗≠0\vec a\ne 0, it does NOT necessarily follow that b⃗=c⃗\vec b=\vec c. Using the distributive law, a⃗⋅(b⃗−c⃗)=0\vec a\cdot(\vec b-\vec c)=0, which only tells you that either b⃗=c⃗\vec b=\vec c OR a⃗\vec a is perpendicular to (b⃗−c⃗)(\vec b-\vec c) — the dot product cannot be 'cancelled' the way ordinary multiplication can. …

Figure 2.9Projection of vectors

What this figure shows. Vector P⃗\vec P is drawn horizontally from a point O. Vector Q⃗\vec Q is drawn from the same point O at angle θ\theta above P⃗\vec P. A dashed perpendicular line drops from the head of Q⃗\vec Q down onto the line containing P⃗\vec P (or its extension), marking the projection of Q⃗\vec Q onto P⃗\vec P's direction — this projected length is labelled 'Q cos θ' along the line of P. A separate labelled segment 'P cos θ' is also marked near the direction of Q, representing the projection of P onto Q's direction. The angle θ between the two vectors is marked at O. This construction visually demonstrates that the dot product P⃗⋅Q⃗=PQcos⁡θ\vec P\cdot\vec Q = PQ\cos\theta equals the magnitude of one vector times the projected …

Misc Ex.2.6Computing a scalar (dot) product

Worked out. Given two vectors v⃗1=i^+2j^+3k^\vec v_1 = \hat i+2\hat j+3\hat k and v⃗2=3i^+4j^−5k^\vec v_2 = 3\hat i+4\hat j-5\hat k, the problem asks for their scalar (dot) product. The method applies the component formula for the dot product directly, multiplying and summing corresponding components (v1xv2x+v1yv2y+v1zv2zv_{1x}v_{2x}+v_{1y}v_{2y}+v_{1z}v_{2z}), using the fact that i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1 while all cross terms betwee …

Misc Ex.2.8Finding the angle between two vectors using the dot product

Worked out. Given two vectors A⃗=5i^+6j^+4k^\vec A = 5\hat i+6\hat j+4\hat k and B⃗=2i^−2j^+3k^\vec B = 2\hat i-2\hat j+3\hat k, the problem asks for the angle between them. The method uses the dot-product definition A⃗⋅B⃗=ABcos⁡θ\vec A\cdot\vec B = AB\cos\theta, computing A⃗⋅B⃗\vec A\cdot\vec B via the component formula, then the individual magnitudes ∣A⃗∣|\vec A| and ∣B⃗∣|\vec B| via the magnitude formula, and finally solving for $\theta = \cos^{-1}!\left(\dfrac{\vec A\cdot\vec B}{|\vec A||\vec B …

Start your 14-day free trial to unlock the full solution →
← Previous · 2.5Multiplication of VectorsNext · 2.5.2 →Vector Product (cross product)